OP Malhotra Class 12 Maths Solutions Chapter 18 Probability Exercise 18 (B)

Question 1. Given that E and F are events such that P(E) = 0.6, P(F) = 0.3 and P(E \( \cap \) F) = 0.2, find P(E/F) and P(F/E).
Answer: We are given the probabilities of events E and F, and their intersection.
Given: \( P(E) = 0.6 \)
\( P(F) = 0.3 \)
\( P(E \cap F) = 0.2 \)
We use the formula for conditional probability: \( P(A/B) = \frac{P(A \cap B)}{P(B)} \).
So, for \( P(E/F) \):
\( P(E/F) = \frac{P(E \cap F)}{P(F)} \)
\( \implies P(E/F) = \frac{0.2}{0.3} \)
\( \implies P(E/F) = \frac{2}{3} \)
Next, for \( P(F/E) \):
\( P(F/E) = \frac{P(E \cap F)}{P(E)} \)
\( \implies P(F/E) = \frac{0.2}{0.6} \)
\( \implies P(F/E) = \frac{1}{3} \)
Conditional probability helps us understand how the occurrence of one event affects the probability of another event happening.
In simple words: We know the chances of event E, event F, and both happening. We used a special rule to find the chance of E happening if F already happened, and then the chance of F happening if E already happened.

🎯 Exam Tip: Remember the formula for conditional probability \( P(A/B) = \frac{P(A \cap B)}{P(B)} \) and be careful to use the correct denominator for each calculation.

Question 2. If A and B are two events such that P(A) = 0.5, P(B) = 0.6 and P(A \( \cup \) B) = 0.8, find P(A/B) and P(B/A).
Answer: We are given the probabilities for events A, B, and their union.
Given: \( P(A) = 0.5 \)
\( P(B) = 0.6 \)
\( P(A \cup B) = 0.8 \)
First, we need to find the probability of the intersection, \( P(A \cap B) \), using the addition rule for probabilities:
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
Substitute the given values:
\( 0.8 = 0.5 + 0.6 - P(A \cap B) \)
\( \implies 0.8 = 1.1 - P(A \cap B) \)
\( \implies P(A \cap B) = 1.1 - 0.8 \)
\( \implies P(A \cap B) = 0.3 \)
Now, we can find the conditional probabilities:
For \( P(A/B) \):
\( P(A/B) = \frac{P(A \cap B)}{P(B)} \)
\( \implies P(A/B) = \frac{0.3}{0.6} \)
\( \implies P(A/B) = \frac{1}{2} \)
For \( P(B/A) \):
\( P(B/A) = \frac{P(A \cap B)}{P(A)} \)
\( \implies P(B/A) = \frac{0.3}{0.5} \)
\( \implies P(B/A) = \frac{3}{5} \)
The relationship between union, intersection, and individual probabilities is fundamental in probability theory.
In simple words: We had the chances of A, B, and A or B happening. We first figured out the chance of A and B both happening. Then, we used that to find the chance of A happening if B already did, and the chance of B happening if A already did.

🎯 Exam Tip: Always calculate \( P(A \cap B) \) using the addition rule if \( P(A \cup B) \) is given, before finding conditional probabilities.

Question 3. If A and B are two events such that P(A) = 0.3, P(B) = 0.6 and P(B/A) = 0.5, find P(A/B) and P(A U B).
Answer: We are given the probabilities of events A, B, and the conditional probability of B given A.
Given: \( P(A) = 0.3 \)
\( P(B) = 0.6 \)
\( P(B/A) = 0.5 \)
First, we use the conditional probability formula to find \( P(A \cap B) \):
\( P(B/A) = \frac{P(A \cap B)}{P(A)} \)
\( \implies 0.5 = \frac{P(A \cap B)}{0.3} \)
\( \implies P(A \cap B) = 0.5 \times 0.3 \)
\( \implies P(A \cap B) = 0.15 \)
Now we can find \( P(A/B) \):
\( P(A/B) = \frac{P(A \cap B)}{P(B)} \)
\( \implies P(A/B) = \frac{0.15}{0.6} \)
\( \implies P(A/B) = \frac{15}{60} \)
\( \implies P(A/B) = \frac{1}{4} \)
Next, we find \( P(A \cup B) \) using the addition rule:
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
\( \implies P(A \cup B) = 0.3 + 0.6 - 0.15 \)
\( \implies P(A \cup B) = 0.9 - 0.15 \)
\( \implies P(A \cup B) = 0.75 \)
It's interesting to note how knowing one conditional probability allows us to derive others and the union.
In simple words: We knew the chances of A, B, and the chance of B happening if A did. First, we worked backward to find the chance of both A and B happening. Then, we used that to find the chance of A happening if B did, and the chance of A or B happening.

🎯 Exam Tip: If you are given a conditional probability, always use it first to calculate the probability of the intersection, which is a key value for other calculations.

Question 4. If P(not A) = 0.7, P(B) = 0.7 and P(B/A) = 0.5, find P(A/B) and P(AUB).
Answer: We are given the probability of "not A", the probability of B, and the conditional probability of B given A.
Given: \( P(\text{not } A) = 0.7 \)
\( P(B) = 0.7 \)
\( P(B/A) = 0.5 \)
First, find \( P(A) \) from \( P(\text{not } A) \):
\( P(A) = 1 - P(\text{not } A) \)
\( \implies P(A) = 1 - 0.7 \)
\( \implies P(A) = 0.3 \)
Next, use the conditional probability \( P(B/A) \) to find \( P(A \cap B) \):
\( P(B/A) = \frac{P(B \cap A)}{P(A)} \)
\( \implies 0.5 = \frac{P(B \cap A)}{0.3} \)
\( \implies P(B \cap A) = 0.5 \times 0.3 \)
\( \implies P(B \cap A) = 0.15 \)
Remember that \( P(B \cap A) \) is the same as \( P(A \cap B) \).
Now, we can find \( P(A/B) \):
\( P(A/B) = \frac{P(A \cap B)}{P(B)} \)
\( \implies P(A/B) = \frac{0.15}{0.7} \)
\( \implies P(A/B) = \frac{15}{70} \)
\( \implies P(A/B) = \frac{3}{14} \)
Finally, find \( P(A \cup B) \) using the addition rule:
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
\( \implies P(A \cup B) = 0.3 + 0.7 - 0.15 \)
\( \implies P(A \cup B) = 1.0 - 0.15 \)
\( \implies P(A \cup B) = 0.85 \)
Probabilities always range from 0 to 1, representing impossible to certain events.
In simple words: We started with the chance of "not A" and B, plus B happening if A did. We first found the chance of A, then the chance of A and B both happening. Using these, we found the chance of A happening if B did, and the chance of A or B happening.

🎯 Exam Tip: If \( P(\text{not } A) \) is given, always calculate \( P(A) \) first, as it's often needed in subsequent steps.

Question 5. Given that P(A) = 0.8, P(A/B) = 0.8, P(A\( \cap \)B) = 0.5, find (i) P(B) (ii) P(B/A) (iii) P(AUB) (iv) P (A \( \cap \) B/A \( \cup \) B) (v) P [(A \( \cap \) B)/B'] (vi) P(A \( \cap \) B/A)
(Note: The problem statement has a typo in part (iv) and (v) - it seems to be P(A \( \cup \) B/ A \( \cap \) B) and P(A \( \cap \) B / A \( \cup \) B). Assuming it should be (iv) P(A \( \cup \) B/A \( \cap \) B) and (v) P(A \( \cap \) B/A \( \cup \) B) as indicated by the solution steps.)
(Another clarification: the question lists "(v)" for P[(A \( \cap \) B)/B'] and "(vi)" for P(A \( \cap \) B/A). This seems to be a numbering error in the source, where (iv) is listed and then directly (v) and (vi). The solution follows (i), (ii), (iii), (iv), (v), (vi), (vii). I will follow the solution's numbering and structure by finding (iv) P(A \( \cup \) B / A \( \cap \) B), (v) P(A \( \cap \) B / A \( \cup \) B), (vi) P[(A \( \cap \) B)/B'], (vii) P(A \( \cap \) B/A). )
Answer: We are given some probabilities and need to find several others using probability rules.
Given: \( P(A) = 0.8 \)
\( P(A/B) = 0.8 \)
\( P(A \cap B) = 0.5 \)

(i) Find \( P(B) \):
We know the conditional probability formula: \( P(A/B) = \frac{P(A \cap B)}{P(B)} \)
Substitute the given values:
\( 0.8 = \frac{0.5}{P(B)} \)
\( \implies P(B) = \frac{0.5}{0.8} \)
\( \implies P(B) = \frac{5}{8} \)

(ii) Find \( P(B/A) \):
Using the conditional probability formula:
\( P(B/A) = \frac{P(A \cap B)}{P(A)} \)
\( \implies P(B/A) = \frac{0.5}{0.8} \)
\( \implies P(B/A) = \frac{5}{8} \)

(iii) Find \( P(A \cup B) \):
Using the addition rule for probabilities:
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
\( \implies P(A \cup B) = 0.8 + \frac{5}{8} - 0.5 \)
Convert decimals to fractions for easier calculation: \( 0.8 = \frac{8}{10} = \frac{4}{5} \), \( 0.5 = \frac{1}{2} \)
\( P(A \cup B) = \frac{4}{5} + \frac{5}{8} - \frac{1}{2} \)
Find a common denominator, which is 40:
\( P(A \cup B) = \frac{32}{40} + \frac{25}{40} - \frac{20}{40} \)
\( \implies P(A \cup B) = \frac{32 + 25 - 20}{40} \)
\( \implies P(A \cup B) = \frac{57 - 20}{40} \)
\( \implies P(A \cup B) = \frac{37}{40} \)

(iv) Find \( P(A \cap B / A \cup B) \) (Interpreting \( P(A/\text{A} \cup \text{B}) \) from OCR based on solution step):
\( P(A \cap B / A \cup B) = \frac{P((A \cap B) \cap (A \cup B))}{P(A \cup B)} \)
We know that \( (A \cap B) \cap (A \cup B) = (A \cap B) \) because \( (A \cap B) \) is a subset of \( (A \cup B) \).
So, \( P(A \cap B / A \cup B) = \frac{P(A \cap B)}{P(A \cup B)} \)
\( \implies P(A \cap B / A \cup B) = \frac{0.5}{\frac{37}{40}} \)
\( \implies P(A \cap B / A \cup B) = 0.5 \times \frac{40}{37} \)
\( \implies P(A \cap B / A \cup B) = \frac{1}{2} \times \frac{40}{37} \)
\( \implies P(A \cap B / A \cup B) = \frac{20}{37} \)

(v) Find \( P((A \cap B)/B') \):
First, find \( P(B') \):
\( P(B') = 1 - P(B) \)
\( \implies P(B') = 1 - \frac{5}{8} \)
\( \implies P(B') = \frac{3}{8} \)
Now, we need \( P((A \cap B) \cap B') \). The event \( (A \cap B) \) means A and B both happen. The event \( B' \) means B does not happen. It is impossible for B to happen and not happen at the same time. So, \( (A \cap B) \cap B' = \emptyset \).
Therefore, \( P((A \cap B) \cap B') = P(\emptyset) = 0 \).
\( P((A \cap B)/B') = \frac{P((A \cap B) \cap B')}{P(B')} \)
\( \implies P((A \cap B)/B') = \frac{0}{\frac{3}{8}} \)
\( \implies P((A \cap B)/B') = 0 \)

(vi) Find \( P((A \cup B)/A) \) (Interpreting \( P(A \cup B / A) \) from OCR based on solution step):
\( P((A \cup B)/A) = \frac{P((A \cup B) \cap A)}{P(A)} \)
We know that \( (A \cup B) \cap A = A \) because A is a subset of \( (A \cup B) \).
So, \( P((A \cup B)/A) = \frac{P(A)}{P(A)} \)
\( \implies P((A \cup B)/A) = \frac{0.8}{0.8} \)
\( \implies P((A \cup B)/A) = 1 \)

(vii) Find \( P(A \cap B / A) \):
This is the probability of the intersection of A and B given A.
\( P(A \cap B / A) = \frac{P((A \cap B) \cap A)}{P(A)} \)
We know that \( (A \cap B) \cap A = (A \cap B) \).
So, \( P(A \cap B / A) = \frac{P(A \cap B)}{P(A)} \)
\( \implies P(A \cap B / A) = \frac{0.5}{0.8} \)
\( \implies P(A \cap B / A) = \frac{5}{8} \)
Conditional probability can reveal dependencies between events, which is very useful in risk assessment.
In simple words: We used different rules to find many probabilities based on the given information. We calculated the chance of B, B given A, A or B, A and B given A or B, A and B given B not happening, A or B given A, and finally, A and B given A.

🎯 Exam Tip: When dealing with multiple conditional probabilities and set operations, break down each part and clearly identify the intersection needed for the numerator and the conditioning event for the denominator.

Question 6. A die is rolled. If the outcome is an odd number, what is the probability that it is prime?
Answer: We are looking for the conditional probability that a number is prime, given that it is odd, when a die is rolled.
When a die is rolled, the sample space S (all possible outcomes) is:
\( S = \{1, 2, 3, 4, 5, 6\} \)
The total number of outcomes is \( n(S) = 6 \).
Let A be the event that the outcome is an odd number:
\( A = \{1, 3, 5\} \)
The number of outcomes in A is \( n(A) = 3 \).
Let B be the event that the outcome is a prime number (prime numbers are greater than 1 and only divisible by 1 and themselves):
\( B = \{2, 3, 5\} \)
The number of outcomes in B is \( n(B) = 3 \).
We need to find the intersection of A and B, which means numbers that are both odd and prime:
\( A \cap B = \{3, 5\} \)
The number of outcomes in \( A \cap B \) is \( n(A \cap B) = 2 \).
Now we calculate the probabilities:
\( P(A) = \frac{n(A)}{n(S)} = \frac{3}{6} = \frac{1}{2} \)
\( P(B) = \frac{n(B)}{n(S)} = \frac{3}{6} = \frac{1}{2} \)
\( P(A \cap B) = \frac{n(A \cap B)}{n(S)} = \frac{2}{6} = \frac{1}{3} \)
We want to find the probability that the number is prime given it's an odd number, which is \( P(B/A) \):
\( P(B/A) = \frac{P(A \cap B)}{P(A)} \)
\( \implies P(B/A) = \frac{\frac{1}{3}}{\frac{1}{2}} \)
\( \implies P(B/A) = \frac{1}{3} \times \frac{2}{1} \)
\( \implies P(B/A) = \frac{2}{3} \)
This shows that knowing the number is odd changes our expectation of it being prime from \( 1/2 \) to \( 2/3 \).
In simple words: We rolled a die. We only cared about odd numbers. Out of those odd numbers (1, 3, 5), we checked how many were prime (2, 3, 5). The numbers that were both odd and prime were 3 and 5. So, the chance of getting a prime number, knowing it was odd, is 2 out of 3.

🎯 Exam Tip: Clearly list the sample space, events A and B, and their intersection to avoid errors in counting outcomes for conditional probability problems.

Question 7. A die is thrown twice and the sum of the numbers appearing is observed to be 6. What is the conditional probability that the number 4 has appeared at least once?
Answer: We are throwing a die twice and checking the sum, then looking for the presence of the number 4.
When a die is thrown twice, the total number of outcomes is \( 6 \times 6 = 36 \). So, \( n(S) = 36 \).
Let A be the event that the sum of the numbers appearing is 6.
The pairs that sum to 6 are: \( A = \{(1, 5), (5, 1), (2, 4), (4, 2), (3, 3)\} \).
The number of outcomes in A is \( n(A) = 5 \).
Let B be the event that the number 4 has appeared at least once.
The pairs where 4 appears at least once are:
\( B = \{(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (1, 4), (2, 4), (3, 4), (5, 4), (6, 4)\} \).
The number of outcomes in B is \( n(B) = 11 \).
Next, we find the intersection \( A \cap B \), which means the sum is 6 AND 4 appears at least once:
\( A \cap B = \{(2, 4), (4, 2)\} \)
The number of outcomes in \( A \cap B \) is \( n(A \cap B) = 2 \).
Now we can calculate the probabilities needed for conditional probability:
\( P(A \cap B) = \frac{n(A \cap B)}{n(S)} = \frac{2}{36} \)
\( P(A) = \frac{n(A)}{n(S)} = \frac{5}{36} \)
We want to find the conditional probability that 4 has appeared at least once, given that the sum is 6. This is \( P(B/A) \).
\( P(B/A) = \frac{P(A \cap B)}{P(A)} \)
\( \implies P(B/A) = \frac{\frac{2}{36}}{\frac{5}{36}} \)
\( \implies P(B/A) = \frac{2}{5} \)
Conditional probability helps us narrow down the possibilities based on known information.
In simple words: We rolled two dice. We only looked at results where the numbers added up to 6. Out of those results, we wanted to know the chance that a '4' showed up at least once. We found there were 5 ways to get a sum of 6, and 2 of those ways included a '4'. So, the chance is 2 out of 5.

🎯 Exam Tip: When dealing with two dice, carefully list all possible outcomes for the given conditions (sum, presence of a number) to accurately find the intersection and total events.

Question 8. Two dice are thrown. Find the probability that numbers appeared have a sum 8, if it is known, that the second die always exhibits 4.
Answer: We are throwing two dice. We are given that the second die is always a 4, and we want to find the probability that the sum is 8.
When two dice are thrown, the total number of outcomes is \( n(S) = 6^2 = 36 \).
Let B be the event that the second die always exhibits 4.
The outcomes where the second die is 4 are: \( B = \{(1, 4), (2, 4), (3, 4), (4, 4), (5, 4), (6, 4)\} \).
The number of outcomes in B is \( n(B) = 6 \).
Let A be the event that the numbers appeared have a sum of 8.
The outcomes that sum to 8 are: \( A = \{(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)\} \).
The number of outcomes in A is \( n(A) = 5 \).
Now, we find the intersection \( A \cap B \), which means the sum is 8 AND the second die is 4.
\( A \cap B = \{(4, 4)\} \)
The number of outcomes in \( A \cap B \) is \( n(A \cap B) = 1 \).
We need to find the conditional probability of event A (sum is 8) given event B (second die is 4), which is \( P(A/B) \).
\( P(A/B) = \frac{P(A \cap B)}{P(B)} \)
To use this formula, we need the probabilities \( P(A \cap B) \) and \( P(B) \):
\( P(A \cap B) = \frac{n(A \cap B)}{n(S)} = \frac{1}{36} \)
\( P(B) = \frac{n(B)}{n(S)} = \frac{6}{36} \)
Now, substitute these into the conditional probability formula:
\( P(A/B) = \frac{\frac{1}{36}}{\frac{6}{36}} \)
\( \implies P(A/B) = \frac{1}{6} \)
This means if we already know the second die is a 4, there's only one outcome (4,4) that gives a sum of 8 among the 6 possibilities for the second die being 4.
In simple words: We rolled two dice, but we already knew the second die was a 4. So, we only looked at pairs like (1,4), (2,4), (3,4), (4,4), (5,4), (6,4). Out of these 6 pairs, only one pair, (4,4), adds up to 8. So, the chance of getting a sum of 8, given the second die is 4, is 1 out of 6.

🎯 Exam Tip: In conditional probability problems where one event is "known," make sure to reduce your sample space to only include outcomes where the known event has occurred.

Question 9. Assume that each born child is equally likely to be a boy or a girl. If a family has two children then what is the conditional probability that both are girls ? Given that (i) the youngest is a girl? (ii) at least one is a girl?
Answer: We are looking at probabilities for a family with two children, where each child can be a boy (B) or a girl (G).
The sample space S for two children (order matters, e.g., oldest first) is:
\( S = \{BB, BG, GB, GG\} \)
The total number of outcomes is \( n(S) = 4 \).
Let A be the event that both children are girls:
\( A = \{GG\} \)
The number of outcomes in A is \( n(A) = 1 \).

(i) Conditional probability that both are girls, given the youngest is a girl.
Let B be the event that the youngest child is a girl.
If we assume the first letter is the oldest and the second is the youngest, then:
\( B = \{BG, GG\} \)
The number of outcomes in B is \( n(B) = 2 \).
The intersection \( A \cap B \) (both are girls AND youngest is a girl) is:
\( A \cap B = \{GG\} \)
The number of outcomes in \( A \cap B \) is \( n(A \cap B) = 1 \).
Now, we calculate the probabilities:
\( P(A \cap B) = \frac{n(A \cap B)}{n(S)} = \frac{1}{4} \)
\( P(B) = \frac{n(B)}{n(S)} = \frac{2}{4} = \frac{1}{2} \)
The required conditional probability \( P(A/B) \) is:
\( P(A/B) = \frac{P(A \cap B)}{P(B)} \)
\( \implies P(A/B) = \frac{\frac{1}{4}}{\frac{1}{2}} \)
\( \implies P(A/B) = \frac{1}{4} \times 2 \)
\( \implies P(A/B) = \frac{1}{2} \)

(ii) Conditional probability that both are girls, given at least one is a girl.
Let C be the event that at least one child is a girl.
\( C = \{BG, GB, GG\} \)
The number of outcomes in C is \( n(C) = 3 \).
The intersection \( A \cap C \) (both are girls AND at least one is a girl) is:
\( A \cap C = \{GG\} \)
The number of outcomes in \( A \cap C \) is \( n(A \cap C) = 1 \).
Now, we calculate the probabilities:
\( P(A \cap C) = \frac{n(A \cap C)}{n(S)} = \frac{1}{4} \)
\( P(C) = \frac{n(C)}{n(S)} = \frac{3}{4} \)
The required conditional probability \( P(A/C) \) is:
\( P(A/C) = \frac{P(A \cap C)}{P(C)} \)
\( \implies P(A/C) = \frac{\frac{1}{4}}{\frac{3}{4}} \)
\( \implies P(A/C) = \frac{1}{3} \)
These results show how different pieces of information can change the probability of an event.
In simple words: For a family with two children, we want to know the chance they are both girls. In the first case, we knew the youngest was a girl, so the chance became 1 out of 2. In the second case, we knew at least one was a girl, and the chance became 1 out of 3.

🎯 Exam Tip: When dealing with genetic or family-based probability, clearly define your sample space and events, paying attention to whether the order of children (e.g., oldest/youngest) matters for the definition of the event.

Question 10. A coin is tossed and if the coin shows head it is tossed again but if it shows a tail then a die is tossed. If 8 possible outcomes are equally likely, find the probability that the die shows a number greater than 4 if it is known that the first throw of the coin results in a tail.
Answer: We have a two-stage experiment with different actions based on the first coin toss.
The given sample space S (all equally likely outcomes) is:
\( S = \{HH, HT, T1, T2, T3, T4, T5, T6\} \)
The total number of outcomes is \( n(S) = 8 \).
Let A be the event that the first throw of the coin results in a tail.
\( A = \{T1, T2, T3, T4, T5, T6\} \)
The number of outcomes in A is \( n(A) = 6 \).
Let B be the event that the die shows a number greater than 4.
The outcomes where the die shows a number greater than 4 are: \( B = \{T5, T6\} \)
The number of outcomes in B is \( n(B) = 2 \).
We need to find the intersection \( A \cap B \), which means the first throw is a tail AND the die shows a number greater than 4.
\( A \cap B = \{T5, T6\} \)
The number of outcomes in \( A \cap B \) is \( n(A \cap B) = 2 \).
Now, we calculate the probabilities:
\( P(A) = \frac{n(A)}{n(S)} = \frac{6}{8} \)
\( P(A \cap B) = \frac{n(A \cap B)}{n(S)} = \frac{2}{8} \)
We want to find the conditional probability that the die shows a number greater than 4, given that the first throw of the coin results in a tail. This is \( P(B/A) \).
\( P(B/A) = \frac{P(A \cap B)}{P(A)} \)
\( \implies P(B/A) = \frac{\frac{2}{8}}{\frac{6}{8}} \)
\( \implies P(B/A) = \frac{2}{6} \)
\( \implies P(B/A) = \frac{1}{3} \)
This situation highlights how the sample space effectively shrinks when a condition is known.
In simple words: We had a coin toss that led to either another coin toss or a die roll. We knew for sure that the first coin toss was a tail. So, we only looked at the die roll outcomes (T1, T2, T3, T4, T5, T6). Out of these 6 possibilities, only T5 and T6 show a number greater than 4. So, the chance is 2 out of 6, which simplifies to 1 out of 3.

🎯 Exam Tip: When the experiment has multiple stages or conditions, always identify the *actual* sample space for the conditional event. Here, "first throw is a tail" changed the sample space from 8 total outcomes to 6 specific outcomes related to the die.

Question 11. In a class 40% students read Statistics, 25% Mathematics and 15% both Mathematics and Statistics. One student is selected at random. Find the probability:
(i) that he reads Statistics, if it is known that he reads Mathematics.
(ii) that he reads Mathematics, if it is known that he reads Statistics.
Answer: We are given percentages of students studying Statistics and Mathematics, including those studying both.
Let S be the event that a student reads Statistics.
Let M be the event that a student reads Mathematics.
From the given information:
\( P(S) = 40\% = \frac{40}{100} = 0.40 \)
\( P(M) = 25\% = \frac{25}{100} = 0.25 \)
\( P(S \cap M) = 15\% = \frac{15}{100} = 0.15 \)

(i) Find the probability that a student reads Statistics, if it is known that he reads Mathematics. This is \( P(S/M) \).
Using the conditional probability formula:
\( P(S/M) = \frac{P(S \cap M)}{P(M)} \)
\( \implies P(S/M) = \frac{0.15}{0.25} \)
\( \implies P(S/M) = \frac{15}{25} \)
\( \implies P(S/M) = \frac{3}{5} \)

(ii) Find the probability that a student reads Mathematics, if it is known that he reads Statistics. This is \( P(M/S) \).
Using the conditional probability formula:
\( P(M/S) = \frac{P(S \cap M)}{P(S)} \)
\( \implies P(M/S) = \frac{0.15}{0.40} \)
\( \implies P(M/S) = \frac{15}{40} \)
\( \implies P(M/S) = \frac{3}{8} \)
Conditional probabilities are often used in educational settings to understand subject choices and performance.
In simple words: We know the chances of students studying Statistics, Math, or both. For the first part, we found the chance a student studies Statistics, knowing they already study Math. For the second part, we found the chance a student studies Math, knowing they already study Statistics.

🎯 Exam Tip: Always clearly define your events (S and M in this case) and write down the given probabilities as fractions or decimals before applying the conditional probability formula.

Question 12. In a certain school, 20% students failed in English, 15% students failed in Mathematics and 10% students failed in both English and Mathematics. A student is selected at random. If he failed in English, what is the probability that he also failed in Mathematics?
Answer: We are given failure rates in English and Mathematics.
Let E be the event that a student failed in English.
Let M be the event that a student failed in Mathematics.
From the given information:
\( P(E) = 20\% = \frac{20}{100} = 0.20 \)
\( P(M) = 15\% = \frac{15}{100} = 0.15 \)
\( P(E \cap M) = 10\% = \frac{10}{100} = 0.10 \)
We want to find the probability that a student also failed in Mathematics, given that he failed in English. This is \( P(M/E) \).
Using the conditional probability formula:
\( P(M/E) = \frac{P(E \cap M)}{P(E)} \)
\( \implies P(M/E) = \frac{0.10}{0.20} \)
\( \implies P(M/E) = \frac{1}{2} \)
This type of analysis can help schools identify areas where students might struggle in multiple subjects.
In simple words: We know how many students failed English, Math, and both. If a student failed English, we want to know the chance they also failed Math. We used a rule that says we divide the chance of failing both by the chance of failing English.

🎯 Exam Tip: Remember that "failed in both" means the intersection of the two events (E \( \cap \) M), which is crucial for the numerator in conditional probability.

Question 13. One card is drawn from a well-shuffled pack of 52 cards. If E is the event “the card drawn is a king or an ace” and F is the event “the card drawn is an ace or a jack”, then find the probability of the conditional event (E/F).
Answer: We are drawing one card from a standard deck and defining two events based on its face value.
The total number of cards in a pack is 52. So, \( n(S) = 52 \).
Let E be the event that the card drawn is a king or an ace.
There are 4 kings and 4 aces in a deck.
So, \( n(E) = 4 (\text{kings}) + 4 (\text{aces}) = 8 \).
\( P(E) = \frac{n(E)}{n(S)} = \frac{8}{52} = \frac{2}{13} \)
Let F be the event that the card drawn is an ace or a jack.
There are 4 aces and 4 jacks in a deck.
So, \( n(F) = 4 (\text{aces}) + 4 (\text{jacks}) = 8 \).
\( P(F) = \frac{n(F)}{n(S)} = \frac{8}{52} = \frac{2}{13} \)
Now, we need to find the intersection \( E \cap F \), which means the card is (king or ace) AND (ace or jack). This means the card must be an ace.
There are 4 aces in a deck.
So, \( n(E \cap F) = 4 \).
\( P(E \cap F) = \frac{n(E \cap F)}{n(S)} = \frac{4}{52} = \frac{1}{13} \)
We want to find the conditional probability \( P(E/F) \).
Using the conditional probability formula:
\( P(E/F) = \frac{P(E \cap F)}{P(F)} \)
\( \implies P(E/F) = \frac{\frac{1}{13}}{\frac{2}{13}} \)
\( \implies P(E/F) = \frac{1}{2} \)
Understanding overlaps between events is crucial in probability, especially with playing cards.
In simple words: We picked one card. Event E was getting a King or an Ace. Event F was getting an Ace or a Jack. We first found the chance of E, F, and both happening (which meant getting an Ace). Then we calculated the chance of E happening if F already happened.

🎯 Exam Tip: When defining events involving card types, ensure you count each distinct type only once for non-overlapping categories (like kings, aces, jacks) and correctly identify the shared outcomes for the intersection.

Question 14. Two coins are tossed once, where
(i) E: tail appears on one coin F: one coin shows head
(ii) E; no tail appears F: no head appears. Find P(E/F).
Answer: We are tossing two coins and defining events based on the outcomes.
The sample space S for tossing two coins once is:
\( S = \{HH, HT, TH, TT\} \)
The total number of outcomes is \( n(S) = 4 \).

(i) For E: tail appears on one coin and F: one coin shows head.
Let E be the event that a tail appears on exactly one coin.
\( E = \{HT, TH\} \)
The number of outcomes in E is \( n(E) = 2 \).
Let F be the event that one coin shows a head (meaning exactly one head).
\( F = \{HT, TH\} \)
The number of outcomes in F is \( n(F) = 2 \).
The intersection \( E \cap F \) (a tail appears on one coin AND one coin shows a head) is:
\( E \cap F = \{HT, TH\} \)
The number of outcomes in \( E \cap F \) is \( n(E \cap F) = 2 \).
Now, we calculate the probabilities:
\( P(E) = \frac{n(E)}{n(S)} = \frac{2}{4} \)
\( P(F) = \frac{n(F)}{n(S)} = \frac{2}{4} \)
\( P(E \cap F) = \frac{n(E \cap F)}{n(S)} = \frac{2}{4} \)
We want to find the conditional probability \( P(E/F) \):
\( P(E/F) = \frac{P(E \cap F)}{P(F)} \)
\( \implies P(E/F) = \frac{\frac{2}{4}}{\frac{2}{4}} \)
\( \implies P(E/F) = 1 \)

(ii) For E: no tail appears and F: no head appears.
Let E be the event that no tail appears. This means both are heads.
\( E = \{HH\} \)
The number of outcomes in E is \( n(E) = 1 \).
Let F be the event that no head appears. This means both are tails.
\( F = \{TT\} \)
The number of outcomes in F is \( n(F) = 1 \).
The intersection \( E \cap F \) (no tail AND no head) is impossible, as the outcome must have either tails or heads.
\( E \cap F = \emptyset \)
The number of outcomes in \( E \cap F \) is \( n(E \cap F) = 0 \).
Now, we calculate the probabilities:
\( P(E \cap F) = \frac{n(E \cap F)}{n(S)} = \frac{0}{4} = 0 \)
\( P(F) = \frac{n(F)}{n(S)} = \frac{1}{4} \)
We want to find the conditional probability \( P(E/F) \):
\( P(E/F) = \frac{P(E \cap F)}{P(F)} \)
\( \implies P(E/F) = \frac{0}{\frac{1}{4}} \)
\( \implies P(E/F) = 0 \)
These examples illustrate how specific conditions can drastically change event probabilities, even making them impossible.
In simple words: We tossed two coins. In the first case, we looked at the chance of getting one tail, knowing one coin showed a head. This turned out to be 1, meaning it's certain. In the second case, we looked at the chance of no tails, knowing no heads appeared. This was impossible, so the chance was 0.

🎯 Exam Tip: Pay close attention to the wording "on one coin" versus "at least one coin" or "no coin" when defining events, as it affects the outcomes included in your sets.

Question 15. A black and a red die are rolled.
(i) Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5.
(ii) Find the conditional probability of obtaining sum 8, given that the red die resulted in a number less than 4.
Answer: We are rolling two distinct dice, a black one and a red one. The total number of outcomes is \( n(S) = 6 \times 6 = 36 \).

(i) Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5.
Let E be the event of obtaining a sum greater than 9.
The pairs that sum to more than 9 are: \( E = \{(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)\} \).
The number of outcomes in E is \( n(E) = 6 \).
Let F be the event that the black die resulted in a 5.
Assuming the first number in the pair is from the black die:
\( F = \{(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)\} \).
The number of outcomes in F is \( n(F) = 6 \).
The intersection \( E \cap F \) (sum > 9 AND black die is 5) is:
\( E \cap F = \{(5, 5), (5, 6)\} \)
The number of outcomes in \( E \cap F \) is \( n(E \cap F) = 2 \).
Now, we calculate the probabilities:
\( P(E \cap F) = \frac{n(E \cap F)}{n(S)} = \frac{2}{36} \)
\( P(F) = \frac{n(F)}{n(S)} = \frac{6}{36} \)
The required conditional probability \( P(E/F) \) is:
\( P(E/F) = \frac{P(E \cap F)}{P(F)} \)
\( \implies P(E/F) = \frac{\frac{2}{36}}{\frac{6}{36}} \)
\( \implies P(E/F) = \frac{2}{6} \)
\( \implies P(E/F) = \frac{1}{3} \)

(ii) Find the conditional probability of obtaining sum 8, given that the red die resulted in a number less than 4.
Let G be the event of obtaining a sum of 8.
The pairs that sum to 8 are: \( G = \{(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)\} \).
The number of outcomes in G is \( n(G) = 5 \).
Let H be the event that the red die resulted in a number less than 4.
Assuming the second number in the pair is from the red die, and numbers less than 4 are {1, 2, 3}:
\( H = \{(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (5, 3), (6, 1), (6, 2), (6, 3)\} \).
The number of outcomes in H is \( n(H) = 6 \times 3 = 18 \).
The intersection \( G \cap H \) (sum = 8 AND red die < 4) is:
From G, find pairs where the second number is 1, 2, or 3:
\( G \cap H = \{(5, 3), (6, 2)\} \).
The number of outcomes in \( G \cap H \) is \( n(G \cap H) = 2 \).
Now, we calculate the probabilities:
\( P(G \cap H) = \frac{n(G \cap H)}{n(S)} = \frac{2}{36} \)
\( P(H) = \frac{n(H)}{n(S)} = \frac{18}{36} \)
The required conditional probability \( P(G/H) \) is:
\( P(G/H) = \frac{P(G \cap H)}{P(H)} \)
\( \implies P(G/H) = \frac{\frac{2}{36}}{\frac{18}{36}} \)
\( \implies P(G/H) = \frac{2}{18} \)
\( \implies P(G/H) = \frac{1}{9} \)
Rolling two distinguishable dice, like a black and a red one, makes outcomes like (1,2) and (2,1) distinct.
In simple words: We rolled two dice, one black and one red. For the first part, we wanted the chance of getting a sum more than 9, knowing the black die was a 5. This was 1 out of 3. For the second part, we wanted the chance of getting a sum of 8, knowing the red die was less than 4. This was 1 out of 9.

🎯 Exam Tip: When dice are distinguishable (e.g., black and red), always treat (x,y) as distinct from (y,x) if x \( \neq \) y. This affects the counting of outcomes for your events and their intersections.

Question 16. Given that the two numbers appearing on throwing two dice are different, find the probability of the events the sum of numbers on the dice is 4.
Answer: We are throwing two dice, and we are given a condition about the outcomes.
The total number of outcomes when throwing two dice is \( n(S) = 6 \times 6 = 36 \).
Let E be the event that the sum of the numbers on the dice is 4.
The pairs that sum to 4 are: \( E = \{(1, 3), (2, 2), (3, 1)\} \).
The number of outcomes in E is \( n(E) = 3 \).
Let F be the event that the two numbers appearing on the dice are different.
The total outcomes are 36. The outcomes where the numbers are the same (doubles) are: \( \{(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)\} \). There are 6 such outcomes.
So, the number of outcomes where the numbers are different is \( n(F) = 36 - 6 = 30 \).
The intersection \( E \cap F \) (sum is 4 AND numbers are different) is:
From E, we exclude the pair where numbers are the same, which is (2, 2).
\( E \cap F = \{(1, 3), (3, 1)\} \).
The number of outcomes in \( E \cap F \) is \( n(E \cap F) = 2 \).
Now, we calculate the probabilities:
\( P(E \cap F) = \frac{n(E \cap F)}{n(S)} = \frac{2}{36} \)
\( P(F) = \frac{n(F)}{n(S)} = \frac{30}{36} \)
We want to find the conditional probability \( P(E/F) \).
Using the conditional probability formula:
\( P(E/F) = \frac{P(E \cap F)}{P(F)} \)
\( \implies P(E/F) = \frac{\frac{2}{36}}{\frac{30}{36}} \)
\( \implies P(E/F) = \frac{2}{30} \)
\( \implies P(E/F) = \frac{1}{15} \)
This problem demonstrates how to adjust probabilities when part of the sample space is excluded by a condition.
In simple words: We rolled two dice. We only looked at results where the two numbers were different. Out of these specific results, we wanted to find the chance that the numbers added up to 4. We found there were 30 ways to get different numbers, and only 2 of those ways also added up to 4. So, the chance is 2 out of 30, which is 1 out of 15.

🎯 Exam Tip: When a condition like "numbers are different" is given, it's often easier to first find the total outcomes and subtract the unwanted outcomes (like doubles) to determine the size of the new sample space.

Question 17. Mother, father and son line up at random for a family picture.
E: son on one end
F: father in the middle
Find P{E/F)
Answer: We are arranging three family members (Mother, Father, Son) in a line for a picture.
Let M = Mother, F = Father, S = Son.
The total number of ways to arrange 3 people is \( 3! = 3 \times 2 \times 1 = 6 \).
The sample space S (all possible arrangements) is:
\( S = \{MFS, MSF, FMS, FSM, SMF, SFM\} \)
The total number of outcomes is \( n(S) = 6 \).
Let E be the event that the son is on one end.
The arrangements where the son is on one end (first or last position) are:
\( E = \{SMF, SFM, MFS, FMS\} \)
The number of outcomes in E is \( n(E) = 4 \).
Let F be the event that the father is in the middle.
The arrangements where the father is in the middle are:
\( F = \{MFS, SFM\} \)
The number of outcomes in F is \( n(F) = 2 \).
The intersection \( E \cap F \) (son on one end AND father in the middle) is:
\( E \cap F = \{MFS, SFM\} \)
The number of outcomes in \( E \cap F \) is \( n(E \cap F) = 2 \).
Now, we calculate the probabilities:
\( P(E \cap F) = \frac{n(E \cap F)}{n(S)} = \frac{2}{6} \)
\( P(F) = \frac{n(F)}{n(S)} = \frac{2}{6} \)
We want to find the conditional probability \( P(E/F) \).
Using the conditional probability formula:
\( P(E/F) = \frac{P(E \cap F)}{P(F)} \)
\( \implies P(E/F) = \frac{\frac{2}{6}}{\frac{2}{6}} \)
\( \implies P(E/F) = 1 \)
This outcome makes sense because if the father is in the middle, then the son *must* be on one of the ends (since there are only two other positions), making event E certain.
In simple words: We lined up three family members. We wanted the chance that the son was at one end, knowing that the father was in the middle. If the father is in the middle, the son and mother must be on the ends. So, the son is always at an end in this situation, making the chance 100%.

🎯 Exam Tip: For small sample spaces involving arrangements, listing all permutations can be helpful to accurately identify events and their intersections.

Question 18. Consider the experiment of throwing a die, if a multiple of 3 comes up throw the die again and if any other number comes, toss a coin. Find the conditional probability of the event the coin shows a tail, given that at least one die shows a 3.
Answer: This is a multi-stage experiment depending on the first die roll.
The sample space S includes outcomes from two scenarios:
1. If a multiple of 3 (3 or 6) comes up on the first die, throw the die again: \( \{(3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\} \). (12 outcomes)
2. If any other number (1, 2, 4, 5) comes up, toss a coin (H or T): \( \{(1,H), (1,T), (2,H), (2,T), (4,H), (4,T), (5,H), (5,T)\} \). (8 outcomes)
The total number of outcomes in S is \( n(S) = 12 + 8 = 20 \).
Let E be the event that the coin shows a tail.
The outcomes where the coin shows a tail are: \( E = \{(1,T), (2,T), (4,T), (5,T)\} \).
The number of outcomes in E is \( n(E) = 4 \).
Let F be the event that at least one die shows a 3.
The outcomes where at least one die shows a 3 are:
From scenario 1 (die-die): \( \{(3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (6,3)\} \). (7 outcomes)
From scenario 2 (die-coin): No die shows a 3 (because if a 3 came up first, it would be scenario 1).
So, \( F = \{(3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (6,3)\} \).
The number of outcomes in F is \( n(F) = 7 \).
The intersection \( E \cap F \) (coin shows a tail AND at least one die shows a 3).
For a coin to show a tail, the first die roll must be 1, 2, 4, or 5 (scenario 2).
However, if the first die roll is a 3, it goes to scenario 1 (die-die), where no coin is tossed.
Therefore, it is impossible for both events E and F to happen at the same time.
\( E \cap F = \emptyset \)
The number of outcomes in \( E \cap F \) is \( n(E \cap F) = 0 \).
Now, we calculate the probabilities:
\( P(E \cap F) = \frac{n(E \cap F)}{n(S)} = \frac{0}{20} = 0 \)
\( P(F) = \frac{n(F)}{n(S)} = \frac{7}{20} \)
We want to find the conditional probability \( P(E/F) \).
Using the conditional probability formula:
\( P(E/F) = \frac{P(E \cap F)}{P(F)} \)
\( \implies P(E/F) = \frac{0}{\frac{7}{20}} \)
\( \implies P(E/F) = 0 \)
This problem demonstrates how mutually exclusive conditions in a multi-stage experiment can lead to a zero probability.
In simple words: We had a two-step game: roll a die, then maybe roll it again or flip a coin. We wanted the chance that the coin showed a tail, knowing that at least one die showed a 3. If a die showed a 3, we would roll the die again, not flip a coin. So, it's impossible for a coin to show a tail if a die shows a 3. The chance is 0.

🎯 Exam Tip: In multi-stage experiments, carefully map out all possible outcomes and how the different stages interact. An event in one stage might make an event in another stage impossible.

Question 19. One ticket is selected at random from 50 tickets numbered 00, 01, 02,..., 49. Then, the probability that the sum of the digits on the selected ticket is 8, given that the product of these digits is zero, equals
(a) \( \frac { 1 }{7} \)
(b) \( \frac { 1 }{ 14 } \)
(c) \( \frac { 1 }{ 25 } \)
(d) \( \frac { 1 }{ 50 } \)
Answer: (b) \( \frac { 1 }{ 14 } \)
The sample space S consists of 50 tickets numbered from 00 to 49. So, \( n(S) = 50 \).
Let A be the event that the sum of the digits on the selected ticket is 8.
The tickets where the sum of digits is 8 are: \( A = \{08, 17, 26, 35, 44\} \).
The number of outcomes in A is \( n(A) = 5 \).
Let B be the event that the product of these digits is zero.
For the product of digits to be zero, at least one digit must be zero. This applies to numbers like 00, 01, 02, ..., 09 (where the tens digit is 0) and 10, 20, 30, 40 (where the units digit is 0).
\( B = \{00, 01, 02, 03, 04, 05, 06, 07, 08, 09, 10, 20, 30, 40\} \).
The number of outcomes in B is \( n(B) = 10 + 4 = 14 \).
Now, we find the intersection \( A \cap B \), which means the sum of digits is 8 AND the product of digits is zero.
From event A, the only ticket where a digit is zero is 08.
\( A \cap B = \{08\} \).
The number of outcomes in \( A \cap B \) is \( n(A \cap B) = 1 \).
Now, we calculate the probabilities:
\( P(A \cap B) = \frac{n(A \cap B)}{n(S)} = \frac{1}{50} \)
\( P(B) = \frac{n(B)}{n(S)} = \frac{14}{50} \)
We want to find the conditional probability \( P(A/B) \).
\( P(A/B) = \frac{P(A \cap B)}{P(B)} \)
\( \implies P(A/B) = \frac{\frac{1}{50}}{\frac{14}{50}} \)
\( \implies P(A/B) = \frac{1}{14} \)
This type of number theory combined with probability is common in competitive exams.
In simple words: We picked a ticket from 00 to 49. We wanted the chance that its digits added up to 8, knowing that the digits multiplied to zero. Only one ticket (08) fits both conditions out of the 14 tickets whose digits multiply to zero. So the chance is 1 out of 14.

🎯 Exam Tip: For problems involving digits, systematically list out the outcomes for each event (sum, product) to correctly identify the intersection and the conditioning event's sample space.

Question 20. If A and B are events such that P(A/B) = P(B/A), then
(a) A \( \subset \) B but A \( \neq \) B
(b) A = B
(c) A \( \cap \) B = \( \phi \)
(d) P(A) = P(B)
Answer: (d) P(A) = P(B)
Given that \( P(A/B) = P(B/A) \).
We know the formula for conditional probability:
\( P(A/B) = \frac{P(A \cap B)}{P(B)} \)
And \( P(B/A) = \frac{P(B \cap A)}{P(A)} \)
Since \( P(A \cap B) = P(B \cap A) \), let's call this \( P(A \cap B) \).
Substituting these into the given equality:
\( \frac{P(A \cap B)}{P(B)} = \frac{P(A \cap B)}{P(A)} \)
Now, we can cross-multiply (assuming \( P(A) \neq 0 \) and \( P(B) \neq 0 \), which must be true for the conditional probabilities to be defined):
\( P(A) \times P(A \cap B) = P(B) \times P(A \cap B) \)
If \( P(A \cap B) \neq 0 \), we can divide both sides by \( P(A \cap B) \):
\( P(A) = P(B) \)
If \( P(A \cap B) = 0 \), then the condition \( P(A/B) = P(B/A) \) means \( \frac{0}{P(B)} = \frac{0}{P(A)} \), which is \( 0 = 0 \). In this case, A and B are mutually exclusive. It does not automatically imply \( P(A) = P(B) \) unless both \( P(A) \) and \( P(B) \) are also zero. However, in most probability contexts, events with non-zero probabilities are considered. For the equality to hold for general events where \( P(A \cap B) \) might be non-zero, \( P(A) \) must equal \( P(B) \).
For example, if \( P(A) = 0.5 \), \( P(B) = 0.5 \), and \( P(A \cap B) = 0.2 \), then \( P(A/B) = 0.2/0.5 = 0.4 \) and \( P(B/A) = 0.2/0.5 = 0.4 \). Here \( P(A) = P(B) \).
This result is a key property that relates conditional probabilities to the marginal probabilities of events.
In simple words: If the chance of A happening given B already happened is the same as the chance of B happening given A already happened, then the overall chance of A must be equal to the overall chance of B.

🎯 Exam Tip: Always start by writing down the definitions of the conditional probabilities \( P(A/B) \) and \( P(B/A) \) and then use algebraic manipulation to find the relationship between \( P(A) \) and \( P(B) \).