GSEB Class 12 Statistics Solutions Chapter 1 Probability

Gujarat Board Textbook Solutions Class 12 Statistics Part 2 Chapter 1 Probability Ex 1

Answer The Following Questions By Selecting A Correct Option From The Given Options:

Question 1. Which event is given by a special subset \( \Phi \) of the sample space U?
(a) Certain event
(b) Complementary event of \( \Phi \)
(c) Union of events U and \( \Phi \)
(d) Impossible event
Answer: (d) Impossible event
In simple words: An impossible event is one that cannot happen. In probability, it is represented by the empty set \( \Phi \).

🎯 Exam Tip: Understanding the basic definitions of probability terms like impossible event, certain event, and sample space is crucial for foundational knowledge.

 

Question 2. What is the value of P(A \( \cap \) A') for events A and A' ?
(b) 0
Answer: (b) 0
In simple words: Event A and its complement A' cannot happen at the same time, so the probability of both occurring (their intersection) is zero.

🎯 Exam Tip: The intersection of an event and its complement is always an empty set, resulting in a probability of 0. This is a fundamental property in probability theory.

 

Question 3. Which of the following options is true for any event of the sample space ?
(a) P(A) < 0 (b) 0 \( \le \) P(A) \( \ge \) 1 (c) 0 \( \le \) P(A) \( \le \) 1 (d) P(A) > 1
Answer: (c) 0 \( \le \) P(A) \( \le \) 1
In simple words: The probability of any event always falls between 0 and 1, inclusive. It cannot be negative or greater than 1.

🎯 Exam Tip: Always remember that probability values must be within the range [0, 1]. Any calculated probability outside this range indicates an error.

 

Question 4. Which of the following options is not true for any two events A and B in the sample space U; where, A \( \subset \) B?
(a) P(A \( \cap \) B) = P(B)
(b) P(A \( \cap \) B) = P(A)
(c) P(A \( \cup \) B) \( \ge \) P(A)
(d) P(B-A) = P(B)-P(A)
Answer: (a) P(A \( \cap \) B) = P(B)
In simple words: If event A is a subset of event B, then the outcomes common to A and B are simply the outcomes of A. So, P(A \( \cap \) B) should be P(A), not P(B).

🎯 Exam Tip: For subset relationships (A \( \subset \) B), the intersection A \( \cap \) B equals A, and the union A \( \cup \) B equals B. Use these properties to simplify probability expressions.

 

Question 5. What is the other name of the classical definition of probability?
(a) Mathematical definition
(b) Axiomatic definition
(c) Statistical definition
(d) Geometric definition
Answer: (a) Mathematical definition
In simple words: The classical definition of probability, which relies on favorable outcomes divided by total outcomes, is also known as the mathematical definition.

🎯 Exam Tip: Knowing alternative names for definitions helps in broader understanding and recalling concepts. The classical definition is fundamental for many problems.

 

Question 6. Which of the following statement for probability of elementary events H and T of random experiment of tossing a balanced coin is not true ?
(a) P(T) = 0.5
(b) P(H) + P(T) = 1
(c) P(H \( \cap \) T) = 0.5
(d) P(H) = 0.5
Answer: (c) P(H \( \cap \) T) = 0.5
In simple words: When tossing a coin, getting a Head (H) and a Tail (T) at the same time is impossible. Therefore, the probability of their intersection must be 0, not 0.5.

🎯 Exam Tip: Elementary events from a single trial of a random experiment are always mutually exclusive, meaning their intersection probability is 0.

 

Question 7. Which random experiment from the following random experiments has an infinite sample space ?
(a) Throwing two dice
(b) Selecting two employees from an office
(c) To measure the life of electric bulb
(d) Select a card from 52 cards
Answer: (c) To measure the life of electric bulb
In simple words: Measuring the life of an electric bulb can result in any positive real number, meaning there are infinitely many possible outcomes, thus an infinite sample space.

🎯 Exam Tip: An infinite sample space occurs when outcomes can be any value within a continuous range, like time, length, or temperature.

 

Question 8. If A \( \cup \) A' = U, then what type of events are A and A'?
(a) Independent events
(b) Complementary events
(c) Certain events
(d) Impossible events
Answer: (b) Complementary events
In simple words: When the union of two events equals the entire sample space (U) and one event is the opposite of the other, they are called complementary events.

🎯 Exam Tip: Complementary events cover all possible outcomes of the sample space, meaning one or the other must occur. Their probabilities sum to 1.

 

Question 9. If P(A|B) = P(A) and P(B|A) = P(B), then what type of events are A and B ?
(a) Independent events
(b) Complementary events
(c) Certain events
(d) Impossible events
Answer: (a) Independent events
In simple words: If the probability of an event happening does not change even if another event has occurred, then the two events are independent.

🎯 Exam Tip: The defining characteristic of independent events is that the occurrence of one does not affect the probability of the other. This is a key concept for solving complex probability problems.

 

Question 10. Two events A and B of a sample space are mutually exclusive. Which of the following will be equal to P(B-A)?
(b) P(B)
(c) P(A \( \cap \) B)
(d) P(A \( \cup \) B)
Answer: (b) P(B)
In simple words: If events A and B are mutually exclusive, they cannot happen at the same time. So, the event "B-A" (B occurring without A) is simply event B itself, and its probability is P(B).

🎯 Exam Tip: For mutually exclusive events, A \( \cap \) B is an empty set. Thus, P(B-A) simplifies to P(B) since A does not overlap with B.

 

Question 11. What is the total number of sample points in the sample space formed by throwing three six-faced balanced dice simultaneously?
(a) 6\(^2\)
(b) 36
(c) 6 \( \times \) 3
(d) 6\(^3\)
Answer: (d) 6\(^3\)
In simple words: When rolling multiple dice, you multiply the number of outcomes for each die together to find the total number of sample points. For three 6-sided dice, it's 6 x 6 x 6.

🎯 Exam Tip: For independent trials (like rolling multiple dice or flipping multiple coins), the total number of outcomes is the product of the number of outcomes for each individual trial.

 

Question 12. If one number is randomly selected between 1 and 20, what is the probability that the number is a multiple of 5 ?
(a) \( \frac{1}{2} \)
(b) \( \frac{1}{6} \)
(c) \( \frac{1}{5} \)
(d) \( \frac{1}{3} \)
Answer: (c) \( \frac{1}{5} \)
In simple words: From 1 to 20, the multiples of 5 are 5, 10, 15, 20. There are 4 such numbers. Since there are 20 total numbers, the probability is 4/20, which simplifies to 1/5.

🎯 Exam Tip: To calculate simple probabilities, always identify the total number of possible outcomes and the number of favorable outcomes. Simplify the fraction if possible.

 

Question 13. If events A and B are independent, which of the following options is true?
(a) P(A \( \cap \) B) = P(A) \( \times \) P(B)
(b) P(A \( \cup \) B) = P(A) + P(B)
(c) P(A \( \cup \) B) = P(A) \( \times \) P(B)
(d) P(A \( \cap \) B) = P(A) + P(B)
Answer: (a) P(A \( \cap \) B) = P(A) \( \times \) P(B)
In simple words: For independent events, the probability that both events A and B happen is found by multiplying their individual probabilities.

🎯 Exam Tip: The multiplication rule for independent events (P(A \( \cap \) B) = P(A)P(B)) is a fundamental concept. Do not confuse it with the addition rule for mutually exclusive events.

 

Question 14. What is the probability of having 5 Thursdays in the month of February in a year which is not a leap year?
(a) 0
(b) \( \frac{1}{7} \)
(c) \( \frac{2}{7} \)
(d) \( \frac{3}{7} \)
Answer: (a) 0
In simple words: February in a non-leap year has 28 days, which is exactly 4 weeks. This means every day of the week appears exactly 4 times. So, having 5 Thursdays is impossible.

🎯 Exam Tip: Remember that a common year has 365 days (52 weeks and 1 day), and February has 28 days (exactly 4 weeks), making it impossible to have 5 of any specific weekday in a non-leap year.

 

Question 15. If P(A) = 0.4 and P(B') = 0.3 for two independent events A and B of a sample space, then state the value of P(A \( \cap \) B).
(a) 0.12
(b) 0.42
(c) 0.28
(d) 0.18
Answer: (c) 0.28
In simple words: First, find P(B) using P(B) = 1 - P(B'). Then, since A and B are independent, multiply P(A) by P(B) to get P(A \( \cap \) B). So, P(B) = 1 - 0.3 = 0.7, and P(A \( \cap \) B) = 0.4 * 0.7 = 0.28.

🎯 Exam Tip: When events are independent, the complement of one event is also independent of the other event. Always calculate P(B) from P(B') before applying the multiplication rule.

 

Question 16. For two events A and B of a sample space, state the event (A \( \cap \) B) \( \cup \) (A \( \cap \) B').
(а) \( \Phi \)
(b) B
(c) A
(d) U
Answer: (c) A
In simple words: The union of outcomes where A and B both occur, and outcomes where A occurs but B does not, covers all instances where A occurs, regardless of B. This simplifies to event A.

🎯 Exam Tip: Use Venn diagrams or distributive laws of set theory. (A \( \cap \) B) \( \cup \) (A \( \cap \) B') = A \( \cap \) (B \( \cup \) B') = A \( \cap \) U = A.

 

Question 17. According to the mathematical definition of probability, what is the probability of each outcome among the n outcomes of a random experiment ?
(a) 0
(b) \( \frac{1}{n} \)
Answer: (b) \( \frac{1}{n} \)
In simple words: In a random experiment with 'n' equally likely outcomes, each individual outcome has a probability of 1 divided by 'n'.

🎯 Exam Tip: The mathematical (classical) definition assumes all outcomes are equally probable. This means each elementary event has a probability of 1/n.

 

Section B

Answer The Following Questions In One Sentence:

Question 1. Give two examples of random experiment.
Answer: Two examples of random experiment are:
1. The experiment of throwing a balanced die and
2. The experiment of finding defective units from a lot of units produced.
In simple words: A random experiment is one where you know all possible results, but you cannot predict which one will happen next. Examples include rolling a die or checking products for defects.

🎯 Exam Tip: When asked for examples, choose simple, common scenarios where the outcomes are clear but uncertain beforehand.

 

Question 2. Draw the Venn diagram for A - B, the difference event of A and B.
Answer: The Venn diagram for difference event A-B:
ℹ️ चित्र व्याख्या (Diagram Explanation): इस वेन डायग्राम में, दो वृत्त A और B एक बड़े आयत (जो सैंपल स्पेस U को दर्शाता है) के भीतर हैं। इवेंट A - B को वृत्त A के उस भाग के रूप में दर्शाया गया है जो वृत्त B के साथ ओवरलैप नहीं करता है। यह A के उन तत्वों को दिखाता है जो B में नहीं हैं।
In simple words: The Venn diagram for A-B shows the part of event A that does not overlap with event B, meaning outcomes that are in A but not in B.

🎯 Exam Tip: Clearly shading the specific region representing the difference event (A-B) in a Venn diagram is crucial for correct interpretation and scoring.

 

Question 3. Define an event.
Answer: Any subset of the sample space of a random experiment is called an event. It is denoted by A, B, C ....
In simple words: An event is simply a collection of one or more possible outcomes from a random experiment.

🎯 Exam Tip: Remember that an event is always a subset of the sample space. This relationship is fundamental to understanding probability.

 

Question 4. Write the sample space of a random experiment of throwing one balanced die and a balanced coin simultaneously.
Answer: The sample space of a random experiment of throwing one balanced die and a balanced coin simultaneously is obtained as follows:
U = {(1, H), (2, H), (3, H), (4, H), (5, H), (6, H), (1, T), (2, T), (3, T), (4, T), (5, T), (6, T)}
Where, H = Head; T = Tail;
1, 2, 3, 4, 5, 6 = Numbers on die.
In simple words: When you roll a die and flip a coin at the same time, the sample space lists all possible pairs of outcomes, like getting a '1' on the die and 'Head' on the coin, and so on.

🎯 Exam Tip: To construct a sample space for multiple independent actions, list all possible combinations systematically, often using ordered pairs or a tree diagram.

 

Question 5. Define conditional probability.
Answer: U is a finite sample space and A and B are any two events of U. The probability of event B, under the condition that event A is happened, is called conditional probability of the event B.
In simple words: Conditional probability is the chance of an event happening, given that another event has already occurred.

🎯 Exam Tip: The notation P(B|A) means "the probability of B given A". This concept is crucial for understanding how events influence each other.

 

Question 6. State the formula for the probability of occurrence of at least one event out of three events A, B and C.
Answer: The formula for the probability of occurrence of at least one event out of three events A, B and C is as follows:
P(A \( \cup \) B \( \cup \) C) = P(A) + P(B) + P(C) - P(A \( \cap \) B) - P(A \( \cap \) C) - P(B \( \cap \) C) + P(A \( \cap \) B \( \cap \) C)
In simple words: To find the probability of at least one of three events happening, you add their individual probabilities, subtract the probabilities of their pairwise intersections, and then add back the probability of their triple intersection.

🎯 Exam Tip: This formula, known as the Principle of Inclusion-Exclusion for three events, is vital for problems involving the union of multiple non-mutually exclusive events.

 

Question 7. Define independent events.
Answer: U is a finite sample space and A and B are any two events of U. If the probability of occurrence of the event A does not depend on the occurrence or non-occurrence of the event B, then the events A and B are called independent events.
In simple words: Two events are independent if what happens in one event does not change the likelihood of the other event happening.

🎯 Exam Tip: Understanding independence is key; it implies that P(A|B) = P(A) and P(B|A) = P(B), and the multiplication rule P(A \( \cap \) B) = P(A)P(B) applies.

 

Question 8. Write the law of multiplication of probability for two independent events A and B in a sample space.
Answer: The law of multiplication of probability for two independent events A and B in a sample space is as follows:
P(A \( \cap \) B) = P(A) \( \times \) P(B)
In simple words: For events that do not affect each other, the chance of both happening is simply the product of their individual chances.

🎯 Exam Tip: This formula is a direct test of independence. If the formula holds true for two events, they are independent.

 

Question 9. Interpret P(A|B) and P(B|A).
Answer: Interpretation of P(A|B): The conditional probability of the event A under the condition that the event B is occurred.
Interpretation of P(B|A): The conditional probability of the event B under the condition that the event A has occurred.
In simple words: P(A|B) tells us the probability of event A happening, knowing that event B has already happened. P(B|A) tells us the probability of event B happening, knowing that event A has already happened.

🎯 Exam Tip: Differentiate clearly between P(A \( \cap \) B) (both A and B happen) and P(A|B) (A happens given B). Conditional probability restricts the sample space to the condition.

 

Question 10. When can we say that three events A, B and C in a sample space are exhaustive?
Answer: When P(A) + P(B) + P(C) = 1, we can say that three events A, B and C in a sample space are exhaustive.
In simple words: Three events are exhaustive if at least one of them must happen, meaning their probabilities collectively cover all possibilities in the sample space.

🎯 Exam Tip: For exhaustive events, the union of the events covers the entire sample space (A \( \cup \) B \( \cup \) C = U). If they are also mutually exclusive, their individual probabilities sum to 1.

 

Question 11. Arrange P(A \( \cup \) B), P(A), P(A \( \cap \) B), 0, P (A) + P (B) in the ascending order.
Answer: 0, P(A \( \cap \) B), P(A), P(A \( \cup \) B), P(A) + P(B) are in the ascending order.
In simple words: The order goes from the lowest possible probability (0) to the highest, considering how events relate: intersection is smaller than individual events, which are smaller than their union, and the sum of individual probabilities can be largest.

🎯 Exam Tip: Remember these fundamental relationships: P(A \( \cap \) B) \( \le \) P(A) \( \le \) P(A \( \cup \) B) and P(A \( \cap \) B) \( \le \) P(B) \( \le \) P(A \( \cup \) B). Also, P(A \( \cup \) B) \( \le \) P(A) + P(B).

 

Question 12. Define:
Answer:
1. Random Experiment: The experiment which can be independently repeated under identical conditions and all its possible outcomes are known but it cannot be predicted with certainty which of the outcomes will appear is called a random experiment.
2. Sample Space: The set of all possible outcomes of a random experiment is called a sample space of that random experiment. It is denoted by U or S.
3. Equi-probable Events: If there is no apparent reason to believe that out of one or more random experiment, any one event is more or less likely to occur than the other events, then those events are called as equi-probable events.
4. Favourable Outcomes: If some outcomes out of all the elementary outcomes in the sample space of a random experiment indicate the occurrence of a certain event A, then these outcomes are called the favourable outcomes of the event A.
5. Probability (Mathematical Definition): If out of n outcomes of the finite sample space of a random experiment which are mutually exclusive, exhaustive and equi-probable, m outcomes are favourable for an event A, then the probability of the event A, P(A) = \( \frac{m}{n} \).
6. Probability (Statistical Definition) : Suppose, a random experiment is repeated n times under identical conditions. If an 'event A occurs in m trials then the relative frequency \( \frac{m}{n} \) of the event A gives the estimate of the probability of the event A. When n tends to infinity, the limiting value of \( \frac{m}{n} \) is called the probability of the event A. Thus,
P(A) = \( \lim_{n \to \infty} \frac{m}{n} \).
7. Impossible Event: The special subset \( \Phi \) or { } of the sample space of a random experiment is called an impossible event.
8. Certain Event: The special subset U of the sample space of a random experiment is called a certain event.
In simple words: This section defines key probability terms: a random experiment has unpredictable results; a sample space lists all possible results; equi-probable events have equal chances; favorable outcomes are desired results; mathematical probability uses 'm/n'; statistical probability uses observed frequencies over many trials; an impossible event cannot happen; and a certain event must happen.

🎯 Exam Tip: Memorize these core definitions as they form the backbone of probability theory. Clarity in defining each term is essential for conceptual understanding.

 

Question 13. For two events A and B in a sample space, A \( \cap \) B = \( \Phi \) and A \( \cup \) B = U. State the values of P(A \( \cap \) B) and P(A \( \cup \) B).
Answer: A \( \cap \) B = \( \Phi \) and A \( \cup \) B = U
.. P(A \( \cap \) B) = 0 and P(A \( \cup \) B) = 1
In simple words: If two events have no common outcomes (empty intersection) and together cover all possible outcomes (their union is the whole sample space), then the probability of their intersection is 0, and the probability of their union is 1.

🎯 Exam Tip: These conditions (A \( \cap \) B = \( \Phi \) and A \( \cup \) B = U) describe events that are both mutually exclusive and exhaustive. These are crucial properties.

 

Question 14. If two events A and B in a sample space are independent, then state the formula for P(A \( \cup \) B).
Answer: A and B are independent events.
.. P(A \( \cup \) B) = P(A) + P(B) - P(A \( \cap \) B)
= P(A) + P(B) - [P(A) \( \cdot \) P(B)]
In simple words: For independent events, the probability of A or B happening is the sum of their individual probabilities minus the product of their probabilities.

🎯 Exam Tip: Always start with the general addition rule P(A \( \cup \) B) = P(A) + P(B) - P(A \( \cap \) B), then substitute P(A \( \cap \) B) with P(A)P(B) if events are independent.

 

Question 15. If A = {x|0 < x < 1} and B = {x|\( \frac{1}{4} \) \( \le \) x \( \le \) 3}, then find A \( \cap \) B.
Answer: A = {x|0 < x < 1} = { \( \frac{1}{4} \), \( \frac{1}{2} \), \( \frac{3}{4} \) }
B = {x|\( \frac{1}{4} \) \( \le \) x \( \le \) 3} = { \( \frac{1}{4} \), \( \frac{1}{2} \), \( \frac{3}{4} \), ... , 1, 2, 3}
.. A \( \cap \) B = {x|\( \frac{1}{4} \) < x < 1}
In simple words: To find the intersection of two sets, look for the range of values that are common to both. For A, x is between 0 and 1 (not including 0 or 1). For B, x is between 1/4 and 3 (including 1/4 and 3). The common range is where x is greater than 1/4 but less than 1.

🎯 Exam Tip: When dealing with intervals for sets, carefully note whether the endpoints are inclusive or exclusive. The intersection will be the stricter (more restrictive) common range.

 

Question 16. For two independent events A and B, P (A) = 0.5 and P(B) = 0.7. Find P(A' \( \cap \) Β').
Answer: A and B are independent events. So A' and B' are also independent events. P(A) = 0.5, P(B) = 0.7 are given.
.. P(A' \( \cap \) B') = P(A') \( \cdot \) P(B')
= [1 - P(A)] [1 - P(B)]
= [1 - 0.5] [1 - 0.7]
= [0.5 \( \times \) 0.3]
= 0.15
In simple words: Since A and B are independent, their complements (A' and B') are also independent. First, calculate the probabilities of the complements, P(A') and P(B'). Then, multiply these values to find the probability of both A' and B' occurring.

🎯 Exam Tip: The property that complements of independent events are also independent is crucial. This often simplifies calculations involving A' and B'.

 

Question 17. If P(A) = 0.8 and P (A \( \cap \) B) = 0.25, find P(A - B).
Answer: P(A) = 0.8, P(A \( \cap \) B) = 0.25 are given.
.. P (A - B) = P(A) - P(A \( \cap \) B)
= 0.8 - 0.25 = 0.55
In simple words: The probability of A happening without B is found by taking the probability of A and subtracting the probability that both A and B happen.

🎯 Exam Tip: The formula P(A - B) = P(A) - P(A \( \cap \) B) is always true. It represents the part of A that does not overlap with B.

 

Question 18. If P(A) = 0.3 and P(A \( \cap \) B) = 0.03, find P(B|A).
Answer: P(A) = 0.3, P(A \( \cap \) B) = 0.03 are given.
.. P(B|A) = \( \frac{P(A \cap B)}{P(A)} \) = \( \frac{0.03}{0.3} \) = 0.1
In simple words: To find the probability of B given A, divide the probability of both A and B happening by the probability of A happening.

🎯 Exam Tip: Always use the definition of conditional probability: P(X|Y) = P(X \( \cap \) Y) / P(Y). Make sure to use the correct probability in the denominator.

 

Question 19. If P(A) = P(B) = K for two mutually exclusive events A and B, find P(A \( \cup \) B).
Answer: P (A) = P(B) = k.
A, B are mutually exclusive events
.. P(A \( \cap \) B) = 0
Now, P(A \( \cup \) B) = P(A) + P(B) .
= k + k = 2k
In simple words: Since A and B cannot happen together, the probability of either A or B happening is simply the sum of their individual probabilities.

🎯 Exam Tip: For mutually exclusive events, P(A \( \cap \) B) = 0. This simplifies the general addition rule P(A \( \cup \) B) = P(A) + P(B) - P(A \( \cap \) B) to just P(A) + P(B).

 

Question 20. If P(A' \( \cap \) B) = 0.45 and A \( \cap \) B = \( \Phi \), find P(B).
Answer: P (A' \( \cap \) B) = 0.45 and A \( \cap \) B = \( \Phi \) are given.
.. P(A \( \cap \) B) = 0
Now, P(A' \( \cap \) B) = P(B) - P(A \( \cap \) B)
.. 0.45 = P(B) - 0
.. P(B) = 0.45
In simple words: If A and B are mutually exclusive, then the event "B and not A" is simply the event B itself. So, the probability of B is 0.45.

🎯 Exam Tip: When events are mutually exclusive, P(A \( \cap \) B) = 0. This simplifies expressions like P(A' \( \cap \) B) which becomes P(B) - P(A \( \cap \) B) = P(B) - 0 = P(B).

 

Question 21. Two events A and B in a sample space are mutually exclusive and exhaustive. If P(A) = \( \frac{1}{3} \), find P(B).
Answer: A and B are mutually exclusive and exhaustive events.
.. P(A) + P(B) = 1 (Putting P(A) = \( \frac{1}{3} \))
\( \frac{1}{3} \) + P(B) = 1
.. P(B) = 1 - \( \frac{1}{3} \) = \( \frac{2}{3} \)
In simple words: Since A and B are mutually exclusive (no overlap) and exhaustive (cover all possibilities), their probabilities must add up to 1. If P(A) is 1/3, then P(B) must be 1 - 1/3, which is 2/3.

🎯 Exam Tip: For two events that are both mutually exclusive and exhaustive, they are complements of each other. Therefore, P(A) + P(B) = 1.

 

Question 22. 2 % items in a lot are defective. What is the probability that an item randomly selected from this lot is non-defective?
Answer: A = Event that an item is defective
.. P(A) = 2 % = \( \frac{2}{100} \) = 0.02
A' = Event that an item is non-defective
.. P(A') = 1 – P(A) = 1 – 0.02 = 0.98
In simple words: If 2% of items are bad, then the rest are good. So, 100% minus 2% gives 98% good items.

🎯 Exam Tip: The probability of an event's complement is always 1 minus the probability of the event itself. This is useful for finding the probability of "not A".

 

Question 23. State the number of sample points in the random experiment of tossing five balanced coins.
Answer: The number of sample points in the random experiment of tossing five balanced coins is
n = 2\(^5\) = 32.
In simple words: For each coin, there are 2 possible outcomes (heads or tails). Since there are five coins, you multiply the number of outcomes for each coin: 2 x 2 x 2 x 2 x 2.

🎯 Exam Tip: For 'k' identical independent trials, each with 'n' outcomes, the total number of sample points is n\(^k\).

 

Question 24. State the number of sample points in the random experiment of tossing one balanced coin and two balanced dice simultaneously.
Answer: The number of sample points in the random experiment of tossing one balanced coin and two balanced dice simultaneously is n = 2\(^1\) \( \times \) 6\(^2\) = 2 \( \times \) 36 = 72.
In simple words: For the coin, there are 2 outcomes. For each die, there are 6 outcomes. Since there are two dice, that's 6 times 6. Multiply all these possibilities together to get the total number of outcomes.

🎯 Exam Tip: When combining different independent random experiments, the total number of sample points is the product of the number of outcomes for each individual experiment.

 

Question 25. Is it possible that P (A) = 0.7 and P (A \( \cup \) B) = 0.45 for two events A and B in a sample space ?
Answer: P (A) = 0.7 and P(A \( \cup \) B) = 0.45 is not possible because P(A \( \cup \) B) < P(A).
In simple words: The probability of the union of two events (A or B) must always be greater than or equal to the probability of any single event within that union. Here, P(A \( \cup \) B) is smaller than P(A), which is impossible.

🎯 Exam Tip: A fundamental property of probability is that P(A) \( \le \) P(A \( \cup \) B) and P(B) \( \le \) P(A \( \cup \) B). If this condition is violated, the probabilities are invalid.

 

Question 26. Two cards are selected one by one with replacement from 52 cards. State the number of elements in the sample space of the random experiment.
Answer: The number of elements in the sample space of this random experiment of drawing two cards one by one with replacement from 52 cards is n = 52C\(_{1}\) \( \times \) 52C\(_{1}\) = 52 \( \times \) 52 = 2704
In simple words: Since the first card is replaced before drawing the second, the choice for the second card is independent of the first. So, you have 52 choices for the first card and 52 choices for the second, making a total of 52 times 52 possible outcomes.

🎯 Exam Tip: "With replacement" implies that the outcomes of successive selections are independent, and the number of choices remains constant for each selection. This leads to multiplication of possibilities.

 

Question 27. For two independent events A and B, P(B|A) = \( \frac{1}{2} \) and P(A \( \cap \) B) = \( \frac{1}{5} \). Find P(A).
Answer: P(B|A) = \( \frac{1}{2} \), P(A \( \cap \) B) = \( \frac{1}{5} \), P(A) = ?
P(B|A) = \( \frac{P(A \cap B)}{P(A)} \)
.: \( \frac{1}{2} \) = \( \frac{\frac{1}{5}}{P(A)} \)
.. \( \frac{1}{2} \)p(A) = \( \frac{1}{5} \)
.. P(A) = \( \frac{1}{5} \) x 2 = \( \frac{2}{5} \)
In simple words: If events are independent, the conditional probability P(B|A) is just P(B). However, the question gives P(B|A) and P(A \( \cap \) B). Using the conditional probability formula, we can solve for P(A) by dividing P(A \( \cap \) B) by P(B|A).

🎯 Exam Tip: Even if events are independent, you can still use the conditional probability formula. For independent events, P(B|A) will simplify to P(B), which means \( \frac{P(A \cap B)}{P(A)} \) = P(B), leading to P(A \( \cap \) B) = P(A)P(B).

 

Question 28. 1998 tickets out of 2000 tickets do not have a prize. If a person randomly selects one ticket from 2000 tickets, then what is the probability that the ticket selected is eligible for prize ?
Answer: Total number of tickets = 2000
No. of tickets do not have a prize = 1998
No. of tickets eligible for prize = 2000 - 1998
= 2
Total number of outcomes of selecting a ticket is n = 2000C\(_{1}\) = 2000
A = Event that the selected ticket is eligible for prize
.. m = 2c\(_{1}\) = 2
.. P(A) = \( \frac{m}{n} \)
= \( \frac{2}{2000} \)
= \( \frac{1}{1000} \)
In simple words: First, find how many tickets have a prize by subtracting the non-prize tickets from the total. Then, divide the number of prize tickets by the total number of tickets to get the probability.

🎯 Exam Tip: Carefully identify the number of favorable outcomes and the total number of possible outcomes. Always simplify the probability fraction to its lowest terms.

 

Section C

Answer The Following Questions As Required:

Question 1. Define the following events and draw their Venn diagram:
1. Mutually exclusive events
2. Union of events
3. Intersection of events
4. Difference event
Answer:
1. Mutually exclusive events: Suppose, U is a finite sample space. A and B are any two events of U. If A \( \cap \) B = \( \Phi \), then A and B are mutually exclusive events.
ℹ️ चित्र व्याख्या (Diagram Explanation): इस वेन डायग्राम में, दो वृत्त A और B एक बड़े आयत (U) के भीतर हैं, और वे आपस में ओवरलैप नहीं करते हैं। इसका मतलब है कि इवेंट A और इवेंट B में कोई भी परिणाम सामान्य नहीं है, जिससे पता चलता है कि वे परस्पर अनन्य घटनाएँ हैं।
In simple words: Mutually exclusive events are events that cannot happen at the same time, meaning they have no common outcomes. Their Venn diagrams show no overlap.

 

2. Union of events: Suppose, U is a finite sample space. A and B are any two events of U. Event A occurs or event B occurs or events A and B occur togethers, i.e., at least one of the events A and B occurs is called the union of events A and B. It is denoted by A \( \cup \) B. Thus,
A \( \cup \) B = {x; x \( \in \) A or x \( \in \) B or x \( \in \) A \( \cap \) B}
ℹ️ चित्र व्याख्या (Diagram Explanation): इस वेन डायग्राम में, दो वृत्त A और B एक बड़े आयत (U) के भीतर हैं, और उनका कुछ हिस्सा ओवरलैप करता है। यूनियन A \( \cup \) B को A और B दोनों वृत्तों के पूरे छायांकित क्षेत्र के रूप में दर्शाया गया है, जिसमें उनका सामान्य ओवरलैप क्षेत्र भी शामिल है। यह A, B, या दोनों में किसी भी परिणाम को दर्शाता है।
In simple words: The union of events A and B includes all outcomes that are in A, or in B, or in both A and B. It represents "A or B".

 

3. Intersection of Events: Suppose, U is a finite sample space. A and B are any two events of U. The event that A and B occur together is called the intersection of events A and B. It is denoted by A \( \cap \) B. Thus,
A \( \cap \) B = {x; x \( \in \) A and x \( \in \) B}
ℹ️ चित्र व्याख्या (Diagram Explanation): इस वेन डायग्राम में, दो वृत्त A और B एक बड़े आयत (U) के भीतर हैं और वे एक-दूसरे को काटते हैं। इंटरसेक्शन A \( \cap \) B को वृत्तों के बीच के ओवरलैप वाले क्षेत्र के रूप में दर्शाया गया है। यह उन परिणामों को दिखाता है जो इवेंट A और इवेंट B दोनों में सामान्य हैं।
In simple words: The intersection of events A and B includes only the outcomes that are common to both A and B. It represents "A and B".

 

4. Difference event: Suppose, U is a finite sample space. A and B are any two events of U. The event that A occurs but B does not occur is called the difference event of A and B. It is denoted by A - B or A \( \cap \) B'. Similarly, the event that A does not occur but B occurs is called the difference event of B and A. It is denoted by B - A or A' \( \cap \) B. Thus,
A - B = {x; x \( \in \) A and x \( \notin \) B}
ℹ️ चित्र व्याख्या (Diagram Explanation): इस वेन डायग्राम में, दो वृत्त A और B एक बड़े आयत (U) के भीतर हैं। इवेंट A - B को वृत्त A के उस हिस्से के रूप में दिखाया गया है जो वृत्त B के साथ ओवरलैप नहीं करता है। यह उन परिणामों को दर्शाता है जो A में तो हैं लेकिन B में नहीं हैं। इसी तरह, B - A को वृत्त B के उस हिस्से के रूप में दिखाया गया है जो वृत्त A के साथ ओवरलैप नहीं करता है।
In simple words: The difference event A-B means that event A occurs, but event B does not occur. It includes all outcomes of A that are not also in B.

🎯 Exam Tip: For definitions requiring Venn diagrams, always ensure the sample space (U) is represented, and the shaded regions accurately depict the defined event. Pay attention to labels and clear boundaries.

Gujarat Board Textbook Solutions Class 12 Statistics Part 2 Chapter 1 Probability Ex 1

Section B

Question 2.Draw the Venn diagram for A - B, the difference event of A and B.
Answer:The Venn diagram for difference event A-B is shown below, illustrating the region where A occurs but B does not.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख दो घटनाओं A और B के लिए वेन डायग्राम दिखाता है। A-B क्षेत्र वह हिस्सा है जहाँ घटना A होती है लेकिन घटना B नहीं होती है। यह उन तत्वों को दर्शाता है जो केवल A में हैं।

Class of Employees	Gender	Total
	Males	Females
Class 3	3600	900	4500
Class 4	400	1100	1500
Total	4000	2000	6000

In simple words: This Venn diagram shows the region representing "A minus B", which means the outcomes that are in event A but not in event B.

🎯 Exam Tip: Clearly label the sets and the shaded region in Venn diagrams to accurately represent set operations like difference.

Question 3.Define an event.
Answer:An event is any subset of the sample space of a random experiment. It is commonly represented by letters like A, B, C, and so on.
In simple words: An event is a specific outcome or a group of outcomes from a random experiment.

🎯 Exam Tip: Remember that an event must always be a subset of the sample space.

Question 4.Write the sample space of a random experiment of throwing one balanced die and a balanced coin simultaneously.
Answer:When a fair die and a fair coin are tossed together, the sample space is: \(U = \{(1, H), (2, H), (3, H), (4, H), (5, H), (6, H), (1, T), (2, T), (3, T), (4, T), (5, T), (6, T)\}\) Here, H stands for Head, T for Tail, and 1, 2, 3, 4, 5, 6 are the numbers on the die.
In simple words: When you roll a die and flip a coin at the same time, the sample space lists all possible pairs, like getting a 1 and a Head, or a 6 and a Tail.

🎯 Exam Tip: Ensure all possible combinations are listed systematically to form a complete sample space.

Question 5.Define conditional probability.
Answer:For a finite sample space U, and any two events A and B within U, the conditional probability of event B is the probability that event B occurs given that event A has already occurred.
In simple words: Conditional probability is finding the chance of one event happening, knowing that another event has already happened.

🎯 Exam Tip: Clearly state the condition under which the probability is being calculated when defining conditional probability.

Question 6.State the formula for the probability of occurrence of at least one event out of three events A, B and C.
Answer:The formula for the probability of at least one event occurring from three events A, B, and C is:
\(P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C)\)
In simple words: To find the chance that A, B, or C happens, you add their individual chances, subtract the chances of any two happening together, and then add back the chance of all three happening together.

🎯 Exam Tip: This formula is an extension of the addition rule for two events and is crucial for problems involving three intersecting events.

Question 7.Define independent events.
Answer:Given a finite sample space U, two events A and B are considered independent if the probability of event A occurring does not depend on whether event B has occurred or not.
In simple words: Two events are independent if what happens in one event does not change the chances of the other event happening.

🎯 Exam Tip: Independent events are often characterized by \(P(A \cap B) = P(A) \times P(B)\) or \(P(A|B) = P(A)\).

Question 8.Write the law of multiplication of probability for two independent events A and B in a sample space.
Answer:The law of multiplication for two independent events A and B in a sample space is:
\(P(A \cap B) = P(A) \times P(B)\)
In simple words: If two events don't affect each other, the chance of both happening is simply the chance of the first multiplied by the chance of the second.

🎯 Exam Tip: This formula is fundamental for calculating the probability of two independent events occurring together.

Question 9.Interpret P(A|B) and P(B|A).
Answer:The interpretation of \(P(A|B)\) is the conditional probability of event A occurring, given that event B has already occurred. The interpretation of \(P(B|A)\) is the conditional probability of event B occurring, given that event A has already occurred.
In simple words: P(A|B) means the probability of A happening if we know B has already happened. P(B|A) means the probability of B happening if we know A has already happened.

🎯 Exam Tip: Understanding conditional probability notation is key to solving problems where the outcome of one event influences the probability of another.

Question 10.When can we say that three events A, B and C in a sample space are exhaustive?
Answer:Three events A, B, and C in a sample space are considered exhaustive if their probabilities sum up to 1, meaning that at least one of these events must occur.
\(P(A) + P(B) + P(C) = 1\)
In simple words: Events A, B, and C are exhaustive if they cover all possible outcomes, so one of them is sure to happen.

🎯 Exam Tip: For exhaustive events, their union represents the entire sample space.

Question 11.Arrange \(P(A \cup B)\), \(P(A)\), \(P(A \cap B)\), 0, \(P(A) + P(B)\) in the ascending order.
Answer:The ascending order of the probabilities is:
\(0, P(A \cap B), P(A), P(A \cup B), P(A) + P(B)\)
In simple words: The smallest is zero. Then comes the chance of both A and B happening. Next is the chance of A. Then the chance of A or B happening. The largest is the sum of chances of A and B, which can be more than 1 if they overlap.

🎯 Exam Tip: Remember the relationships between probabilities: \(0 \le P(E) \le 1\), \(P(A \cap B) \le P(A)\), \(P(A) \le P(A \cup B)\), and \(P(A \cup B) \le P(A) + P(B)\).

Question 12.Define:
Answer:1. Random Experiment: A random experiment is one that can be repeated many times under the same conditions, where all possible outcomes are known, but the specific outcome of any single trial cannot be predicted with certainty.
2. Sample Space: The sample space of a random experiment is the collection of all possible outcomes. It is denoted by U or S.
3. Equi-probable Events: Equi-probable events are those where there is no clear reason to believe that one event is more or less likely to occur than any other event in a random experiment.
4. Favourable Outcomes: Favourable outcomes are the specific results from a random experiment that indicate the occurrence of a particular event A.
5. Probability (Mathematical Definition): If a finite sample space has 'n' mutually exclusive, exhaustive, and equi-probable outcomes, and 'm' of these outcomes are favourable for event A, then the probability of A is \(P(A) = \frac{m}{n}\).
6. Probability (Statistical Definition): If a random experiment is repeated 'n' times under identical conditions, and an event A occurs 'm' times, then the relative frequency \(\frac{m}{n}\) estimates the probability of event A. As 'n' approaches infinity, this limiting value is called the probability of event A.
\[P(A) = \lim_{n \to \infty} \frac{m}{n}\]
7. Impossible Event: An impossible event is a special subset \(\Phi\) or \(\{\}\) of the sample space that contains no outcomes and therefore cannot occur.
8. Certain Event: A certain event is a special subset U of the sample space that includes all possible outcomes and is guaranteed to occur.
In simple words: This question asks for definitions of key probability terms. A random experiment has unknown results but known possibilities. A sample space lists all possible results. Equi-probable events have equal chances. Favorable outcomes are the ones we want. Mathematical probability uses counts to find chances, while statistical probability uses observed frequencies. An impossible event never happens, and a certain event always happens.

🎯 Exam Tip: Learn these definitions precisely as they form the foundation of probability theory. Pay attention to the conditions for each definition (e.g., mutually exclusive, exhaustive, equi-probable for mathematical definition).

Question 13.For two events A and B in a sample space, \(A \cap B = \Phi\) and \(A \cup B = U\). State the values of \(P(A \cap B)\) and \(P(A \cup B)\).
Answer:Given \(A \cap B = \Phi\) and \(A \cup B = U\). Therefore, \(P(A \cap B) = 0\) and \(P(A \cup B) = 1\).
In simple words: If events A and B cannot happen at the same time (their intersection is empty), the chance of both happening is zero. If A and B together cover all possibilities (their union is the whole sample space), the chance of either one happening is one.

🎯 Exam Tip: Remember that an impossible event has a probability of 0, and a certain event (or the entire sample space) has a probability of 1.

Question 14.If two events A and B in a sample space are independent, then state the formula for P(A ∪ B).
Answer:If A and B are independent events, the formula for \(P(A \cup B)\) is:
\(P(A \cup B) = P(A) + P(B) - P(A \cap B)\) Since A and B are independent, \(P(A \cap B) = P(A) \cdot P(B)\).
So, \(P(A \cup B) = P(A) + P(B) - [P(A) \cdot P(B)]\)
In simple words: For independent events, the chance of A or B happening is found by adding their individual chances and then subtracting the product of their chances.

🎯 Exam Tip: Always use the specific multiplication rule for independent events, \(P(A \cap B) = P(A) \cdot P(B)\), when asked to find \(P(A \cup B)\) for independent events.

Question 15.If A = {x|0 < x < 1} and B = {x|\(\frac{1}{4}\) ≤ x ≤ 3}, then find \(A \cap B\).
Answer:Given A = {x | 0 < x < 1} and B = {x | \(\frac{1}{4}\) ≤ x ≤ 3}.
To find \(A \cap B\), we need to find the values of x that are common to both sets. For A, x is strictly between 0 and 1. For B, x is between \(\frac{1}{4}\) (inclusive) and 3 (inclusive).
The common interval will be where x is greater than 0 and greater than or equal to \(\frac{1}{4}\), and also less than 1 and less than or equal to 3.
So, the lower bound is \(\max(0, \frac{1}{4}) = \frac{1}{4}\). The upper bound is \(\min(1, 3) = 1\).
Since A is strictly less than 1, the intersection will also be strictly less than 1. Therefore, \(A \cap B = \{x | \frac{1}{4} \le x < 1\}\).
In simple words: We are looking for numbers that are in both set A and set B. Set A has numbers between 0 and 1. Set B has numbers between 1/4 and 3. The numbers that fit both rules are those from 1/4 up to (but not including) 1.

🎯 Exam Tip: When finding the intersection of intervals, determine the maximum of the lower bounds and the minimum of the upper bounds to define the new interval.

Question 16.For two independent events A and B, \(P(A) = 0.5\) and \(P(B) = 0.7\). Find \(P(A' \cap B')\).
Answer:Given that A and B are independent events. This means \(P(A) = 0.5\) and \(P(B) = 0.7\). If A and B are independent, then their complements, A' and B', are also independent. First, find \(P(A')\) and \(P(B')\):
\(P(A') = 1 - P(A) = 1 - 0.5 = 0.5\)
\(P(B') = 1 - P(B) = 1 - 0.7 = 0.3\) Since A' and B' are independent, the probability of their intersection is the product of their individual probabilities:
\(P(A' \cap B') = P(A') \cdot P(B')\)
\(P(A' \cap B') = 0.5 \times 0.3\)
\(P(A' \cap B') = 0.15\)
In simple words: If A and B don't affect each other, then the chances of A not happening and B not happening also don't affect each other. So, we find the chance of A not happening (1 - P(A)) and the chance of B not happening (1 - P(B)), and then multiply these two chances.

🎯 Exam Tip: Remember De Morgan's Law: \(P(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B)\). This provides an alternative method to verify your answer if you calculate \(P(A \cup B)\) first for independent events.

Question 17.If \(P(A) = 0.8\) and \(P(A \cap B) = 0.25\), find \(P(A - B)\).
Answer:Given \(P(A) = 0.8\) and \(P(A \cap B) = 0.25\). The probability of event A occurring but not event B, denoted as \(P(A - B)\), is calculated as:
\(P(A - B) = P(A) - P(A \cap B)\) Substitute the given values:
\(P(A - B) = 0.8 - 0.25\)
\(P(A - B) = 0.55\)
In simple words: To find the chance that only A happens and B does not, we take the total chance of A happening and subtract the chance that both A and B happen together.

🎯 Exam Tip: The formula \(P(A - B) = P(A) - P(A \cap B)\) is valid for any two events A and B. Visualizing it with a Venn diagram can aid understanding.

Question 18.If \(P(A) = 0.3\) and \(P(A \cap B) = 0.03\), find \(P(B|A)\).
Answer:Given \(P(A) = 0.3\) and \(P(A \cap B) = 0.03\). The formula for conditional probability \(P(B|A)\) is:
\[P(B|A) = \frac{P(A \cap B)}{P(A)}\] Substitute the given values:
\[P(B|A) = \frac{0.03}{0.3}\]
\[P(B|A) = 0.1\]
In simple words: To find the chance of B happening when we already know A has happened, we divide the chance of both A and B happening by the chance of A happening.

🎯 Exam Tip: Ensure that the probability of the conditioning event (the denominator) is not zero when applying the conditional probability formula.

Question 19.If \(P(A) = P(B) = K\) for two mutually exclusive events A and B, find \(P(A \cup B)\).
Answer:Given \(P(A) = P(B) = K\). A and B are mutually exclusive events, which means they cannot occur at the same time. For mutually exclusive events, the probability of their intersection is zero:
\(P(A \cap B) = 0\) The addition rule for probability states:
\(P(A \cup B) = P(A) + P(B) - P(A \cap B)\) Substitute the given values and the fact that they are mutually exclusive:
\(P(A \cup B) = K + K - 0\)
\(P(A \cup B) = 2K\)
In simple words: If two events cannot happen together and have the same chance 'K', then the chance of either one happening is just the sum of their individual chances, which is K plus K, or 2K.

🎯 Exam Tip: The key property for mutually exclusive events is \(P(A \cap B) = 0\), which simplifies the addition rule significantly.

Question 20.If \(P(A' \cap B) = 0.45\) and \(A \cap B = \Phi\), find \(P(B)\).
Answer:Given \(P(A' \cap B) = 0.45\) and \(A \cap B = \Phi\). When \(A \cap B = \Phi\), it means events A and B are mutually exclusive. The event \(A' \cap B\) represents the outcomes that are in B but not in A. Since \(A \cap B = \Phi\), there is no overlap between A and B. This implies that if an outcome is in B, it cannot be in A. Therefore, an outcome in B is automatically in \(A' \cap B\). So, \(P(A' \cap B) = P(B) - P(A \cap B)\). Since \(P(A \cap B) = 0\) for mutually exclusive events:
\(P(A' \cap B) = P(B) - 0\)
\(P(A' \cap B) = P(B)\) Substitute the given value:
\(0.45 = P(B)\) Therefore, \(P(B) = 0.45\).
In simple words: If A and B can't happen at the same time, then the chance of B happening but not A is simply the chance of B happening, because if B happens, A automatically doesn't.

🎯 Exam Tip: Recognize that for mutually exclusive events, the outcomes in B are entirely separate from A, so \(A' \cap B\) is equivalent to B itself.

Question 21.Two events A and B in a sample space are mutually exclusive and exhaustive. If \(P(A) = \frac{1}{3}\), find \(P(B)\).
Answer:Given that A and B are mutually exclusive events. This means \(P(A \cap B) = 0\). Given that A and B are exhaustive events. This means their union covers the entire sample space, so \(P(A \cup B) = 1\). The addition rule for probability is:
\(P(A \cup B) = P(A) + P(B) - P(A \cap B)\) Substitute the values for mutually exclusive and exhaustive events:
\(1 = P(A) + P(B) - 0\)
\(1 = P(A) + P(B)\) Given \(P(A) = \frac{1}{3}\):
\(1 = \frac{1}{3} + P(B)\) Now, solve for \(P(B)\):
\(P(B) = 1 - \frac{1}{3}\)
\(P(B) = \frac{3 - 1}{3}\)
\(P(B) = \frac{2}{3}\)
In simple words: If two events cannot happen together but one of them must happen, and we know the chance of the first event, then the chance of the second event is 1 minus the chance of the first event.

🎯 Exam Tip: For events that are both mutually exclusive and exhaustive, the sum of their probabilities is always 1, i.e., \(P(A) + P(B) = 1\).

Question 22.2% items in a lot are defective. What is the probability that an item randomly selected from this lot is non-defective?
Answer:Let A be the event that an item is defective. Given that 2% of items are defective, the probability of event A is:
\(P(A) = 2\% = \frac{2}{100} = 0.02\) Let A' be the event that an item is non-defective. The probability of a non-defective item is the complement of a defective item:
\(P(A') = 1 - P(A)\)
\(P(A') = 1 - 0.02\)
\(P(A') = 0.98\)
In simple words: If 2 out of 100 items are faulty, then the remaining 98 out of 100 items are not faulty. So, the chance of picking a non-faulty item is 98%.

🎯 Exam Tip: The probability of an event and its complement must always sum to 1. This is a fundamental concept in probability.

Question 23.State the number of sample points in the random experiment of tossing five balanced coins.
Answer:For a single balanced coin, there are 2 possible outcomes (Head or Tail). When tossing five balanced coins simultaneously, the number of sample points in the sample space is calculated by raising the number of outcomes per coin (2) to the power of the number of coins (5).
Number of sample points = \(2^5 = 32\).
In simple words: Each coin flip has two results. If you flip five coins, you multiply 2 by itself five times to get the total number of different possible outcomes.

🎯 Exam Tip: For 'n' independent trials, each with 'k' possible outcomes, the total number of sample points is \(k^n\).

Question 24.State the number of sample points in the random experiment of tossing one balanced coin and two balanced dice simultaneously.
Answer:For one balanced coin, there are 2 possible outcomes. For one balanced die, there are 6 possible outcomes. For two balanced dice, the number of outcomes is \(6 \times 6 = 6^2 = 36\). When tossing one balanced coin and two balanced dice simultaneously, the total number of sample points is the product of the number of outcomes for each independent event.
Number of sample points = (Outcomes of coin) × (Outcomes of first die) × (Outcomes of second die)
Number of sample points = \(2^1 \times 6^2 = 2 \times 36 = 72\).
In simple words: You multiply the number of results from the coin (2) by the number of results from the first die (6) and then by the number of results from the second die (6) to get all possible combinations.

🎯 Exam Tip: For multiple independent experiments, the total number of outcomes in the sample space is the product of the number of outcomes for each individual experiment.

Question 25.Is it possible that \(P(A) = 0.7\) and \(P(A \cup B) = 0.45\) for two events A and B in a sample space?
Answer:Given \(P(A) = 0.7\) and \(P(A \cup B) = 0.45\). We know that the probability of the union of two events must always be greater than or equal to the probability of each individual event. This is because \(A\) is a subset of \(A \cup B\), and \(B\) is a subset of \(A \cup B\). Therefore, it must be true that \(P(A \cup B) \ge P(A)\) and \(P(A \cup B) \ge P(B)\). In this case, \(P(A \cup B) = 0.45\) and \(P(A) = 0.7\). Since \(0.45 < 0.7\), it contradicts the fundamental rule that \(P(A \cup B) \ge P(A)\). Thus, it is not possible for \(P(A) = 0.7\) and \(P(A \cup B) = 0.45\) simultaneously for two events A and B in a sample space.
In simple words: The chance of A or B happening together can never be less than the chance of just A happening alone. Since 0.45 is less than 0.7, this situation is not possible.

🎯 Exam Tip: Always remember that the probability of a union of events must be at least as large as the probability of any single event within that union.

Question 26.Two cards are selected one by one with replacement from 52 cards. State the number of elements in the sample space of the random experiment.
Answer:When selecting cards with replacement, the outcome of the first selection does not affect the possible outcomes of the second selection. Number of cards in a standard deck = 52. For the first card selection, there are 52 possible outcomes. Since the card is replaced, for the second card selection, there are still 52 possible outcomes. The total number of elements in the sample space is the product of the number of outcomes for each selection:
Number of elements = (Outcomes for 1st card) × (Outcomes for 2nd card)
Number of elements = \(52 \times 52 = 2704\). This can also be expressed using combinations if we consider ordered selections with replacement, which is essentially permutation with repetition, \(n^r\). Here, \(n=52\) and \(r=2\).
Alternatively, using combinations notation for ordered selections (which is essentially what \(n \times n\) implies for two selections):
\(n = 52C_1 \times 52C_1 = 52 \times 52 = 2704\).
In simple words: When you pick a card, put it back, and then pick another, the first pick doesn't change what you can pick the second time. So, you multiply the number of choices for the first pick by the number of choices for the second pick.

🎯 Exam Tip: The phrase "with replacement" indicates that each selection is an independent event, and the total number of outcomes is found by multiplying the number of options at each step.

Question 27.For two independent events A and B, \(P(B|A) = \frac{1}{2}\) and \(P(A \cap B) = \frac{1}{5}\). Find \(P(A)\).
Answer:Given \(P(B|A) = \frac{1}{2}\) and \(P(A \cap B) = \frac{1}{5}\). The formula for conditional probability is:
\[P(B|A) = \frac{P(A \cap B)}{P(A)}\] Substitute the given values into the formula:
\[\frac{1}{2} = \frac{\frac{1}{5}}{P(A)}\] To find \(P(A)\), rearrange the equation:
\[P(A) = \frac{\frac{1}{5}}{\frac{1}{2}}\]
\[P(A) = \frac{1}{5} \times 2\]
\[P(A) = \frac{2}{5}\]
In simple words: We know the chance of B happening if A already happened, and we know the chance of both A and B happening. Using the rule for conditional probability, we can work backward to find the chance of A happening.

🎯 Exam Tip: This problem tests your ability to rearrange the conditional probability formula to solve for an unknown probability. Be careful with fractions.

Question 28.1998 tickets out of 2000 tickets do not have a prize. If a person randomly selects one ticket from 2000 tickets, then what is the probability that the ticket selected is eligible for prize?
Answer:Total number of tickets = 2000. Number of tickets without a prize = 1998. To find the number of tickets eligible for a prize, subtract the non-prize tickets from the total:
Number of tickets eligible for prize = Total tickets - Tickets without a prize
Number of tickets eligible for prize = \(2000 - 1998 = 2\). Let A be the event that the selected ticket is eligible for a prize. The total number of primary outcomes (selecting one ticket from 2000) is \(n = {}^{2000}C_1 = 2000\). The number of favourable outcomes for event A (selecting one prize-eligible ticket) is \(m = {}^{2}C_1 = 2\). The probability of event A is:
\[P(A) = \frac{m}{n} = \frac{\text{Number of prize tickets}}{\text{Total number of tickets}}\]
\[P(A) = \frac{2}{2000}\]
\[P(A) = \frac{1}{1000}\]
In simple words: First, figure out how many tickets actually have a prize. If 1998 out of 2000 have no prize, then 2 tickets have a prize. The chance of picking a prize ticket is these 2 tickets divided by the total 2000 tickets.

🎯 Exam Tip: Clearly identify the number of favourable outcomes and the total number of possible outcomes before applying the basic probability formula.

Section C

Question 1.Define the following events and draw their Venn diagram: 1. Mutually exclusive events 2. Union of events 3. Intersection of events 4. Difference event 5. Exhaustive events 6. Complementary event
Answer:1. Mutually exclusive events: Suppose U is a finite sample space. Two events A and B are mutually exclusive if they cannot occur at the same time, meaning their intersection is empty (\(A \cap B = \Phi\)).
ℹ️ चित्र व्याख्या (Diagram Explanation): इस वेन डायग्राम में, दो वृत्त A और B एक दूसरे को काटते नहीं हैं, जो दर्शाता है कि उनमें कोई सामान्य परिणाम नहीं है। इसका मतलब है कि घटनाएँ A और B एक साथ घटित नहीं हो सकती हैं।
2. Union of events: Suppose U is a finite sample space. The union of two events A and B is the event where A occurs, or B occurs, or both A and B occur. It represents at least one of the events occurring and is denoted by \(A \cup B\).
Thus, \(A \cup B = \{x; x \in A \text{ or } x \in B \text{ or } x \in A \cap B\}\).
ℹ️ चित्र व्याख्या (Diagram Explanation): इस वेन डायग्राम में, वृत्त A और वृत्त B दोनों के सभी क्षेत्र को छायांकित किया गया है, जो उनकी 'यूनियन' को दर्शाता है। यह उन सभी परिणामों का प्रतिनिधित्व करता है जो घटना A, घटना B, या दोनों में हैं।
3. Intersection of Events: Suppose U is a finite sample space. The intersection of two events A and B is the event where both A and B occur together. It is denoted by \(A \cap B\).
Thus, \(A \cap B = \{x; x \in A \text{ and } x \in B\}\).
ℹ️ चित्र व्याख्या (Diagram Explanation): इस वेन डायग्राम में, वृत्त A और वृत्त B के बीच का ओवरलैपिंग क्षेत्र छायांकित है, जो उनकी 'इंटरसेक्शन' को दर्शाता है। यह उन परिणामों का प्रतिनिधित्व करता है जो घटना A और घटना B दोनों में सामान्य हैं।
4. Difference event: Suppose U is a finite sample space. The difference event \(A - B\) is the event where A occurs but B does not occur. It is also denoted by \(A \cap B'\). Similarly, \(B - A\) is the event where B occurs but A does not, denoted by \(B \cap A'\).
Thus, \(A - B = \{x; x \in A \text{ and } x \notin B\}\).
ℹ️ चित्र व्याख्या (Diagram Explanation): इस वेन डायग्राम में, केवल वृत्त A का वह भाग छायांकित है जो वृत्त B के साथ ओवरलैप नहीं करता है। यह 'A माइनस B' को दर्शाता है, यानी घटना A के वे परिणाम जो घटना B में नहीं हैं। दूसरे डायग्राम में, केवल B का वह भाग छायांकित है जो A में नहीं है, जो 'B माइनस A' को दर्शाता है।
5. Exhaustive events: Suppose U is a finite sample space. Two events A and B are exhaustive if their union covers the entire sample space, meaning \(A \cup B = U\). This implies that at least one of them must occur.
ℹ️ चित्र व्याख्या (Diagram Explanation): इस वेन डायग्राम में, वृत्त A और वृत्त B मिलकर पूरे नमूना स्थान U को कवर करते हैं। कोई भी क्षेत्र बिना छायांकित नहीं है, जो दर्शाता है कि घटना A या घटना B (या दोनों) का घटित होना निश्चित है, और उनके बाहर कोई परिणाम नहीं है।
6. Complementary event: Suppose U is a finite sample space. The complementary event of A, denoted by \(A'\) or \(A^c\), is the event that A does not occur. It consists of all elements in U that are not in A.
Thus, \(A' = \{x; x \notin A, x \in U\}\).
ℹ️ चित्र व्याख्या (Diagram Explanation): इस वेन डायग्राम में, नमूना स्थान U के अंदर वृत्त A के बाहर का पूरा क्षेत्र छायांकित है। यह 'A के पूरक' को दर्शाता है, यानी नमूना स्थान के वे सभी परिणाम जो घटना A में नहीं हैं।
In simple words: This question asks to define different types of events in probability and show them with pictures (Venn diagrams). Mutually exclusive means events can't happen together. Union means either event happens. Intersection means both events happen. Difference means one event happens but the other doesn't. Exhaustive means events cover all possible results. Complementary means an event does not happen.

🎯 Exam Tip: For each definition, provide a clear, concise explanation and ensure the Venn diagram accurately illustrates the concept. Focus on the relationships between the events within the sample space.

Question 2.Give the illustrations of finite and infinite sample space.
Answer:Finite sample space: An example of a finite sample space is the set of outcomes when throwing a fair six-faced die. The sample space U = {1, 2, 3, 4, 5, 6} contains a definite, countable number of outcomes. Infinite sample space: An example of an infinite sample space is in a random experiment where you select a card from a deck of 52 cards repeatedly (with replacement) until the ace of hearts is drawn. The number of trials required could be any positive integer (1, 2, 3, ...), making the sample space countably infinite.
In simple words: A finite sample space has a limited number of possible results, like rolling a die (1 to 6). An infinite sample space has endless possible results, like continuously drawing cards until a specific one appears, as you might never pick it on your first try.

🎯 Exam Tip: For illustrations, choose simple, clear examples that distinctly show the difference between a countable, finite set of outcomes and a potentially unlimited set of outcomes.

Question 3.Give the illustrations of impossible and certain event.
Answer:Impossible event: An illustration of an impossible event is getting a number greater than 6 on the upper face of a standard six-sided die when rolled. Since a standard die only has faces numbered 1 to 6, rolling a 7 or higher is impossible. Certain event: An illustration of a certain event is getting either a head or a tail when tossing a balanced coin. These are the only two possible outcomes, so it's guaranteed that one of them will occur.
In simple words: An impossible event is something that can never happen, like rolling a 7 on a normal die. A certain event is something that will always happen, like getting a head or a tail when flipping a coin.

🎯 Exam Tip: Choose simple, everyday examples that clearly demonstrate the concepts of impossibility (probability = 0) and certainty (probability = 1).

Question 4.State the characteristics of random experiment.
Answer:The characteristics of a random experiment are as follows: - It can be repeated under identical conditions multiple times. - All its possible outcomes are known beforehand. - It is impossible to predict with certainty which specific outcome will occur in any given trial. - It always results in a definite outcome after each trial.
In simple words: A random experiment is one you can do over and over, you know all the possible results, but you can't guess the exact result before it happens, and it always gives some result.

🎯 Exam Tip: Focus on the four core traits: repeatability, known outcomes, unpredictability of specific outcome, and definite result. These distinguish random experiments from deterministic ones.

Question 5.State the assumptions of mathematical definition of probability.
Answer:The assumptions underlying the mathematical definition of probability are: - The total number of outcomes in the sample space must be finite. - All possible outcomes in the sample space are known. - All outcomes in the sample space are equi-probable (equally likely to occur).
In simple words: For math probability to work, we assume there are a limited number of results, we know all of them, and each result has an equal chance of happening.

🎯 Exam Tip: These assumptions are crucial. If any of them are not met, the mathematical definition of probability may not be applicable.

Question 6.State the limitations of mathematical definition of probability.
Answer:The limitations of the mathematical definition of probability are: - It cannot be used to find the probability of an event if the number of outcomes in the sample space is infinite. - It cannot be applied if the total number of outcomes in the sample space is not known.
In simple words: This method doesn't work if there are endless possible results or if we don't know how many total results there are.

🎯 Exam Tip: Understand that these limitations often lead to the use of other definitions of probability, such as the statistical or axiomatic definitions.

Question 7.State the limitations of statistical definition of probability.
Answer:The limitations of the statistical definition of probability are: - It is not practical to perform an experiment an infinite number of times (\(n \to \infty\)). - The exact value of probability can only be approximated and cannot be known precisely through this method.
In simple words: This method has limits because we can't truly repeat an experiment forever, so we can only get close to the actual probability, not find it exactly.

🎯 Exam Tip: Statistical probability relies on large numbers of trials; its main drawback is the impracticality of achieving a truly infinite number of repetitions.

Question 8.Explain the equi-probable events with illustration.
Answer:Equi-probable events are those events in a random experiment where there is no apparent reason to believe that one event is more or less likely to occur than any other event. In simpler terms, each possible outcome has an equal chance of happening. Illustration: Consider the random experiment of tossing a balanced coin. There are two possible outcomes: getting a Head (H) or getting a Tail (T). Since the coin is balanced, there is no reason to assume that one side is more likely to land up than the other. Therefore, the event of getting a Head and the event of getting a Tail are equi-probable, with \(P(H) = P(T) = \frac{1}{2}\).
In simple words: Equi-probable events are outcomes that have an equal chance of happening. For example, when you flip a fair coin, getting heads or getting tails both have an equal 1/2 chance.

🎯 Exam Tip: Use simple examples like coin tosses or die rolls to illustrate equi-probable events, as they clearly show equal likelihood.

Question 9.State the law of addition of probability for two events A and B. Write the law of addition of probability if these two events are mutually exclusive.
Answer:The general law of addition of probability for two events A and B is:
\(P(A \cup B) = P(A) + P(B) - P(A \cap B)\) If events A and B are mutually exclusive, it means they cannot occur simultaneously, so their intersection is empty (\(A \cap B = \Phi\)) and thus \(P(A \cap B) = 0\). In this case, the law of addition of probability simplifies to:
\(P(A \cup B) = P(A) + P(B)\)
In simple words: The general rule for the chance of A or B happening is to add their individual chances and subtract the chance of both happening. If A and B cannot happen at the same time, then the chance of both happening is zero, so the rule simplifies to just adding their individual chances.

🎯 Exam Tip: Distinguish between the general addition rule and its simplified form for mutually exclusive events. This distinction is critical for correct calculations.

Question 10.State the law of multiplication of probability for two events A and B. Write the law of multiplication of probability if these two events are independent.
Answer:The general law of multiplication of probability for two events A and B is:
\(P(A \cap B) = P(A|B) \cdot P(B)\) OR \(P(A \cap B) = P(B|A) \cdot P(A)\) If events A and B are independent, the occurrence of one does not affect the probability of the other. In this case, \(P(A|B) = P(A)\) and \(P(B|A) = P(B)\). So, for independent events A and B, the law of multiplication simplifies to:
\(P(A \cap B) = P(A) \cdot P(B)\)
In simple words: The general rule for the chance of both A and B happening is to multiply the chance of A by the chance of B given that A happened. But if A and B are independent (meaning one doesn't affect the other), then you simply multiply their individual chances.

🎯 Exam Tip: The general multiplication rule always holds, but for independent events, it simplifies significantly, making calculations easier. Understand when to apply each version.

Question 11.State the following results for two independent events A and B: 1. \(P(A \cap B)\) 2. \(P(A' \cap B')\) 3. \(P(A \cap B')\) 4. \(P(A' \cap B)\)
Answer:For two independent events A and B, the following results are obtained: 1. If A and B are independent, then \(P(A \cap B) = P(A) \times P(B)\). 2. If A and B are independent, their complements A' and B' are also independent. So, \(P(A' \cap B') = P(A') \times P(B')\). 3. If A and B are independent, then A and B' are also independent. So, \(P(A \cap B') = P(A) \times P(B')\). 4. If A and B are independent, then A' and B are also independent. So, \(P(A' \cap B) = P(A') \times P(B)\).
In simple words: If A and B don't affect each other, then: 1. The chance of both A and B is P(A) times P(B). 2. The chance of neither A nor B happening is P(not A) times P(not B). 3. The chance of A happening but not B is P(A) times P(not B). 4. The chance of B happening but not A is P(not A) times P(B).

🎯 Exam Tip: It is important to know that if two events are independent, then any combination of the events and their complements are also independent.

Question 12.If \(P(A) = \frac{1}{3}\), \(P(B) = \frac{2}{3}\) and \(P(A \cap B) = \frac{1}{6}\), then find \(P(A' \cap B')\).
Answer:Given \(P(A) = \frac{1}{3}\), \(P(B) = \frac{2}{3}\), and \(P(A \cap B) = \frac{1}{6}\). We need to find \(P(A' \cap B')\). Using De Morgan's Law, we know that \(A' \cap B' = (A \cup B)'\). Therefore, \(P(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B)\). First, calculate \(P(A \cup B)\) using the addition law of probability:
\(P(A \cup B) = P(A) + P(B) - P(A \cap B)\) Substitute the given values:
\(P(A \cup B) = \frac{1}{3} + \frac{2}{3} - \frac{1}{6}\)
\(P(A \cup B) = \frac{3}{3} - \frac{1}{6}\)
\(P(A \cup B) = 1 - \frac{1}{6}\)
To subtract, find a common denominator:
\(P(A \cup B) = \frac{6}{6} - \frac{1}{6}\)
\(P(A \cup B) = \frac{5}{6}\) Now, substitute this value back into the formula for \(P(A' \cap B')\):
\(P(A' \cap B') = 1 - P(A \cup B)\)
\(P(A' \cap B') = 1 - \frac{5}{6}\)
\(P(A' \cap B') = \frac{6}{6} - \frac{5}{6}\)
\(P(A' \cap B') = \frac{1}{6}\)
In simple words: To find the chance that neither A nor B happens, we first find the chance that A or B happens. We add the chances of A and B, then subtract the chance of both. Once we have the chance of A or B, we subtract that from 1 to get the chance of neither happening.

🎯 Exam Tip: De Morgan's Law \((A' \cap B' = (A \cup B)')\) is a very useful tool for solving problems involving complements of unions or intersections. Always start by calculating \(P(A \cup B)\) if needed.

Question 13.If \(P(B) = 2P(A|B) = 0.4\), then find \(P(A \cap B)\).
Answer:Given \(P(B) = 0.4\) and \(2P(A|B) = 0.4\). From \(2P(A|B) = 0.4\), we can find \(P(A|B)\):
\(P(A|B) = \frac{0.4}{2} = 0.2\) Now, use the conditional probability formula to find \(P(A \cap B)\):
\[P(A|B) = \frac{P(A \cap B)}{P(B)}\] Rearrange the formula to solve for \(P(A \cap B)\):
\(P(A \cap B) = P(A|B) \cdot P(B)\) Substitute the values:
\(P(A \cap B) = 0.2 \times 0.4\)
\(P(A \cap B) = 0.08\)
In simple words: We are given the chance of B and the chance of A happening if B has already happened. To find the chance of both A and B happening, we multiply the chance of A given B by the chance of B.

🎯 Exam Tip: Break down complex given information into simpler parts, like finding \(P(A|B)\) first, before applying the main formula.

Question 14.If the events A and B are independent and \(3P(A) = 2P(B) = 0.12\), then find \(P(A \cap B)\).
Answer:Given that A and B are independent events. Also given \(3P(A) = 0.12\) and \(2P(B) = 0.12\). First, find \(P(A)\) and \(P(B)\): From \(3P(A) = 0.12\):
\(P(A) = \frac{0.12}{3} = 0.04\) From \(2P(B) = 0.12\):
\(P(B) = \frac{0.12}{2} = 0.06\) Since A and B are independent events, the probability of their intersection is the product of their individual probabilities:
\(P(A \cap B) = P(A) \cdot P(B)\) Substitute the calculated values:
\(P(A \cap B) = 0.04 \times 0.06\)
\(P(A \cap B) = 0.0024\)
In simple words: First, we find the individual chances of A and B from the given information. Since A and B don't affect each other, the chance of both happening is simply the chance of A multiplied by the chance of B.

🎯 Exam Tip: When events are independent, the multiplication rule \(P(A \cap B) = P(A) \times P(B)\) is a direct and efficient way to find the probability of their joint occurrence.

Question 15.If \(5P(A) = 3P(B) = 2P(A \cup B) = \frac{3}{2}\) for two events A and B, then find \(P(A' \cup B')\).
Answer:Given \(5P(A) = \frac{3}{2}\), \(3P(B) = \frac{3}{2}\), and \(2P(A \cup B) = \frac{3}{2}\). First, find the individual probabilities: From \(5P(A) = \frac{3}{2}\):
\(P(A) = \frac{3}{2 \times 5} = \frac{3}{10}\) From \(3P(B) = \frac{3}{2}\):
\(P(B) = \frac{3}{2 \times 3} = \frac{1}{2}\) From \(2P(A \cup B) = \frac{3}{2}\):
\(P(A \cup B) = \frac{3}{2 \times 2} = \frac{3}{4}\) We need to find \(P(A' \cup B')\). Using De Morgan's Law, \(P(A' \cup B') = P((A \cap B)') = 1 - P(A \cap B)\). First, calculate \(P(A \cap B)\) using the addition law of probability:
\(P(A \cup B) = P(A) + P(B) - P(A \cap B)\) Rearrange to solve for \(P(A \cap B)\):
\(P(A \cap B) = P(A) + P(B) - P(A \cup B)\) Substitute the calculated probabilities:
\(P(A \cap B) = \frac{3}{10} + \frac{1}{2} - \frac{3}{4}\) To sum these fractions, find a common denominator, which is 20:
\(P(A \cap B) = \frac{3 \times 2}{20} + \frac{1 \times 10}{20} - \frac{3 \times 5}{20}\)
\(P(A \cap B) = \frac{6}{20} + \frac{10}{20} - \frac{15}{20}\)
\(P(A \cap B) = \frac{6 + 10 - 15}{20} = \frac{1}{20}\) Now, find \(P(A' \cup B')\):
\(P(A' \cup B') = 1 - P(A \cap B)\)
\(P(A' \cup B') = 1 - \frac{1}{20}\)
\(P(A' \cup B') = \frac{20 - 1}{20} = \frac{19}{20}\)
In simple words: First, we find the individual chances of A, B, and A or B happening. Then, using a probability rule, we find the chance of both A and B happening. Finally, to get the chance of neither A nor B happening, we subtract the chance of both A and B from 1.

🎯 Exam Tip: This problem requires careful step-by-step calculation, especially with fractions. De Morgan's Law is a key shortcut, but don't forget the addition rule to find \(P(A \cap B)\) first.

Question 16.If \(P(A \cap B) = 0.12\) and \(P(B) = 0.3\) for two independent events A and B, then find \(P(A \cup B)\).
Answer:Given that A and B are independent events. We have \(P(A \cap B) = 0.12\) and \(P(B) = 0.3\). Since A and B are independent, we know that \(P(A \cap B) = P(A) \times P(B)\). We can use this to find \(P(A)\):
\(0.12 = P(A) \times 0.3\)
\(P(A) = \frac{0.12}{0.3} = 0.4\) Now, use the addition law of probability to find \(P(A \cup B)\):
\(P(A \cup B) = P(A) + P(B) - P(A \cap B)\) Substitute the values:
\(P(A \cup B) = 0.4 + 0.3 - 0.12\)
\(P(A \cup B) = 0.7 - 0.12\)
\(P(A \cup B) = 0.58\)
In simple words: We know the chance of both A and B happening, and the chance of B. Since they are independent, we can find the chance of A. Then, to get the chance of A or B happening, we add the chances of A and B, and subtract the chance of both.

🎯 Exam Tip: For independent events, remember the specific relationship \(P(A \cap B) = P(A) \times P(B)\), which is often used to find missing probabilities like \(P(A)\) in this case.

Question 17.If \(A = \{x | 1 < x < 3\}\) and \(B = \{x | \frac{1}{2} \le x < 2\}\), then find \(A \cup B\) and \(A \cap B\).
Answer:Given sets are: \(A = \{x | 1 < x < 3\}\) (This means x is between 1 and 3, not including 1 and 3) \(B = \{x | \frac{1}{2} \le x < 2\}\) (This means x is between 0.5 and 2, including 0.5 but not 2) To find \(A \cup B\) (union, meaning elements in A, or in B, or in both): The minimum value in either set is \(\frac{1}{2}\) (from B). The maximum value in either set is 3 (from A). So, \(A \cup B = \{x | \frac{1}{2} \le x < 3\}\). To find \(A \cap B\) (intersection, meaning elements common to both A and B): The lower bound for intersection is the maximum of the lower bounds of A and B: \(\max(1, \frac{1}{2}) = 1\). The upper bound for intersection is the minimum of the upper bounds of A and B: \(\min(3, 2) = 2\). Since A does not include 1, and B does not include 2, the intersection will be strictly between 1 and 2. So, \(A \cap B = \{x | 1 < x < 2\}\).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख संख्या रेखा पर दो अंतराल A और B को दर्शाता है। A का प्रतिनिधित्व 1 और 3 के बीच के बिंदुओं से होता है (1 और 3 को छोड़कर), और B का प्रतिनिधित्व 0.5 और 2 के बीच के बिंदुओं से होता है (0.5 को शामिल करते हुए, 2 को छोड़कर)। यूनियन (A∪B) में 0.5 से 3 तक के सभी बिंदु शामिल हैं (0.5 शामिल है, 3 शामिल नहीं है), जबकि इंटरसेक्शन (A∩B) में 1 से 2 तक के सामान्य बिंदु शामिल हैं (1 और 2 दोनों शामिल नहीं हैं)।
In simple words: For A or B, we take the widest range covered by both. For A and B, we take only the numbers that are in both ranges at the same time.

🎯 Exam Tip: When dealing with intervals, correctly identify whether endpoints are included or excluded using '<', '>', '≤', or '≥' symbols. For union, take the smallest lower bound and largest upper bound. For intersection, take the largest lower bound and smallest upper bound.

Question 18.The probability of occurrence of at least one of the two events A and B is \(\frac{1}{4}\). The probability that event A occurs, but event B does not occur is \(\frac{1}{5}\). Find the probability of event B.
Answer:Given: Probability of at least one of the two events A and B occurring means \(P(A \cup B)\).
\(P(A \cup B) = \frac{1}{4}\) Probability that event A occurs, but event B does not occur, is \(P(A \cap B')\) or \(P(A - B)\).
\(P(A - B) = \frac{1}{5}\) We know the formula for \(P(A - B)\):
\(P(A - B) = P(A) - P(A \cap B)\) So, \(\frac{1}{5} = P(A) - P(A \cap B)\) -------- (Equation 1) Also, the addition rule of probability is:
\(P(A \cup B) = P(A) + P(B) - P(A \cap B)\) We can rearrange this equation to substitute \((P(A) - P(A \cap B))\):
\(P(A \cup B) = (P(A) - P(A \cap B)) + P(B)\) Now, substitute the known values into this rearranged equation:
\(\frac{1}{4} = \frac{1}{5} + P(B)\) To find \(P(B)\), subtract \(\frac{1}{5}\) from \(\frac{1}{4}\):
\(P(B) = \frac{1}{4} - \frac{1}{5}\) Find a common denominator, which is 20:
\(P(B) = \frac{5}{20} - \frac{4}{20}\)
\(P(B) = \frac{1}{20}\)
In simple words: We are given the chance that A or B happens, and the chance that only A happens. We know that "A or B" equals "only A" plus "only B" plus "both A and B". If we think of "A or B" as "only A" plus "B", then we can simply subtract the chance of "only A" from the chance of "A or B" to find the chance of B.

🎯 Exam Tip: Recognize the relationship \(P(A \cup B) = P(A-B) + P(B)\) when events are not necessarily mutually exclusive, but you are given \(P(A-B)\) directly.

Question 19.If \(P(B) = \frac{3}{5}\) and \(P(A' \cap B) = \frac{1}{2}\) for two events A and B, find \(P(A|B)\).
Answer:Given \(P(B) = \frac{3}{5}\) and \(P(A' \cap B) = \frac{1}{2}\). The event \(A' \cap B\) represents the outcomes that are in B but not in A. We know that \(P(A' \cap B) = P(B) - P(A \cap B)\). Substitute the given values into this formula:
\(\frac{1}{2} = \frac{3}{5} - P(A \cap B)\) Now, solve for \(P(A \cap B)\):
\(P(A \cap B) = \frac{3}{5} - \frac{1}{2}\) To subtract, find a common denominator, which is 10:
\(P(A \cap B) = \frac{3 \times 2}{10} - \frac{1 \times 5}{10}\)
\(P(A \cap B) = \frac{6 - 5}{10} = \frac{1}{10}\) Now we have \(P(A \cap B) = \frac{1}{10}\) and \(P(B) = \frac{3}{5}\). The formula for conditional probability \(P(A|B)\) is:
\[P(A|B) = \frac{P(A \cap B)}{P(B)}\] Substitute the calculated values:
\[P(A|B) = \frac{\frac{1}{10}}{\frac{3}{5}}\]
\[P(A|B) = \frac{1}{10} \times \frac{5}{3}\]
\[P(A|B) = \frac{5}{30}\]
\[P(A|B) = \frac{1}{6}\]
In simple words: We know the chance of B happening, and the chance of B happening without A. From this, we first find the chance of both A and B happening. Then, to find the chance of A given B, we divide the chance of both by the chance of B.

🎯 Exam Tip: This problem involves working backwards from \(P(A' \cap B)\) to find \(P(A \cap B)\) before calculating the conditional probability. Be precise with fraction arithmetic.

Question 20.6 persons have a passport in' a group of 10 persons. If 3 persons are randomly selected from this group, find the probability that (i) all the three persons have a passport (ii) two persons among them do not have a passport.
Answer:Total number of persons in the group = 10. Number of persons with a passport = 6. Number of persons without a passport = \(10 - 6 = 4\). We are selecting 3 persons randomly from this group. The total number of ways to select 3 persons from 10 is:
\(n = {}^{10}C_3 = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 10 \times 3 \times 4 = 120\). (i) A = Event that all three selected persons have a passport. For this event, we need to select all 3 persons from the 6 who have passports. The number of favourable outcomes for event A is:
\(m_A = {}^{6}C_3 = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20\). The probability of event A is:
\(P(A) = \frac{m_A}{n} = \frac{20}{120} = \frac{1}{6}\). (ii) B = Event that two persons among the three selected do not have a passport. If two persons do not have a passport, then the remaining one person must have a passport (since a total of 3 are selected). We need to select 2 persons from the 4 without passports AND 1 person from the 6 with passports. The number of favourable outcomes for event B is:
\(m_B = {}^{4}C_2 \times {}^{6}C_1\)
\(m_B = \left(\frac{4 \times 3}{2 \times 1}\right) \times \left(\frac{6}{1}\right)\)
\(m_B = 6 \times 6 = 36\). The probability of event B is:
\(P(B) = \frac{m_B}{n} = \frac{36}{120}\). To simplify the fraction, divide both numerator and denominator by common factors (e.g., 12):
\(P(B) = \frac{36 \div 12}{120 \div 12} = \frac{3}{10}\).
In simple words: Out of 10 people, 6 have passports and 4 don't. If we pick 3 people: (i) To find the chance that all 3 have passports, we calculate the ways to pick 3 from the 6 passport holders and divide by the total ways to pick 3 from 10. (ii) To find the chance that 2 don't have passports, we pick 2 from the 4 without passports AND 1 from the 6 with passports, then divide by the total ways to pick 3 from 10.

🎯 Exam Tip: Clearly define the total sample space (total ways to select) and the favorable outcomes for each sub-question. Use combinations (\({}^nC_r\)) for selection problems where order does not matter.

Question 21.The probability that the tax-limit for income of males increases in the budget of a year is 0.66 and the probability that the tax-limit increases for income of females is 0.72. The probability that the tax-limit increases for income of both the males and females is 0.47. Find the probability that (i) the tax-limit increases for income of only one of the two, males and females, (ii) the tax-limit does not increase for income of males as well as females in the budget of that year.
Answer:Let A be the event that the tax-limit for income of males increases. So, \(P(A) = 0.66\). Let B be the event that the tax-limit for income of females increases. So, \(P(B) = 0.72\). Let \(A \cap B\) be the event that the tax-limit increases for both males and females. So, \(P(A \cap B) = 0.47\). (i) Probability that the tax-limit increases for income of only one of the two (males or females): This event (let's call it C) means (A occurs but B does not) OR (B occurs but A does not). This can be written as \((A \cap B') \cup (A' \cap B)\). The events \((A \cap B')\) and \((A' \cap B)\) are mutually exclusive. So, \(P(C) = P(A \cap B') + P(A' \cap B)\). We know that:
\(P(A \cap B') = P(A) - P(A \cap B)\)
\(P(A' \cap B) = P(B) - P(A \cap B)\) Substitute the values:
\(P(A \cap B') = 0.66 - 0.47 = 0.19\)
\(P(A' \cap B) = 0.72 - 0.47 = 0.25\) Now, add these probabilities:
\(P(C) = 0.19 + 0.25 = 0.44\). (ii) Probability that the tax-limit does not increase for income of males as well as females: This event means neither A nor B occurs, which is \(A' \cap B'\). Using De Morgan's Law, \(A' \cap B' = (A \cup B)'\). So, \(P(A' \cap B') = 1 - P(A \cup B)\). First, calculate \(P(A \cup B)\) using the addition law:
\(P(A \cup B) = P(A) + P(B) - P(A \cap B)\) Substitute the given values:
\(P(A \cup B) = 0.66 + 0.72 - 0.47\)
\(P(A \cup B) = 1.38 - 0.47\)
\(P(A \cup B) = 0.91\) Now, calculate \(P(A' \cap B')\):
\(P(A' \cap B') = 1 - P(A \cup B)\)
\(P(A' \cap B') = 1 - 0.91\)
\(P(A' \cap B') = 0.09\)
In simple words: Let A be male tax increase and B be female tax increase. (i) "Only one" means A happens without B, OR B happens without A. We calculate the chance of A without B (P(A) - P(A and B)) and add it to the chance of B without A (P(B) - P(A and B)). (ii) "Neither" means no increase for males AND no increase for females. This is found by taking 1 minus the chance that either males or females (or both) get an increase.

🎯 Exam Tip: For "only one" of two events, use \(P(A \cap B') + P(A' \cap B)\). For "neither" event, use De Morgan's Law: \(P(A' \cap B') = 1 - P(A \cup B)\).

Question 22.The price of petrol rises in 80% of the cases and the price of diesel rises in 77% of the cases after the rise in price of crude oil. The price of petrol and diesel rises in 68% cases. Find the probability that the price of diesel rises under the condition that there is a rise in the price of petrol.
Answer:Let A be the event that the price of petrol rises.
\(P(A) = 80\% = \frac{80}{100} = 0.80\) Let B be the event that the price of diesel rises.
\(P(B) = 77\% = \frac{77}{100} = 0.77\) Let \(A \cap B\) be the event that the prices of both petrol and diesel rise.
\(P(A \cap B) = 68\% = \frac{68}{100} = 0.68\) We need to find the probability that the price of diesel rises given that there is a rise in the price of petrol. This is \(P(B|A)\). The formula for conditional probability \(P(B|A)\) is:
\[P(B|A) = \frac{P(A \cap B)}{P(A)}\] Substitute the given values:
\[P(B|A) = \frac{0.68}{0.80}\]
\[P(B|A) = \frac{68}{80}\] Simplify the fraction by dividing both numerator and denominator by 4:
\[P(B|A) = \frac{17}{20}\]
In simple words: We want to find the chance of diesel price rising, knowing that petrol price has already risen. We divide the chance of both rising by the chance of petrol rising.

🎯 Exam Tip: Identify the conditional event (the one that has already occurred) correctly, as it forms the denominator in the conditional probability formula.

Question 23.As per the prediction of weather bureau, the probabilities for rains on three days; Thursday, Friday and Saturday in the next week are 0.8, 0.7 and 0.6 respectively. Find the probability that it rains on at least one of the three days in the next week. (Note: The events of rains on three days; Thursday, Friday and Saturday of a week are independent.)
Answer:Let A be the event that it rains on Thursday. So, \(P(A) = 0.8\). Let B be the event that it rains on Friday. So, \(P(B) = 0.7\). Let C be the event that it rains on Saturday. So, \(P(C) = 0.6\). The events A, B, and C are independent. We need to find the probability that it rains on at least one of the three days, which is \(P(A \cup B \cup C)\). For independent events, we can use the complement rule: \(P(A \cup B \cup C) = 1 - P((A \cup B \cup C)') = 1 - P(A' \cap B' \cap C')\). Since A, B, C are independent, their complements A', B', C' are also independent. First, find the probabilities of the complements:
\(P(A') = 1 - P(A) = 1 - 0.8 = 0.2\)
\(P(B') = 1 - P(B) = 1 - 0.7 = 0.3\)
\(P(C') = 1 - P(C) = 1 - 0.6 = 0.4\) Now, find \(P(A' \cap B' \cap C')\) for independent events:
\(P(A' \cap B' \cap C') = P(A') \cdot P(B') \cdot P(C')\)
\(P(A' \cap B' \cap C') = 0.2 \times 0.3 \times 0.4\)
\(P(A' \cap B' \cap C') = 0.06 \times 0.4\)
\(P(A' \cap B' \cap C') = 0.024\) Finally, calculate \(P(A \cup B \cup C)\):
\(P(A \cup B \cup C) = 1 - P(A' \cap B' \cap C')\)
\(P(A \cup B \cup C) = 1 - 0.024\)
\(P(A \cup B \cup C) = 0.976\) Alternatively, using the general addition law for three events:
\(P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C)\) Since A, B, C are independent:
\(P(A \cap B) = P(A)P(B) = 0.8 \times 0.7 = 0.56\)
\(P(A \cap C) = P(A)P(C) = 0.8 \times 0.6 = 0.48\)
\(P(B \cap C) = P(B)P(C) = 0.7 \times 0.6 = 0.42\)
\(P(A \cap B \cap C) = P(A)P(B)P(C) = 0.8 \times 0.7 \times 0.6 = 0.336\) Substitute these values:
\(P(A \cup B \cup C) = 0.8 + 0.7 + 0.6 - 0.56 - 0.48 - 0.42 + 0.336\)
\(P(A \cup B \cup C) = 2.1 - 1.46 + 0.336\)
\(P(A \cup B \cup C) = 0.64 + 0.336\)
\(P(A \cup B \cup C) = 0.976\)
In simple words: To find the chance of rain on at least one of the three independent days, it's easier to find the chance of NO rain on any of the three days and subtract that from 1. If events are independent, the chance of no rain on any day is the chance of no rain on Thursday, times the chance of no rain on Friday, times the chance of no rain on Saturday.

🎯 Exam Tip: For "at least one" problems with independent events, using the complement rule \(1 - P(\text{none occur})\) is often much simpler than using the full addition rule for multiple events.

Section D

Question 1.6 LED televisions and 4 LCD televisions are displayed in digital store A whereas 5 LED televisions and 3 LCD televisions are displayed in digital store B. One of the two stores is randomly selected and one television is selected from that store. Find the probability that it is an LCD television.
Answer:Let's denote the stores and TV types: Store A: 6 LED TVs, 4 LCD TVs. Total TVs in Store A = \(6 + 4 = 10\). Store B: 5 LED TVs, 3 LCD TVs. Total TVs in Store B = \(5 + 3 = 8\). We randomly select one of the two stores.
Probability of selecting Store A, \(P(\text{Store A}) = \frac{1}{2}\).
Probability of selecting Store B, \(P(\text{Store B}) = \frac{1}{2}\). Let L be the event that an LCD television is selected. We want to find \(P(L)\). This can be found using the law of total probability.
\(P(L) = P(L | \text{Store A}) \cdot P(\text{Store A}) + P(L | \text{Store B}) \cdot P(\text{Store B})\) First, find the conditional probabilities of selecting an LCD TV from each store:
\(P(L | \text{Store A}) = \frac{\text{Number of LCD TVs in A}}{\text{Total TVs in A}} = \frac{4}{10}\).
\(P(L | \text{Store B}) = \frac{\text{Number of LCD TVs in B}}{\text{Total TVs in B}} = \frac{3}{8}\). Now, substitute these values into the total probability formula:
\(P(L) = \frac{4}{10} \times \frac{1}{2} + \frac{3}{8} \times \frac{1}{2}\)
\(P(L) = \frac{4}{20} + \frac{3}{16}\) Simplify the first fraction: \(\frac{4}{20} = \frac{1}{5}\).
\(P(L) = \frac{1}{5} + \frac{3}{16}\) Find a common denominator, which is 80:
\(P(L) = \frac{1 \times 16}{80} + \frac{3 \times 5}{80}\)
\(P(L) = \frac{16}{80} + \frac{15}{80}\)
\(P(L) = \frac{16 + 15}{80} = \frac{31}{80}\)
In simple words: To find the total chance of picking an LCD TV, first consider the chance of choosing Store A and then an LCD from it, plus the chance of choosing Store B and then an LCD from it. Add these two chances together.

🎯 Exam Tip: This is a classic application of the Law of Total Probability. Clearly define events and conditional probabilities, and then sum the products of prior and conditional probabilities.

Question 2.One number is randomly selected from the natural numbers 1 to 100. Find the probability that the number selected is either a single digit number or a perfect square.
Answer:Total number of natural numbers from 1 to 100 = 100. So, the total number of primary outcomes \(n = {}^{100}C_1 = 100\). Let A be the event that the selected number is a single digit number. Single digit numbers from 1 to 100 are {1, 2, 3, 4, 5, 6, 7, 8, 9}. Number of favourable outcomes for A, \(m_A = 9\).
\(P(A) = \frac{m_A}{n} = \frac{9}{100}\). Let B be the event that the selected number is a perfect square. Perfect squares from 1 to 100 are {1, 4, 9, 16, 25, 36, 49, 64, 81, 100}. Number of favourable outcomes for B, \(m_B = 10\).
\(P(B) = \frac{m_B}{n} = \frac{10}{100}\). Now, find the intersection \(A \cap B\), which is the event that the selected number is both a single digit number AND a perfect square. Numbers that are both single digit and perfect squares are {1, 4, 9}. Number of favourable outcomes for \(A \cap B\), \(m_{A \cap B} = 3\).
\(P(A \cap B) = \frac{m_{A \cap B}}{n} = \frac{3}{100}\). We need to find the probability that the number selected is either a single digit number OR a perfect square, which is \(P(A \cup B)\). Using the addition law of probability:
\(P(A \cup B) = P(A) + P(B) - P(A \cap B)\) Substitute the calculated probabilities:
\(P(A \cup B) = \frac{9}{100} + \frac{10}{100} - \frac{3}{100}\)
\(P(A \cup B) = \frac{9 + 10 - 3}{100}\)
\(P(A \cup B) = \frac{16}{100}\) Simplify the fraction:
\(P(A \cup B) = \frac{4}{25}\)
In simple words: First, list numbers that are single digits (event A) and numbers that are perfect squares (event B) from 1 to 100. Then, find the numbers that are in both lists (A and B). To get the chance of A or B, add the chances of A and B, then subtract the chance of A and B happening together.

🎯 Exam Tip: Clearly list the elements of each set and their intersection to avoid miscounting. The addition rule for probability is essential for "either/or" scenarios.

Question 3.A fair coin is tossed thrice. If the first two tosses have resulted in tail, find the probability that tail appears on the coin in all the three trials.
Answer:Let C be the event that the first two tosses have resulted in tail. The possible outcomes for two tosses are HH, HT, TH, TT. So, C = {TT}. Let D be the event that tail appears on the coin in all three trials. The possible outcomes for three tosses are HHH, HHT, HTH, THH, HTT, THT, TTH, TTT. So, D = {TTT}. We need to find the probability that tail appears on the coin in all three trials, given that the first two tosses resulted in tail. This is \(P(D|C)\). First, let's explicitly list the sample space for three tosses: U = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}. Total outcomes = 8. Event C: first two tosses are tail. From the sample space, C = {TTT, TTH}. Event D: tail appears in all three trials. From the sample space, D = {TTT}. Now, find the intersection \(D \cap C\): The outcomes where both D and C occur. \(D \cap C = \{TTT\}\). The probabilities:
\(P(C) = \frac{\text{Number of outcomes in C}}{\text{Total outcomes}} = \frac{2}{8} = \frac{1}{4}\).
\(P(D \cap C) = \frac{\text{Number of outcomes in D \cap C}}{\text{Total outcomes}} = \frac{1}{8}\). Using the conditional probability formula:
\[P(D|C) = \frac{P(D \cap C)}{P(C)}\]
\[P(D|C) = \frac{\frac{1}{8}}{\frac{1}{4}}\]
\[P(D|C) = \frac{1}{8} \times \frac{4}{1}\]
\[P(D|C) = \frac{4}{8} = \frac{1}{2}\] Alternatively, we can reason that since the first two tosses are already tails, we only need the third toss to be a tail. For a fair coin, the probability of getting a tail on the third toss is \(\frac{1}{2}\), independent of the previous tosses.
In simple words: We are given that the first two flips were tails. We want the chance that all three flips are tails. Since coin flips are independent, the first two don't affect the third. So, we just need the third flip to be a tail, which has a 1/2 chance.

🎯 Exam Tip: When events are independent, prior information (like the first two tosses being tails) only affects what you need for future events if the question focuses on *all* trials. For conditional probability on independent trials, the condition often simplifies the problem by fixing some outcomes.

Question 4.If events A, B and C are independent events and \(P(A) = P(B) = P(C) = p\), then find the value of \(P(A \cup B \cup C)\) in terms of p.
Answer:Given that A, B, and C are independent events. Also given \(P(A) = P(B) = P(C) = p\). We need to find \(P(A \cup B \cup C)\). Using the general addition law for three events:
\(P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C)\) Since A, B, C are independent events:
\(P(A \cap B) = P(A)P(B) = p \times p = p^2\)
\(P(A \cap C) = P(A)P(C) = p \times p = p^2\)
\(P(B \cap C) = P(B)P(C) = p \times p = p^2\)
\(P(A \cap B \cap C) = P(A)P(B)P(C) = p \times p \times p = p^3\) Substitute these into the addition law:
\(P(A \cup B \cup C) = p + p + p - p^2 - p^2 - p^2 + p^3\)
\(P(A \cup B \cup C) = 3p - 3p^2 + p^3\) This can also be factored as:
\(P(A \cup B \cup C) = p(3 - 3p + p^2)\) Alternatively, using the complement rule for independent events:
\(P(A \cup B \cup C) = 1 - P(A' \cap B' \cap C')\) Since A, B, C are independent, A', B', C' are also independent.
\(P(A') = 1 - P(A) = 1 - p\)
\(P(B') = 1 - P(B) = 1 - p\)
\(P(C') = 1 - P(C) = 1 - p\)
\(P(A' \cap B' \cap C') = P(A')P(B')P(C') = (1 - p)(1 - p)(1 - p) = (1 - p)^3\) So, \(P(A \cup B \cup C) = 1 - (1 - p)^3\). Expanding \((1 - p)^3\):
\((1 - p)^3 = 1^3 - 3(1)^2 p + 3(1)p^2 - p^3 = 1 - 3p + 3p^2 - p^3\) Therefore, \(P(A \cup B \cup C) = 1 - (1 - 3p + 3p^2 - p^3)\)
\(P(A \cup B \cup C) = 1 - 1 + 3p - 3p^2 + p^3\)
\(P(A \cup B \cup C) = 3p - 3p^2 + p^3\) Both methods yield the same result.
In simple words: If three events happen with the same chance 'p' and don't affect each other, to find the chance of at least one of them happening, we can use a formula that sums individual chances, subtracts chances of pairs, and adds back the chance of all three. Or, it's 1 minus the chance that none of them happen, where the chance of none happening is (1-p) multiplied by itself three times.

🎯 Exam Tip: For problems involving "at least one" with independent events, using the complement rule \(1 - P(\text{none occur})\) is often more efficient and less prone to calculation errors than the expanded addition rule.

Question 5.The genderwise data of a sample of 6000 employees selected from class 3 and class 4 employees in the government jobs of a state are shown in the following table:

Class of Employees	Gender		Total
	Males	Females
Class 3	3600	900	4500
Class 4	400	1100	1500
Total	4000	2000	6000

One employee is randomly selected from all the class 3 and class 4 employees in government jobs of this state. (1) If the selected employee is a male, find the probability that he belongs to class 3. (2) If it is given that the selected employee belongs to class 3, find the probability that he is a male.
Answer:Let's define the events based on the table: Total number of employees = 6000. Let A be the event that an employee belongs to Class 3. Number of Class 3 employees = 4500.
\(P(A) = \frac{4500}{6000} = \frac{45}{60} = \frac{3}{4}\). Let C be the event that an employee is a male. Number of male employees = 4000.
\(P(C) = \frac{4000}{6000} = \frac{40}{60} = \frac{2}{3}\). Let \(A \cap C\) be the event that an employee is a male AND belongs to Class 3. Number of male employees in Class 3 = 3600.
\(P(A \cap C) = \frac{3600}{6000} = \frac{36}{60} = \frac{3}{5}\). (1) If the selected employee is a male, find the probability that he belongs to class 3. This is asking for \(P(A|C)\), the probability that an employee is from Class 3 given that they are male. Using the conditional probability formula:
\[P(A|C) = \frac{P(A \cap C)}{P(C)}\] Substitute the probabilities:
\[P(A|C) = \frac{\frac{3600}{6000}}{\frac{4000}{6000}} = \frac{3600}{4000}\]
\[P(A|C) = \frac{36}{40} = \frac{9}{10}\] (2) If it is given that the selected employee belongs to class 3, find the probability that he is a male. This is asking for \(P(C|A)\), the probability that an employee is male given that they are from Class 3. Using the conditional probability formula:
\[P(C|A) = \frac{P(A \cap C)}{P(A)}\] Substitute the probabilities:
\[P(C|A) = \frac{\frac{3600}{6000}}{\frac{4500}{6000}} = \frac{3600}{4500}\]
\[P(C|A) = \frac{36}{45} = \frac{4}{5}\]
In simple words: We have a table of employees by class and gender. (1) If we pick a male, we want the chance he is from Class 3. So, we look at the males from Class 3 and divide by the total number of males. (2) If we pick someone from Class 3, we want the chance they are male. So, we look at the males from Class 3 and divide by the total number of people in Class 3.

🎯 Exam Tip: Clearly distinguish between \(P(A|C)\) and \(P(C|A)\). The condition (the event that has already occurred) always goes in the denominator of the conditional probability formula.

GSEB Solutions Class 12 Statistics Part 2 Chapter 1 Probability Ex 1

Section A

 

Question 1. Which event is given by a special subset \( \Phi \) of the sample space U?
(a) Certain event
(b) Complementary event of \( \Phi \)
(c) Union of events U and \( \Phi \)
(d) Impossible event
Answer: (d) Impossible event
In simple words: An impossible event is one that can never happen, and in probability, it is represented by the empty set \( \Phi \).

🎯 Exam Tip: Understanding the basic definitions of events like impossible events is crucial for multiple-choice questions.

 

Question 2. What is the value of P(A ∩ A') for events A and A' ?
(b) 0
(c) 0.5
(d) between 0 and 1
Answer: (b) 0
In simple words: The intersection of an event A and its complement A' means that both A and not A happen at the same time, which is impossible, so its probability is 0.

🎯 Exam Tip: The intersection of an event and its complement is always an empty set, resulting in a probability of zero.

 

Question 3. Which of the following options is true for any event of the sample space ?
(a) P(A) < 0
(b) 0 ≤ P(A) ≥ 1
(c) 0 ≤ P(A) ≤ 1
(d) P(A) > 1
Answer: (c) \( 0 \leq P(A) \leq 1 \)
In simple words: The probability of any event must be a number between 0 and 1, including 0 and 1.

🎯 Exam Tip: Always remember that probability values range from 0 (impossible event) to 1 (certain event).

 

Question 4. Which of the following options is not true for any two events A and B in the sample space U; where, A \( \subset \) B?
(a) P(A ∩ B) = P(B)
(b) P(A ∩ B) = P(A)
(c) P(A U B) ≥ P(A)
(d) P(B-A) = P(B)-P(A)
Answer: (a) P(A ∩ B) = P(B)
In simple words: If event A is a subset of event B, it means A is entirely contained within B. In this case, the intersection of A and B is A itself, so P(A ∩ B) should be P(A), not P(B).

🎯 Exam Tip: When one event is a subset of another, correctly applying subset properties to probability calculations is key.

 

Question 5. What is the other name of the classical definition of probability?
(a) Mathematical definition
(b) Axiomatic definition
(c) Statistical definition
(d) Geometric definition
Answer: (a) Mathematical definition
In simple words: The classical definition of probability, which uses the ratio of favorable outcomes to total outcomes, is also known as the mathematical definition.

🎯 Exam Tip: Knowing alternative names for definitions helps in understanding different terminologies used in probability theory.

 

Question 6. Which of the following statement for probability of elementary events H and T of random experiment of tossing a balanced coin is not true ?
(a) P(T) = 0.5
(b) P(H) + P(T) = 1
(c) P(H ∩ T) = 0.5
(d) P(H) = 0.5
Answer: (c) P(H ∩ T) = 0.5
In simple words: When you toss a coin, getting heads (H) and tails (T) are mutually exclusive events, meaning they cannot happen at the same time. Therefore, the probability of both happening, P(H ∩ T), must be 0, not 0.5.

🎯 Exam Tip: For mutually exclusive events, the probability of their intersection is always zero.

 

Question 7. Which random experiment from the following random experiments has an infinite sample space ?
(a) Throwing two dice
(b) Selecting two employees from an office
(c) To measure the life of electric bulb
(d) Select a card from 52 cards
Answer: (c) To measure the life of electric bulb
In simple words: Measuring the life of an electric bulb can result in any positive real number (e.g., 100.1 hours, 100.15 hours), leading to an infinite number of possible outcomes.

🎯 Exam Tip: An infinite sample space occurs when outcomes are continuous or unbounded, rather than discrete and limited.

 

Question 8. If A U A' = U, then what type of events are A and A'?
(a) Independent events
(b) Complementary events
(c) Certain events
(d) Impossible events
Answer: (b) Complementary events
In simple words: If the union of two events A and A' covers the entire sample space (U), it means A' is everything that is not A, making them complementary events.

🎯 Exam Tip: Complementary events always sum up to the entire sample space, meaning one or the other must occur.

 

Question 9. If P(A|B) = P(A) and P(B|A) = P(B), then what type of events are A and B ?
(a) Independent events
(b) Complementary events
(c) Certain events
(d) Impossible events
Answer: (a) Independent events
In simple words: When the probability of event A happening is not changed by event B happening (and vice versa), the events are called independent.

🎯 Exam Tip: This condition, P(A|B) = P(A), is the defining characteristic of independent events in conditional probability.

 

Question 10. Two events A and B of a sample space are mutually exclusive. Which of the following will be equal to P(B-A)?
(b) P(B)
(c) P(A ∩ B)
(d) P(A U B)
Answer: (b) P(B)
In simple words: If A and B are mutually exclusive, they cannot happen at the same time. So, the event "B but not A" is simply event B itself, because A is already excluded from B. Therefore, P(B-A) equals P(B).

🎯 Exam Tip: For mutually exclusive events, A and B, \( A \cap B = \Phi \), simplifying many probability formulas.

 

Question 11. What is the total number of sample points in the sample space formed by throwing three six-faced balanced dice simultaneously?
(a) \( 6^2 \)
(b) 36
(c) 6 × 3
(d) \( 6^3 \)
Answer: (d) \( 6^3 \)
In simple words: When throwing three dice, each die has 6 possible outcomes. To find the total outcomes for all three, you multiply the possibilities for each die, which is \( 6 \times 6 \times 6 \), or \( 6^3 \).

🎯 Exam Tip: For multiple independent trials, the total number of outcomes is found by multiplying the number of outcomes for each trial.

 

Question 12. If one number is randomly selected between 1 and 20, what is the probability that the number is a multiple of 5 ?
(a) \( \frac{1}{2} \)
(b) \( \frac{1}{6} \)
(c) \( \frac{1}{5} \)
(d) \( \frac{1}{3} \)
Answer: (c) \( \frac{1}{5} \)
In simple words: Numbers between 1 and 20 (inclusive) are 20 numbers. The multiples of 5 in this range are 5, 10, 15, 20 (4 numbers). So, the probability is 4 out of 20, which simplifies to \( \frac{1}{5} \).

🎯 Exam Tip: To calculate probability, identify the total possible outcomes and the number of favorable outcomes, then form a ratio.

 

Question 13. If events A and B are independent, which of the following options is true?
(a) P(A ∩ B) = P(A) × P(B)
(b) P(A U B) = P(A) + P(B)
(c) P(A U B) = P(A) × P(B)
(d) P(A ∩ B) = P(A) + P(B)
Answer: (a) P(A ∩ B) = P(A) × P(B)
In simple words: For events that do not affect each other (independent events), the chance of both happening is found by multiplying their individual chances.

🎯 Exam Tip: This formula, P(A ∩ B) = P(A)P(B), is a fundamental rule for independent events.

 

Question 14. What is the probability of having 5 Thursdays in the month of February in a year which is not a leap year?
(a) 0
(b) \( \frac{1}{7} \)
(c) \( \frac{2}{7} \)
(d) \( \frac{3}{7} \)
Answer: (a) 0
In simple words: A non-leap year February has 28 days. This means it has exactly 4 weeks and no extra days, so it cannot have 5 Thursdays. The probability is 0.

🎯 Exam Tip: Pay attention to the number of days in the month and whether it's a leap year when calculating probabilities related to calendar days.

 

Question 15. If P(A) = 0.4 and P(B') = 0.3 for two independent events A and B of a sample space, then state the value of P(A ∩ B).
(a) 0.12
(b) 0.42
(c) 0.28
(d) 0.18
Answer: (c) 0.28
In simple words: First, find P(B) from P(B') = 0.3, which is 1 - 0.3 = 0.7. Since A and B are independent, P(A ∩ B) is P(A) multiplied by P(B), which is 0.4 * 0.7 = 0.28.

🎯 Exam Tip: Remember that P(Event) = 1 - P(Complement of Event) and apply the multiplication rule for independent events carefully.

 

Question 16. For two events A and B of a sample space, state the event (A ∩ B) U (A ∩ B').
(а) \( \Phi \)
(b) B
(c) A
(d) U
Answer: (c) A
In simple words: This expression means "elements common to A and B" combined with "elements common to A and not B". Together, these two parts cover all elements that are in A, regardless of B. So the result is event A.

🎯 Exam Tip: Use Venn diagrams or distributive laws of set theory to simplify complex set expressions in probability.

 

Question 17. According to the mathematical definition of probability, what is the probability of each outcome among the n outcomes of a random experiment ?
(a) 0
(b) \( \frac{1}{n} \)
(d) cannot say
Answer: (b) \( \frac{1}{n} \)
In simple words: In a random experiment with 'n' equally likely outcomes, the chance of any single outcome occurring is 1 divided by the total number of outcomes, which is \( \frac{1}{n} \).

🎯 Exam Tip: The classical definition assumes all 'n' outcomes are equally likely, so each has a probability of \( \frac{1}{n} \).

 

Section B

 

Question 1. Give two examples of random experiment.
Answer:
Two examples of random experiments are:
1. The experiment of throwing a balanced die.
2. The experiment of finding defective units from a lot of units produced.
In simple words: A random experiment is an action where the outcome cannot be predicted for sure, but all possible outcomes are known beforehand. Rolling a die and checking items for defects are good examples.

🎯 Exam Tip: A random experiment must have uncertain outcomes but a known set of all possible outcomes.

 

Question 2. Draw the Venn diagram for A – B, the difference event of A and B.
Answer:
The Venn diagram for difference event A-B:

ℹ️ चित्र व्याख्या (Diagram Explanation): इस वेन आरेख में, एक बड़ा आयत नमूना स्थान (U) को दर्शाता है। दो वृत्त, A और B, दो घटनाओं का प्रतिनिधित्व करते हैं। छायांकित क्षेत्र A - B को दिखाता है, जिसका अर्थ है घटना A में वे सभी तत्व जो घटना B में नहीं हैं। यह उन परिणामों को दर्शाता है जो A में होते हैं लेकिन B में नहीं होते हैं।

Class of Employees	Gender		Total
	Males	Females
Class 3	3600	900	4500
Class 4	400	1100	1500
Total	4000	2000	6000

In simple words: A - B represents the elements that are in set A but not in set B. On a Venn diagram, this is the part of circle A that does not overlap with circle B.

🎯 Exam Tip: Be able to visually represent set operations like difference (A-B) using Venn diagrams. The associated table for this question is irrelevant here as it describes a different problem context, but tabular data should always be rendered as an HTML table.

 

Question 3. Define an event.
Answer:
Any subset of the sample space of a random experiment is called an event. It is denoted by A, B, C ....
In simple words: An event is a specific collection of outcomes from a random experiment. It's essentially a part of the total possible outcomes.

🎯 Exam Tip: Events are always subsets of the sample space; understanding this relationship is foundational to probability.

 

Question 4. Write the sample space of a random experiment of throwing one balanced die and a balanced coin simultaneously.
Answer:
The sample space of a random experiment of throwing one balanced die and a balanced coin simultaneously is obtained as follows:
U = {(1, H), (2, H), (3, H), (4, H), (5, H), (6, H), (1, T), (2, T), (3, T), (4, T), (5, T), (6, T)}
Where, H = Head; T = Tail;
1, 2, 3, 4, 5, 6 = Numbers on die.
In simple words: The sample space lists all possible combinations when you roll a die and flip a coin. Each outcome pairs a number from the die (1-6) with either a Head (H) or a Tail (T) from the coin.

🎯 Exam Tip: When combining two experiments, list all possible pairs of outcomes from each experiment to form the combined sample space.

 

Question 5. Define conditional probability.
Answer:
U is a finite sample space and A and B are any two events of U. The probability of event B, under the condition that event A has happened, is called conditional probability of the event B.
In simple words: Conditional probability tells us the chance of an event happening, given that another event has already occurred. It's like updating your prediction based on new information.

🎯 Exam Tip: Conditional probability is written as P(B|A) and means "the probability of B given A."

 

Question 6. State the formula for the probability of occurrence of at least one event out of three events A, B and C.
Answer:
The formula for the probability of occurrence of at least one event out of three events A, B and C is as follows:
P(A U B U C) = P(A) + P(B) + P(C) – P(A ∩ B) – P(A ∩ C) – P(B ∩ C) + P(A ∩ B ∩ C)
In simple words: To find the probability of at least one of three events happening, you add their individual probabilities, subtract the probabilities of any two happening together, and then add back the probability of all three happening together.

🎯 Exam Tip: This is the inclusion-exclusion principle for three events, a key formula for calculating union probabilities.

 

Question 7. Define independent events.
Answer:
U is a finite sample space and A and B are any two events of U. If the probability of occurrence of the event A does not depend on the occurrence or non-occurrence of the event B, then the events A and B are called independent events.
In simple words: Independent events are those where the outcome of one event does not change the chances of another event occurring.

🎯 Exam Tip: The crucial test for independence is whether P(A|B) = P(A) or P(A ∩ B) = P(A)P(B).

 

Question 8. Write the law of multiplication of probability for two independent events A and B in a sample space.
Answer:
The law of multiplication of probability for two independent events A and B in a sample space is as follows:
P(A ∩ B) = P(A) × P(B)
In simple words: For two events that don't affect each other, the chance of both happening is simply the product of their individual chances.

🎯 Exam Tip: This multiplication law is a direct consequence of the definition of independent events.

 

Question 9. Interpret P(A|B) and P(B|A).
Answer:
Interpretation of P(A|B): The conditional probability of the event A under the condition that the event B has occurred.
Interpretation of P(B|A): The conditional probability of the event B under the condition that the event A has occurred.
In simple words: P(A|B) means "the chance of A happening, knowing that B has already happened." P(B|A) means "the chance of B happening, knowing that A has already happened."

🎯 Exam Tip: Clearly distinguishing between P(A|B) and P(B|A) is vital for understanding conditional probability problems.

 

Question 10. When can we say that three events A, B and C in a sample space are exhaustive?
Answer:
When P(A) + P(B) + P(C) = 1, we can say that three events A, B and C in a sample space are exhaustive.
In simple words: Three events are exhaustive if at least one of them must happen, meaning their combined probabilities cover all possibilities in the sample space.

🎯 Exam Tip: Exhaustive events together make up the entire sample space; their union is the sample space U.

 

Question 11. Arrange P(A U B), P(A), P(A ∩ B), 0, P (A) + P (B) in the ascending order.
Answer:
\( 0, P(A \cap B), P(A), P(A \cup B), P(A) + P(B) \) are in the ascending order.
In simple words: Starting from the smallest, the probability of nothing happening (0) comes first. Then the probability of both A and B happening, followed by the probability of A happening alone. Next is the probability of A or B or both happening, and finally, the largest is the simple sum of probabilities of A and B, which may exceed 1 if they overlap.

🎯 Exam Tip: Remember the relationships: \( P(A \cap B) \leq P(A) \leq P(A \cup B) \) and \( P(A \cup B) \leq P(A) + P(B) \).

 

Question 12. Define:
Answer:
1. Random Experiment: The experiment which can be independently repeated under identical conditions and all its possible outcomes are known but it cannot be predicted with certainty which of the outcomes will appear is called a random experiment.
2. Sample Space: The set of all possible outcomes of a random experiment is called a sample space of that random experiment. It is denoted by U or S.
3. Equi-probable Events: If there is no apparent reason to believe that out of one or more random experiments, any one event is more or less likely to occur than the other events, then those events are called as equi-probable events.
4. Favourable Outcomes: If some outcomes out of all the elementary outcomes in the sample space of a random experiment indicate the occurrence of a certain event A, then these outcomes are called the favourable outcomes of the event A.
5. Probability (Mathematical Definition): If out of n outcomes of the finite sample space of a random experiment which are mutually exclusive, exhaustive and equi-probable, m outcomes are favourable for an event A, then the probability of the event A, P(A) = \( \frac{m}{n} \).
6. Probability (Statistical Definition) : Suppose, a random experiment is repeated n times under identical conditions. If an 'event A occurs in m trials then the relative frequency \( \frac{m}{n} \) of the event A gives the estimate of the probability of the event A. When n tends to infinity, the limiting value of \( \frac{m}{n} \) is called the probability of the event A. Thus,
P(A) = \( \lim_{n \to \infty} \frac{m}{n} \)
7. Impossible Event: The special subset \( \Phi \) or { } of the sample space of a random experiment is called an impossible event.
8. Certain Event: The special subset U of the sample space of a random experiment is called a certain event.
In simple words: This question defines core terms in probability. A random experiment has uncertain results but known possibilities. The sample space is all those possibilities. Equi-probable means each outcome has the same chance. Favorable outcomes are what we're looking for. Mathematical probability is counting favorable outcomes out of total. Statistical probability is based on how often an event happens over many trials. An impossible event never happens, and a certain event always happens.

🎯 Exam Tip: Master these fundamental definitions as they form the bedrock of all probability theory and calculations.

 

Question 13. For two events A and B in a sample space, A ∩ B = \( \Phi \) and A U B = U. State the values of P(A ∩ B) and P(A U B).
Answer:
A ∩ B = \( \Phi \) and A U B = U
Therefore, P(A ∩ B) = 0 and P(A U B) = 1
In simple words: If A and B never happen together, their intersection has zero probability. If A or B (or both) always happen and cover all possibilities, their union has a probability of 1.

🎯 Exam Tip: These conditions (A ∩ B = \( \Phi \) and A U B = U) define mutually exclusive and exhaustive events, which are important concepts.

 

Question 14. If two events A and B in a sample space are independent, then state the formula for P(A U B).
Answer:
A and B are independent events.
Therefore, P(A U B) = P(A) + P(B) – P(A ∩ B)
= P(A) + P(B) – [P(A) × P(B)]
In simple words: For independent events, the chance of A or B happening is the sum of their individual chances minus the chance of both happening, where the "both" part is found by multiplying their individual chances.

🎯 Exam Tip: Combine the addition law of probability with the multiplication law for independent events to get this formula.

 

Question 15. If A = {x|0 < x < 1} and B = {x|\( \frac{1}{4} \) ≤ x ≤ 3}, then find A ∩ B.
Answer:
A = {x|0 < x < 1}
B = {x|\( \frac{1}{4} \) ≤ x ≤ 3}
Therefore, A ∩ B = {x|\( \frac{1}{4} \) ≤ x < 1}
In simple words: A contains numbers between 0 and 1 (not including 0 or 1). B contains numbers between \( \frac{1}{4} \) and 3 (including \( \frac{1}{4} \) and 3). The intersection A ∩ B includes numbers that are in both sets, which means numbers from \( \frac{1}{4} \) up to (but not including) 1.

🎯 Exam Tip: When finding the intersection of intervals, look for the overlapping range, taking care with inclusive and exclusive boundaries.

 

Question 16. For two independent events A and B, P (A) = 0.5 and P(B) = 0.7. Find P(A' ∩ Β').
Answer:
A and B are independent events. So A' and B' are also independent events. P(A) = 0.5, P(B) = 0.7 are given.
Therefore, P(A' ∩ B') = P(A') × P(B')
= [1 – P(A)] [1 – P(B)]
= [1 – 0.5] [1 – 0.7]
= [0.5 × 0.3]
= 0.15
In simple words: If A and B are independent, then "not A" (A') and "not B" (B') are also independent. First, find P(A') = 1 - 0.5 = 0.5 and P(B') = 1 - 0.7 = 0.3. Then, multiply these two probabilities: 0.5 * 0.3 = 0.15.

🎯 Exam Tip: A crucial property for independent events is that their complements are also independent.

 

Question 17. If P(A) = 0.8 and P (A ∩ B) = 0.25, find P(A – B).
Answer:
P(A) = 0.8, P(A ∩ B) = 0.25 are given.
Therefore, P (A - B) = P(A) – P(A ∩ B)
= 0.8 - 0.25 = 0.55
In simple words: The probability of event A happening but not event B is found by taking the probability of A and subtracting the probability that both A and B happen.

🎯 Exam Tip: The difference event A-B means A occurs and B does not, and its probability formula is P(A) - P(A ∩ B).

 

Question 18. If P(A) = 0.3 and P(A ∩ B) = 0.03, find P(B|A).
Answer:
P(A) = 0.3, P(A ∩ B) = 0.03 are given.
Therefore, P(B|A) = \( \frac{P(A \cap B)}{P(A)} = \frac{0.03}{0.3} = 0.1 \)
In simple words: To find the chance of B happening given that A has happened, you divide the probability of both A and B happening by the probability of A happening.

🎯 Exam Tip: This is the direct application of the formula for conditional probability, P(B|A).

 

Question 19. If P(A) = P(B) = K for two mutually exclusive events A and B, find P(A U B).
Answer:
P (A) = P(B) = k.
A, B are mutually exclusive events
Therefore, P(A ∩ B) = 0
Now, P(A U B) = P(A) + P(B).
= k + k = 2k
In simple words: If two events cannot happen at the same time and both have the same chance 'k', then the chance of either one happening is simply the sum of their individual chances, which is k + k = 2k.

🎯 Exam Tip: For mutually exclusive events, the addition rule simplifies to P(A U B) = P(A) + P(B) because P(A ∩ B) is 0.

 

Question 20. If P(A' ∩ B) = 0.45 and A ∩ B = \( \Phi \), find P(B).
Answer:
P (A' ∩ B) = 0.45 and A ∩ B = \( \Phi \) are given.
Therefore, P(A ∩ B) = 0
Now, P(A' ∩ B) = P(B) – P(A ∩ B)
Therefore, 0.45 = P(B) – 0
Therefore, P(B) = 0.45
In simple words: If events A and B cannot happen together, then the event "B but not A" is just the event B itself. So, the probability of "B and not A" is the same as the probability of B.

🎯 Exam Tip: When events are mutually exclusive, B = B - A, and therefore P(B) = P(B-A).

 

Question 21. Two events A and B in a sample space are mutually exclusive and exhaustive. If P(A) = \( \frac{1}{3} \), find P(B).
Answer:
A and B are mutually exclusive and exhaustive events.
Therefore, P(A) + P(B) = 1 (Putting P(A) = \( \frac{1}{3} \))
\( \frac{1}{3} + P(B) = 1 \)
Therefore, P(B) = 1 - \( \frac{1}{3} = \frac{2}{3} \)
In simple words: If two events are mutually exclusive (cannot happen together) and exhaustive (cover all possibilities), their probabilities must add up to 1. So if P(A) is \( \frac{1}{3} \), P(B) must be \( 1 - \frac{1}{3} \), which is \( \frac{2}{3} \).

🎯 Exam Tip: For any two events A and B that are both mutually exclusive and exhaustive, P(A) + P(B) = 1.

 

Question 22. 2 % items in a lot are defective. What is the probability that an item randomly selected from this lot is non-defective?
Answer:
A = Event that an item is defective
Therefore, P(A) = 2 % = \( \frac{2}{100} \) = 0.02
A' = Event that an item is non-defective
Therefore, P(A') = 1 – P(A) = 1 – 0.02 = 0.98
In simple words: If 2% of items are faulty, then the remaining percentage must be non-faulty. So, 100% - 2% = 98% of items are good, which is a probability of 0.98.

🎯 Exam Tip: The probability of an event not happening (its complement) is 1 minus the probability of the event happening.

 

Question 23. State the number of sample points in the random experiment of tossing five balanced coins.
Answer:
The number of sample points in the random experiment of tossing five balanced coins is n = \( 2^5 \) = 32.
In simple words: Each coin has 2 possible outcomes (heads or tails). If you toss 5 coins, you multiply the outcomes for each coin: 2 * 2 * 2 * 2 * 2, which equals 32 total possible outcomes.

🎯 Exam Tip: For 'k' independent trials, each with 'n' outcomes, the total sample space size is \( n^k \).

 

Question 24. State the number of sample points in the random experiment of tossing one balanced coin and two balanced dice simultaneously.
Answer:
The number of sample points in the random experiment of tossing one balanced coin and two balanced dice simultaneously is n = \( 2^1 \times 6^2 \) = 2 × 36 = 72.
In simple words: The coin has 2 outcomes. Each die has 6 outcomes, so two dice have \( 6 \times 6 = 36 \) outcomes. To find the total possibilities for all three, multiply 2 by 36, which is 72.

🎯 Exam Tip: When multiple independent experiments occur, multiply the number of outcomes of each experiment to find the total sample space.

 

Question 25. Is it possible that P (A) = 0.7 and P (A U B) = 0.45 for two events A and B in a sample space ?
Answer:
P (A) = 0.7 and P(A U B) = 0.45 is not possible because P(A U B) < P(A).
In simple words: The probability of A or B happening together (A U B) must always be greater than or equal to the probability of A happening alone, because A U B includes A. Since 0.45 is less than 0.7, this situation is not possible.

🎯 Exam Tip: Always remember that the probability of the union of events cannot be less than the probability of any single event within that union.

 

Question 26. Two cards are selected one by one with replacement from 52 cards. State the number of elements in the sample space of the random experiment.
Answer:
The number of elements in the sample space of this random experiment of drawing two cards one by one with replacement from 52 cards is n = \( {}^{52}C_1 \times {}^{52}C_1 \) = 52 × 52 = 2704
In simple words: Since a card is picked and then put back, the first pick has 52 choices, and the second pick also has 52 choices. Multiply these to get the total number of ways to pick two cards, which is 52 times 52, or 2704.

🎯 Exam Tip: "With replacement" means outcomes for each draw are independent, so the number of options remains constant for each selection.

 

Question 27. For two independent events A and B, P(B|A) = \( \frac{1}{2} \) and P(A ∩ B) = \( \frac{1}{5} \). Find P(A).
Answer:
P(B|A) = \( \frac{1}{2} \), P(A ∩ B) = \( \frac{1}{5} \), P(A) = ?
P(B|A) = \( \frac{P(A \cap B)}{P(A)} \)
Therefore, \( \frac{1}{2} = \frac{\frac{1}{5}}{P(A)} \)
Therefore, \( \frac{1}{2}P(A) = \frac{1}{5} \)
Therefore, P(A) = \( \frac{1}{5} \times 2 = \frac{2}{5} \)
In simple words: We know P(B|A) is \( \frac{1}{2} \) and P(A ∩ B) is \( \frac{1}{5} \). Using the formula for conditional probability, we can rearrange it to find P(A). P(A) equals P(A ∩ B) divided by P(B|A), which is \( \frac{1}{5} \) divided by \( \frac{1}{2} \), resulting in \( \frac{2}{5} \).

🎯 Exam Tip: Use the formula P(B|A) = P(A ∩ B) / P(A) and algebraic manipulation to solve for the unknown probability.

 

Question 28. 1998 tickets out of 2000 tickets do not have a prize. If a person randomly selects one ticket from 2000 tickets, then what is the probability that the ticket selected is eligible for prize ?
Answer:
Total number of tickets = 2000
No. of tickets do not have a prize = 1998
No. of tickets eligible for prize = 2000 – 1998 = 2
Total number of outcomes of selecting a ticket is n = \( {}^{2000}C_1 \) = 2000
A = Event that the selected ticket is eligible for prize
Therefore, m = \( {}^{2}C_1 \) = 2
Therefore, P(A) = \( \frac{m}{n} \)
= \( \frac{2}{2000} \)
= \( \frac{1}{1000} \)
In simple words: First, find how many tickets actually have prizes: 2000 total - 1998 no-prize tickets = 2 prize tickets. The probability of picking a prize ticket is the number of prize tickets divided by the total number of tickets, which is 2/2000, or 1/1000.

🎯 Exam Tip: Carefully identify the number of favorable outcomes and the total possible outcomes for accurate probability calculation.

 

Section C

 

Question 1. Define the following events and draw their Venn diagram:
1. Mutually exclusive events
2. Union of events
3. Intersection of events
4. Difference event
5. Exhaustive events
6. Complementary event
Answer:
1. Mutually exclusive events: Suppose, U is a finite sample space. A and B are any two events of U. If A ∩ B = \( \Phi \), then A and B are mutually exclusive events.

ℹ️ चित्र व्याख्या (Diagram Explanation): इस वेन आरेख में, आयत नमूना स्थान (U) को दर्शाता है। दो वृत्त, A और B, दो घटनाओं का प्रतिनिधित्व करते हैं। वृत्त एक-दूसरे को काटते नहीं हैं, जो दर्शाता है कि घटनाओं A और B के बीच कोई सामान्य परिणाम नहीं है, इसलिए वे परस्पर अनन्य हैं।
In simple words: Mutually exclusive events are events that cannot both happen at the same time. Their Venn diagram shows two separate circles with no overlap.

🎯 Exam Tip: Mutually exclusive events have an intersection probability of zero, P(A ∩ B) = 0.

1. Union of events: Suppose, U is a finite sample space. A and B are any two events of U. Event A occurs or event B occurs or events A and B occur together, i.e., at least one of the events A and B occurs is called the union of events A and B. It is denoted by A U B. Thus,
A U B = {x; x \( \in \) A or x \( \in \) B or x \( \in \) A ∩ B}

ℹ️ चित्र व्याख्या (Diagram Explanation): इस वेन आरेख में, आयत नमूना स्थान (U) को दर्शाता है। दो वृत्त, A और B, दो घटनाओं का प्रतिनिधित्व करते हैं। छायांकित क्षेत्र A U B को दिखाता है, जिसका अर्थ है घटना A में या घटना B में या दोनों में होने वाले सभी तत्व। यह दर्शाता है कि कम से कम एक घटना घटित होती है।
In simple words: The union of events A and B (A U B) includes all outcomes that are in A, or in B, or in both. It means at least one of the events happens.

🎯 Exam Tip: The symbol 'U' represents the union of sets, often read as "A or B" in probability contexts.

1. Intersection of Events: Suppose, U is a finite sample space. A and B are any two events of U. The event that A and B occur together is called the intersection of events A and B. It is denoted by A ∩ B. Thus,
A ∩ B = {x; x \( \in \) A and x \( \in \) B}

ℹ️ चित्र व्याख्या (Diagram Explanation): इस वेन आरेख में, आयत नमूना स्थान (U) को दर्शाता है। दो वृत्त, A और B, दो घटनाओं का प्रतिनिधित्व करते हैं। छायांकित क्षेत्र A ∩ B को दिखाता है, जिसका अर्थ है घटना A और घटना B दोनों में सामान्य तत्व। यह उन परिणामों को दर्शाता है जो दोनों घटनाओं में एक साथ होते हैं।
In simple words: The intersection of events A and B (A ∩ B) includes only the outcomes that are common to both A and B. It means both events happen at the same time.

🎯 Exam Tip: The symbol '∩' represents the intersection of sets, often read as "A and B" in probability contexts.

1. Difference event: Suppose, U is a finite sample space. A and B are any two events of U. The event that A occurs but B does not occur is called the difference event of A and B. It is denoted by A – B or A ∩ B'. Similarly, the event that A does not occur but B occurs is called the difference event of B and A. It is denoted by B – A or A' ∩ B. Thus,
A – B = {x; x \( \in \) A and x \( \notin \) B}

ℹ️ चित्र व्याख्या (Diagram Explanation): इस वेन आरेख में, आयत नमूना स्थान (U) को दर्शाता है। दो वृत्त, A और B, दो घटनाओं का प्रतिनिधित्व करते हैं। पहले आरेख में, छायांकित क्षेत्र A - B को दर्शाता है, जिसका अर्थ है A के वे तत्व जो B में नहीं हैं। दूसरे आरेख में, छायांकित क्षेत्र B - A को दर्शाता है, जिसका अर्थ है B के वे तत्व जो A में नहीं हैं।
In simple words: The difference event A - B means that event A happens, but event B does not. It's like taking all elements of A and removing any that are also in B.

🎯 Exam Tip: Understand that A-B is equivalent to A ∩ B', which signifies A occurring and B not occurring.

1. Exhaustive events: Suppose, U is a finite sample space. A and B are any two events of U. If A U B = U, then A and B are called exhaustive events.

ℹ️ चित्र व्याख्या (Diagram Explanation): इस वेन आरेख में, आयत नमूना स्थान (U) को दर्शाता है। दो वृत्त, A और B, दो घटनाओं का प्रतिनिधित्व करते हैं। संपूर्ण नमूना स्थान (U) छायांकित है, जो दर्शाता है कि A और B का संघ (A U B) पूरे नमूना स्थान को कवर करता है। इसका मतलब है कि घटना A या घटना B (या दोनों) निश्चित रूप से घटित होगी।
In simple words: Exhaustive events are a set of events where at least one of them must occur. Their union covers the entire sample space.

🎯 Exam Tip: If events are exhaustive, the sum of their probabilities (after accounting for overlaps) must be 1.

1. Complementary event: Suppose, A is any event of the finite sample space U. The event that A does not occur means the event consists of elements in U but not in A is called the complementary event of A. It is denoted by A'. Thus,
A' = {x; x \( \notin \) A, x \( \in \) U}

ℹ️ चित्र व्याख्या (Diagram Explanation): इस वेन आरेख में, आयत नमूना स्थान (U) को दर्शाता है। एक वृत्त, A, एक घटना का प्रतिनिधित्व करता है। वृत्त A के बाहर का छायांकित क्षेत्र A' को दर्शाता है, जिसका अर्थ है नमूना स्थान U में वे सभी तत्व जो घटना A में नहीं हैं। यह घटना A के घटित न होने को दर्शाता है।
In simple words: A complementary event (A') includes all outcomes in the sample space that are not in event A. It is the opposite of A happening.

🎯 Exam Tip: For any event A, P(A) + P(A') = 1.

 

Question 2. Give the illustrations of finite and infinite sample space.
Answer:
Finite sample space: The sample space obtained for the random experiment throwing a balanced' six face die U = {1, 2, 3, 4, 5, 6} is the illustration of finite sample space.
Infinite sample space: The sample space obtained for the random experiment of selecting a card from a pack of 52 cards till it is ace of heart is the illustration of infinite sample space.
In simple words: A finite sample space has a limited number of outcomes, like the 6 sides of a die. An infinite sample space has an unlimited number of possible outcomes, like continuously drawing cards until a specific one appears, as you don't know how many draws it will take.

🎯 Exam Tip: Finite sample spaces have outcomes that can be counted, while infinite sample spaces involve outcomes that cannot be precisely counted, often related to continuous measurements or unbounded trials.

 

Question 3. Give the illustrations of impossible and certain event.
Answer:
Impossible event: Event to get the number greater than 6 on the upper side of a balanced die.
Certain event: Event to get head or tail in tossing a balanced coin.
In simple words: An impossible event is something that can never happen, like rolling a 7 on a standard six-sided die. A certain event is something that will definitely happen, like getting either heads or tails when flipping a coin.

🎯 Exam Tip: An impossible event has a probability of 0, and a certain event has a probability of 1.

 

Question 4. State the characteristics of random experiment.
Answer:
The characteristics of random experiment are as follows :
- It can be repeated under identical conditions,
- Its all possible outcomes are known,
- It cannot be predicted with certainty which outcome will appear and
- It results into a certain outcome.
In simple words: A random experiment can be done repeatedly in the same way, all its possible results are known, but you can't predict the exact result of any single try. However, you know some outcome will always happen.

🎯 Exam Tip: The unpredictability of the individual outcome, combined with knowledge of all potential outcomes, is key to defining a random experiment.

 

Question 5. State the assumptions of mathematical definition of probability.
Answer:
The assumptions of mathematical definition of probability are as follows:
- The number of outcomes in the sample space is finite.
- The number of all possible outcomes of the sample space is known.
- The outcomes of the sample space are equi-probable.
In simple words: The mathematical way of defining probability assumes that there are a limited, known number of possible results, and each result has an equal chance of happening.

🎯 Exam Tip: The 'equi-probable' assumption is critical for the classical definition; without it, this definition cannot be directly applied.

 

Question 6. State the limitations of mathematical definition of probability.
Answer:
The limitations of mathematical definition of probability are as follows:
- The probability of an event cannot be found if the outcomes are infinite.
- If the total number of outcomes is not known, the probability of an event cannot be determined.
- If the elementary outcomes in the sample space are not equi-probable, the probability of an event cannot be found.
In simple words: This definition can't be used if there are endless possible outcomes, if you don't know all the possible outcomes, or if some outcomes are more likely than others.

🎯 Exam Tip: These limitations highlight why other definitions of probability (like statistical or axiomatic) are necessary for broader applications.

 

Question 7. State the limitations of statistical definition of probability.
Answer:
The limitations of statistical definition of probability are as follows:
- The infinite value of n cannot be taken in practice.
- The exact value of probability cannot be known.
In simple words: This definition, based on doing an experiment many times, is limited because you can't actually do an experiment infinitely many times, and therefore, you can never get a perfectly exact probability value.

🎯 Exam Tip: Statistical probability provides an estimate; it approaches the true probability as the number of trials increases, but never reaches it exactly.

 

Question 8. Explain the equi-probable events with illustration.
Answer:
If there is no apparent reason to believe that out of one or more events of a random experiment, any one event is more or less likely to occur than the other events, then those events are called equi-probable.
Illustration: In the random experiment of tossing a balanced coin, two events of getting head (H) and getting tail (T) are equi-probable, because P (H) = P (T) = \( \frac{1}{2} \)
In simple words: Equi-probable events are those where each outcome has an equal chance of happening. For example, when you flip a fair coin, getting a head has the same chance as getting a tail, both \( \frac{1}{2} \).

🎯 Exam Tip: The concept of equi-probable events is fundamental to the classical definition of probability.

 

Question 9. State the law of addition of probability for two events A and B. Write the law of addition of probability if these two events are mutually exclusive.
Answer:
The law of addition of probability for two events A and B is as follows :
P(A U B) = P(A) + P(B) – P(A ∩ B)
If A and B are mutually exclusive events, A ∩ B = \( \Phi \) and P (A ∩ B) = 0. Hence, the law of addition of probability for two events A and B is written as follows:
P(A U B) = P(A) + P(B)
In simple words: The general rule for the probability of A or B happening is to add their individual probabilities and then subtract the probability of both happening. If they can't happen together, you just add their individual probabilities.

🎯 Exam Tip: Always remember to subtract P(A ∩ B) for non-mutually exclusive events to avoid double-counting the overlap.

 

Question 10. State the law of multiplication of probability for two events A and B. Write the law of multiplication of probability if these two events are independent.
Answer:
The law of multiplication of probability for two events A and B is as follows:
P(A ∩ B) = P(A|B) × P(B) OR
P(A ∩ B) = P(B|A) × P(A)
If two events A and B are independent, P(A|B) = P(A) and P(B|A) = P(B). So the law of multiplication of probability for two events A and B is written as follows:
P(A ∩ B) = P(A) × P(B)
In simple words: The general rule for the probability of both A and B happening is the chance of A, multiplied by the chance of B given that A has already happened. If A and B don't affect each other, then it's simply the chance of A multiplied by the chance of B.

🎯 Exam Tip: Distinguish between the general multiplication rule for any events and the simplified rule for independent events.

 

Question 11. State the following results for two independent events A and B:
1. P(A ∩ B)
2. P (A' ∩ B')
3. P(A ∩ B')
4. P(A' ∩ B)
Answer:
For two independent events A and B, the results are obtained as follows:
1. P(A ∩ B) = P(A) × P(B)
2. P(A' ∩ B') = P(A') × P(B')
3. P(A ∩ B') = P(A) × P(B')
4. P(A' ∩ B) = P(A') × P(B).
In simple words: For events A and B that are independent, the probability of both happening is their probabilities multiplied. This also applies to their complements; for example, the chance of "not A" and "not B" both happening is the chance of "not A" multiplied by the chance of "not B". The same logic applies when one event is a complement and the other is not.

🎯 Exam Tip: Remember that if events A and B are independent, then A and B', A' and B, and A' and B' are also independent.

 

Question 12. If p(A) = \( \frac{1}{3} \), P(B) = \( \frac{2}{3} \) and P(A ∩ B) = \( \frac{1}{6} \), then find P(A' ∩ B').
Answer:
p(A) = \( \frac{1}{3} \), P(B) = \( \frac{2}{3} \) and P(A ∩ B) = \( \frac{1}{6} \) are given.
According to the law of addition of probability,
P(A U B) = P(A) + P(B) – P(A ∩ B)
= \( \frac{1}{3} + \frac{2}{3} - \frac{1}{6} \)
= \( \frac{2+4-1}{6} \)
= \( \frac{5}{6} \)
Now, P(A' ∩ B') = P(A U B)'
= 1 – P(A U B)
= 1 - \( \frac{5}{6} \)
= \( \frac{6-5}{6} \)
= \( \frac{1}{6} \)
In simple words: First, use the addition rule to find the probability of A or B happening (P(A U B)). Then, use De Morgan's Law, which states that "not A and not B" is the same as "not (A or B)". So, P(A' ∩ B') is 1 minus P(A U B).

🎯 Exam Tip: De Morgan's laws (e.g., \( (A \cup B)' = A' \cap B' \)) are powerful tools for simplifying probability calculations involving complements of unions or intersections.

 

Question 13. If P(A|B) = 0.4, P(B) = 0.4, then find P(A ∩ B).
Answer:
P(B) = 2P(A|B) = 0.4
Therefore, P(B) = 0.4, 2P(A|B) = 0.4
Therefore, P(A|B) = \( \frac{0.4}{2} \) = 0.2
Now, P(A|B) = \( \frac{P(A \cap B)}{P(B)} \)
Therefore, P(A ∩ B) = P(A|B) × P(B)
= 0.2 × 0.4 = 0.08
In simple words: We are given that P(A|B) is half of P(B). So P(A|B) is 0.2. To find the probability of both A and B happening, multiply the probability of A given B by the probability of B, which is 0.2 * 0.4 = 0.08.

🎯 Exam Tip: Clearly identify all given values and use the appropriate formula for conditional probability, P(A ∩ B) = P(A|B) × P(B).

 

Question 14. If the events A and B are independent and 3P(A) = 2P(B) = 0.12, then find P(A ∩ B).
Answer:
3P(A) = 2P(B) = 0.12
Therefore, 3P(A) = 0.12 and 2P(B) = 0.12
Therefore, P(A) = \( \frac{0.12}{3} \)
= 0.04
and
P(B) = \( \frac{0.12}{2} \)
= 0.06
Now, A and B are independent events.
Therefore, P(A ∩ B) = P(A) × P(B)
= 0.04 × 0.06
= 0.0024
In simple words: First, calculate P(A) by dividing 0.12 by 3, which is 0.04. Then, calculate P(B) by dividing 0.12 by 2, which is 0.06. Since A and B are independent, multiply P(A) and P(B) to find P(A ∩ B), giving 0.04 * 0.06 = 0.0024.

🎯 Exam Tip: Break down multi-step problems: first find individual probabilities, then apply the correct rule for independent events.

 

Question 15. If 5P (A) = 3P (B) = 2P (A U B) = f for two events A and B, then find P(A' U B').
Answer:
5P (A) = 3P(B) = 2P (A U B) = \( \frac{3}{2} \)
Therefore, 5P(A) = \( \frac{3}{2} \) and 3P(B) = \( \frac{3}{2} \) and 2P(A U B) = \( \frac{3}{2} \)
Therefore, P(A) = \( \frac{3}{2 \times 5} = \frac{3}{10} \)
Therefore, P(B) = \( \frac{3}{2 \times 3} = \frac{1}{2} \)
Therefore, P(A U B) = \( \frac{3}{2 \times 2} = \frac{3}{4} \)
According to the law of addition of probability,
P(A U B) = P(A) + P(B) – P(A ∩ B)
\( \frac{3}{4} = \frac{3}{10} + \frac{1}{2} \) - P(A ∩ B)
P(A ∩ B) = \( \frac{3}{10} + \frac{1}{2} - \frac{3}{4} \)
= \( \frac{6+10-15}{20} = \frac{1}{20} \)
Now, P(A' U B') = P(A ∩ B)'
= 1 – P(A ∩ B)
= 1 - \( \frac{1}{20} = \frac{19}{20} \)
In simple words: First, find P(A), P(B), and P(A U B) from the given ratios. Then, use the addition rule to calculate P(A ∩ B). Finally, apply De Morgan's Law: P(A' U B') is the same as P((A ∩ B)'), which is 1 minus P(A ∩ B).

🎯 Exam Tip: This problem combines ratio interpretation, the addition rule of probability, and De Morgan's laws; solve it step-by-step.

 

Question 16. If P(A ∩ B) = 0.12 and P(B) = 0.3 for two independent events A and B, then find P(A U B).
Answer:
A and B are independent events.
Therefore, P(A ∩ B) = P(A) × P(B)
Therefore, 0.12 = P(A) × 0.3
Therefore, P(A) = \( \frac{0.12}{0.3} \) = 0.4
Now, according to the law of addition of probability,
P(A U B) = P(A) + P(B) – P(A ∩ B)
= 0.4 + 0.3 – 0.12
= 0.58
In simple words: Since A and B are independent, use P(A ∩ B) = P(A) * P(B) to find P(A). Then, use the addition rule P(A U B) = P(A) + P(B) - P(A ∩ B) to find the probability of A or B happening.

🎯 Exam Tip: Leverage the property of independent events to find unknown probabilities before applying other probability laws.

 

Question 17. If A = {x| 1 < x < 3} and B = {x|\( \frac{1}{2} \) < x < 2}, then find A U B and A ∩ B.
Answer:
A = {x| 1 < x < 3} and B = {x|\( \frac{1}{2} \) < x < 2}

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख संख्या रेखा पर दो अंतराल A और B को दर्शाता है। अंतराल A (पीला) 1 और 3 के बीच की संख्याएँ हैं (1 और 3 शामिल नहीं)। अंतराल B (नीला) \( \frac{1}{2} \) और 2 के बीच की संख्याएँ हैं ( \( \frac{1}{2} \) और 2 शामिल नहीं)। आरेख उनके प्रतिच्छेदन (A ∩ B) को 1 और 2 के बीच के अतिव्यापी क्षेत्र के रूप में और उनके संघ (A U B) को \( \frac{1}{2} \) और 3 के बीच के कुल क्षेत्र के रूप में दिखाता है।
From the above diagram, the events are obtained as follows:
(1) A U B = {x|\( \frac{1}{2} \) < x < 3}
(2) A ∩ B = {x|1 < x < 2}
In simple words: Set A includes numbers between 1 and 3. Set B includes numbers between \( \frac{1}{2} \) and 2. The union (A U B) covers all numbers from \( \frac{1}{2} \) up to 3. The intersection (A ∩ B) includes only the numbers that are in both, which are numbers between 1 and 2.

🎯 Exam Tip: Visualize intervals on a number line to easily determine their union (the entire spread) and intersection (the overlap).

 

Question 18. The probability of occurrence of at least one of the two events A and B is \( \frac{1}{4} \). The probability that event A occurs, but event B does not occur is \( \frac{1}{5} \). Find the probability of event B.
Answer:
A U B = Event that at least one of the two events A and B occurs
Therefore, P(A U B) = \( \frac{1}{4} \)
P(A ∩ B') = \( \frac{1}{5} \)
According to the law of addition of probability,
P(A U B) = P(A) + P(B) – P(A ∩ B)
Therefore, P(A U B) = P(A) – P(A ∩ B) + P(B)
[Putting, P(A ∩ B') = P(A) – P(A ∩ B)]
Therefore, P(A U B) = P(A ∩ B') + P(B)
Therefore, P(B) = P(A U B) – P(A ∩ B')
= \( \frac{1}{4} - \frac{1}{5} \)
= \( \frac{5-4}{20} \)
= \( \frac{1}{20} \)
In simple words: We know the chance of A or B happening is \( \frac{1}{4} \). We also know the chance of A happening without B is \( \frac{1}{5} \). Since "A or B" can be thought of as "A without B" plus "B", we can find the probability of B by subtracting "A without B" from "A or B". So, \( \frac{1}{4} - \frac{1}{5} = \frac{1}{20} \).

🎯 Exam Tip: Remember that P(A U B) can also be expressed as P(A ∩ B') + P(B), which simplifies calculations in certain scenarios.

 

Question 19. If P(B) = \( \frac{3}{5} \) and P(A' ∩ B) = \( \frac{1}{2} \) for two events A and B, find P(A|B).
Answer:
P(B) = \( \frac{3}{5} \) and P(A' ∩ B) = \( \frac{1}{2} \) are given.
Therefore, P(A' ∩ B) = P(B) – P(A ∩ B)
Therefore, \( \frac{1}{2} = \frac{3}{5} \) - P(A ∩ B)
Therefore, P(A ∩ B) = \( \frac{3}{5} - \frac{1}{2} = \frac{6-5}{10} \)
= \( \frac{1}{10} \)
Now, P(A|B) = \( \frac{P(A \cap B)}{P(B)} \)
= \( \frac{\frac{1}{10}}{\frac{3}{5}} = \frac{1}{10} \times \frac{5}{3} = \frac{1}{6} \)
In simple words: First, use the formula P(A' ∩ B) = P(B) - P(A ∩ B) to find P(A ∩ B). Then, use the conditional probability formula P(A|B) = P(A ∩ B) / P(B) to find the final answer.

🎯 Exam Tip: Understand the relationship P(A' ∩ B) = P(B) - P(A ∩ B) as it represents the probability of B occurring without A.

 

Question 20. 6 persons have a passport in' a group of 10 persons. If 3 persons are randomly selected from this group, find the probability that
(i) all the three persons have a passport
(ii) two persons among them do not have a passport.
Answer:
Out of 10 persons, 6 persons have a passport. So 4 persons do not have a passport.
Now, 3 persons out of 10 persons are randomly selected.
Therefore, Total number of primary outcomes,
n = \( {}^{10}C_3 = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} \) = 120
(i) A = Event that all three persons have a passport
m = \( {}^{6}C_3 = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} \) = 20
P (A) = \( \frac{m}{n} = \frac{20}{120} = \frac{1}{6} \)
(ii) B = Event that two persons among three persons do not have a passport
Therefore, Favourable outcomes for the event B is
m = \( {}^{6}C_1 \times {}^{4}C_2 \) = 6 × 6 = 36
Therefore, P(B) = \( \frac{m}{n} = \frac{36}{120} = \frac{3}{10} \)
In simple words: There are 10 people, 6 with passports and 4 without. We pick 3 people.
(i) The chance that all 3 picked have passports is found by choosing 3 from the 6 passport holders, divided by choosing any 3 from 10 people.
(ii) The chance that 2 do not have passports means 1 person has a passport and 2 do not. So, we choose 1 from 6 passport holders and 2 from 4 non-passport holders, then divide by the total ways to choose 3 people.

🎯 Exam Tip: For selection problems, use combinations (\( {}^{n}C_k \)) to count the total and favorable outcomes, and clearly identify the conditions for each part of the question.

 

Question 21. The probability that the tax-limit for income of males increases in the budget of a year is 0.66 and the probability that the tax- limit increases for income of females is 0.72. The probability that the tax-limit increases for income of both the males and females is 0.47. Find the probability that
(i) the tax-limit increases for income of only one of the two, males and females,
(ii) the tax-limit does not increase for income of males as well as females in the budget of that year.
Answer:
A = Event that the tax-limit for income of males increases.
B = Event that the tax-limit for income of females increases.
A ∩ B = Event that the tax-limit for income of male and female increases.
Here, P(A) = 0.66, P(B) = 0.72 and P(A ∩ B) =0.47 are given.
(i) C = Event that the tax-limit increases only for one of the two male and female.
Event C occurs in following two ways:
- A ∩ B' = Event that the tax-limit increases for the income of male only and not of female OR
- A' ∩ B = Event that the tax-limit increases for the income of female only and not of male Events A ∩ B' and A' ∩ B are mutually exclusive,
Therefore, P(C) = P(A ∩ B') + P(A' ∩ B)
= [P(A) – P(A ∩ B)] + (P(B) – P(A ∩ B)]
= [0.66 – 0.47] + [0.72 – 0.47]
= 0.19 + 0.25 = 0.44
(ii) A' ∩ B' = Event that the tax-limit does not increase for income of either male or female.
Now, P(A U B) = P(A) + P(B) – P(A ∩ B)
= 0.66 + 0.72 – 0.47 = 0.91
Therefore, P(A' ∩ B') = P(A U B)' = 1 – P(A U B)
= 1-0.91 = 0.09
In simple words: Let A be the event of a tax-limit increase for males and B for females. We are given P(A), P(B), and P(A ∩ B).
(i) "Only one" means (A and not B) OR (B and not A). We find P(A ∩ B') as P(A) - P(A ∩ B) and P(A' ∩ B) as P(B) - P(A ∩ B). Since these are mutually exclusive, we add them up.
(ii) "Neither" means (not A and not B), which is the complement of (A or B). We first find P(A U B) using the addition rule, then subtract it from 1 to get P(A' ∩ B').

🎯 Exam Tip: Carefully translate "only one" and "neither" into set notation (e.g., \( (A \cap B') \cup (A' \cap B) \) and \( A' \cap B' \)) and use the appropriate probability rules.

 

Question 22. The price of petrol rises in 80 % of the cases and the price of diesel rises in 77% of the cases after the rise in price of crude oil. The price of petrol and diesel rises in 68 % cases. Find the probability that the price of diesel rises under the condition that there is a rise in the price of petrol.
Answer:
A = Event that the price of petrol rises
Therefore, P(A) = \( \frac{80}{100} \)
B = Event that the price of diesel rises
Therefore, P(B) = \( \frac{77}{100} \)
A ∩ B = Event that the prices of both petrol and diesel rises
Therefore, P(A ∩ B) = \( \frac{68}{100} \)
B|A = Event that the price of diesel rises knowing that the price of petrol rises
Therefore, P(B|A) = \( \frac{P(A \cap B)}{P(A)} \)
= \( \frac{\frac{68}{100}}{\frac{80}{100}} \)
= \( \frac{68}{100} \times \frac{100}{80} \)
= \( \frac{17}{20} \)
In simple words: Let A be petrol price rise and B be diesel price rise. We are given P(A), P(B), and P(A ∩ B). We want to find the probability of diesel rising given that petrol has risen (P(B|A)). Using the conditional probability formula, this is P(A ∩ B) divided by P(A).

🎯 Exam Tip: Accurately identifying the 'given' event and the event whose probability is sought is crucial for applying conditional probability.

 

Question 23. As per the prediction of weather bureau, the probabilities for rains on three days; Thursday, Friday and Saturday in the next week are 0.8, 0.7 and 0.6 respectively. Find the probability that it rains on at least one of the three days in the next week.
(Note:The events of rains on three days; Thursday, Friday and Saturday of a week are independent.)
Answer:
A = Event that it rains on Thursday
B = Event that it rains on Friday
C = Event that it rains on Saturday
Here, P(A) = 0.8, P(B) = 0.7 and P(C) = 0.6 are given.
A, B and C are independent events.
Therefore, P(A ∩ B) = P(A) × P(B) = 0.8 × 0.7 = 0.56
P(A ∩ C) = P(A) × P(C) = 0.8 × 0.6 = 0.48
P(B ∩ C) = P(B) × P(C) = 0.7 × 0.6 = 0.42
P(A ∩ B ∩ C) = P(A) × P(B) × P(C)
= 0.8 x 0.7 × 0.6 = 0.336
Now, A U B U C = Event that it rains on at least one of the three days in the next week
According to the law of addition of probability,
P(A U B U C) = P(A) + P(B) + P(C) – P(A ∩ B) – P(A ∩ C) – P(B ∩ C) + P(A ∩ B ∩ C)
= 0.8 + 0.7 + 0.6 – 0.56 – 0.48 – 0.42 + 0.336
= 0.976
In simple words: Let A, B, and C be the events of rain on Thursday, Friday, and Saturday, respectively. We are given their individual probabilities. Since these events are independent, we can find the probability of their intersections by multiplying. To find the probability of rain on at least one day, we use the inclusion-exclusion principle for three events.

🎯 Exam Tip: For independent events, calculating P(A ∩ B) and P(A ∩ B ∩ C) is simply multiplying individual probabilities, which simplifies the inclusion-exclusion principle.

 

Section D

 

Question 1. 6 LED televisions and 4 LCD televisions are displayed in digital store A whereas 5 LED televisions and 3 LCD televisions are displayed in digital store B. One of the two stores is randomly selected and one television is selected from that store. Find the probability that it is an LCD television.
Answer:
Digital store A: 6 LED TV + 4 LCD TV = 10 TVs
Digital store B: 5 LED TV + 3 LCD TV = 8 TVs
A₁ = Event that store A is selected.
P(A₁) = \( \frac{1}{2} \)
B₁ = Event that an LCD TV is selected from store A.
P(B₁|A₁) = \( \frac{{}^{4}C_1}{{}^{10}C_1} = \frac{4}{10} \)
P(A₁ ∩ B₁) = P(A₁) P(B₁|A₁) = \( \frac{1}{2} \times \frac{4}{10} = \frac{4}{20} \)
A₂ = Event that store B is selected.
P(A₂) = \( \frac{1}{2} \)
B₂ = Event that an LCD TV is selected from store B.
P(B₂|A₂) = \( \frac{{}^{3}C_1}{{}^{8}C_1} = \frac{3}{8} \)
P(A₂ ∩ B₂) = P(A₂) P(B₂|A₂) = \( \frac{1}{2} \times \frac{3}{8} = \frac{3}{16} \)
C = Event that an LCD TV is selected (This is the union of selecting store A and an LCD TV OR selecting store B and an LCD TV).
Since selecting from store A and selecting from store B are mutually exclusive events for the *overall* selection process:
P(C) = P(A₁ ∩ B₁) + P(A₂ ∩ B₂)
= \( \frac{4}{20} + \frac{3}{16} \)
= \( \frac{16+15}{80} \)
= \( \frac{31}{80} \)
In simple words: We need to find the probability of picking an LCD TV. This can happen in two ways: either we pick store A (which has a \( \frac{1}{2} \) chance) and then an LCD from A ( \( \frac{4}{10} \) chance), OR we pick store B ( \( \frac{1}{2} \) chance) and then an LCD from B ( \( \frac{3}{8} \) chance). We multiply the probabilities for each path and then add them together to get the total probability.

🎯 Exam Tip: This is a classic example of total probability. Break down the problem into mutually exclusive cases (selecting each store), calculate their probabilities, and sum them up.

 

Question 2. One number is randomly selected from the natural numbers 1 to 100. Find the probability that the number selected is either a single digit number or a perfect square.
Answer:
One number is randomly selected from the numbers 1 to 100.
Therefore, Total number of primary outcomes, n = \( {}^{100}C_1 \) = 100
A = Event that the number selected is a single digit number
= {1, 2, 3, 4, 5, 6, 7, 8, 9}
Therefore, m = 9
Therefore, P(A) = \( \frac{m}{n} = \frac{9}{100} \)
B = Event that the number selected is a perfect square
= {1, 4, 9, 16, 25, 36, 49, 64, 81, 100}
Therefore, m = 10
Therefore, P(B) = \( \frac{m}{n} = \frac{10}{100} \)
A ∩ B = Event that the selected number is a single digit number and a perfect square
= {1, 4, 9}
Therefore, m = 3
Therefore, P(A ∩ B) = \( \frac{m}{n} = \frac{3}{100} \)
Now, A U B = Event that the number selected is either a single digit number or a perfect square
Therefore, P(A U B) = P(A) + P(B) – P(A ∩ B)
= \( \frac{9}{100} + \frac{10}{100} - \frac{3}{100} \)
= \( \frac{16}{100} \)
= \( \frac{4}{25} \)
In simple words: Out of 100 numbers, there are 9 single-digit numbers and 10 perfect squares. The numbers that are both (single-digit perfect squares) are 1, 4, 9 (3 numbers). To find the probability of getting a single digit OR a perfect square, add the probabilities of each and subtract the probability of both, to avoid counting the overlapping numbers twice.

🎯 Exam Tip: When using the addition rule P(A U B) = P(A) + P(B) - P(A ∩ B), correctly identifying and subtracting the intersection is crucial for accuracy.

 

Question 3. A balanced coin is tossed thrice. If the first two tosses have resulted in tail, find the probability that tail appears on the coin In all the three trials.
Answer:
Let H be Head and T be Tail.
Total possible outcomes for three tosses = \( 2^3 \) = 8.
Sample Space U = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
A = Event that in the first trial tail is obtained.
B = Event that in the second trial tail is obtained on the coin.
C = Event that in the first two trials tail is obtained on the coin.
C = {TTH, TTT}
D = Event that tail appears on the coin in all three trials.
D = {TTT}
We need to find P(D|C), the probability that tail appears in all three trials, given that the first two tosses resulted in tail.
Using the formula for conditional probability, P(D|C) = \( \frac{P(D \cap C)}{P(C)} \)
The intersection D ∩ C means "all three are tails" AND "the first two are tails". This is simply "all three are tails", so D ∩ C = {TTT}.
P(D ∩ C) = \( \frac{1}{8} \)
P(C) = \( \frac{2}{8} \)
P(D|C) = \( \frac{\frac{1}{8}}{\frac{2}{8}} = \frac{1}{2} \)
In simple words: We are tossing a coin three times. We want to find the chance of getting three tails, given that the first two tosses were already tails. If the first two are already tails, we only need the third toss to be a tail. Since a coin toss is independent, the chance of the third being a tail is \( \frac{1}{2} \).

🎯 Exam Tip: For independent events, conditional probability often simplifies: the past outcomes (first two tails) don't influence future outcomes (third toss). However, always show the formal steps. The original calculation of P(D|C) by direct counting from the reduced sample space {TTH, TTT} is also valid: one favorable outcome (TTT) out of two possible (TTH, TTT), so \( \frac{1}{2} \).

 

Question 4. If events A, B and C are independent events and P(A) = P(B) = P(C) = p, then find the value of P(A U B U C) in terms of p.
Answer:
Events A, B and C are independent events.
P(A) = P(B) = P(C) = p
P(A U B U C) = P(A) + P(B) + P(C) – P (A ∩ B) – P (A ∩ C) – P(B ∩ C) + P(A ∩ B ∩ C)
= P(A) + P(B) + P(C) – P(A) × P(B) – P(A) × P(C) – P(B) × P(C) + P(A) × P(B) × P(C)
= p + p + p - (p × p) – (p × p) – (p × p) + (p × p × p)
= \( 3p – p^2 – p^2 – p^2 + p^3 \)
= \( 3p - 3p^2 + p^3 \)
= \( p (3 – 3p + p^2) \)
In simple words: We want to find the chance of A or B or C happening. Since they are independent, the probability of any two or all three happening together is just the product of their individual probabilities. We use the inclusion-exclusion formula: add individual probabilities, subtract pairs' probabilities, and add back all three's probability. Replacing each P with 'p' gives the final expression.

🎯 Exam Tip: For independent events, the intersection of any subset of events is the product of their individual probabilities. This simplifies the general inclusion-exclusion formula greatly.

 

Question 5. The genderwise data of a sample of 6000 employees selected from class 3 and class 4 employees in the government jobs of a state are shown in the following table:

Class of Employees	Gender		Total
	Males	Females
Class 3	3600	900	4500
Class 4	400	1100	1500
Total	4000	2000	6000

One employee is randomly selected from all the class 3 and class 4 employees in government jobs of this state.
(1) If the selected employee is a male, find the probability that he belongs to class 3.
(2) If it is given that the selected employee belongs to class 3, find the probability that he is a male.
Answer:
Suppose, A = Event that an employee belongs to class 3
B = Event that an employee belongs to class 4
C = Event that an employee is a male
D = Event that an employee is a female
Therefore, P(A) = \( \frac{4500}{6000} \), P(B) = \( \frac{1500}{6000} \), P(C) = \( \frac{4000}{6000} \) and P(D) = \( \frac{2000}{6000} \)
(1) A|C = Event that an employee selected is a male, then he belongs to class 3
P(A|C) = \( \frac{P(A \cap C)}{P(C)} \)
P(A ∩ C) = Probability that an employee is from Class 3 AND is male = \( \frac{3600}{6000} \)
P(A|C) = \( \frac{\frac{3600}{6000}}{\frac{4000}{6000}} \)
= \( \frac{3600}{4000} \)
= \( \frac{9}{10} \)
(ii) C|A = Event that an employee selected belongs to class 3, then he is a male.
P(C|A) = \( \frac{P(A \cap C)}{P(A)} \)
P(A ∩ C) = Probability that an employee is from Class 3 AND is male = \( \frac{3600}{6000} \)
P(A) = Probability that an employee is from Class 3 = \( \frac{4500}{6000} \)
= \( \frac{\frac{3600}{6000}}{\frac{4500}{6000}} \)
= \( \frac{3600}{4500} \)
= \( \frac{4}{5} \)
In simple words: We have employee data by class and gender.
(1) We want the chance that a male employee is from Class 3. We look at the total males (4000) and how many of them are in Class 3 (3600). So, 3600 out of 4000, which is \( \frac{9}{10} \).
(2) We want the chance that a Class 3 employee is male. We look at the total Class 3 employees (4500) and how many of them are male (3600). So, 3600 out of 4500, which is \( \frac{4}{5} \).

🎯 Exam Tip: Carefully read conditional probability questions to determine which event is the 'given' condition (the denominator) and which event's probability is being sought (the numerator, which is the intersection with the given condition).