RBSE Solutions Class 11 Maths Chapter 14 Probability Exercise 14.3

Question 1. Probability of event A is \( \frac{2}{11} \), find the probability of event 'A-not'.
Answer: The probability of event A is given as \( P(A) = \frac{2}{11} \). We need to find the probability of event 'A-not', which is written as \( P(\overline{A}) \). The sum of the probability of an event and the probability of its complement is always 1.
\( P(\overline{A}) = 1 - P(A) \)
\( P(\overline{A}) = 1 - \frac{2}{11} \)
To subtract, we find a common denominator:
\( P(\overline{A}) = \frac{11}{11} - \frac{2}{11} \)
\( P(\overline{A}) = \frac{11 - 2}{11} \)
\( P(\overline{A}) = \frac{9}{11} \)
Thus, the required probability for 'A-not' is \( \frac{9}{11} \). This is a basic rule in probability.
In simple words: If you know the chance of something happening, the chance of it *not* happening is simply 1 minus that chance. Here, if the chance of A is \( \frac{2}{11} \), the chance of not A is \( \frac{9}{11} \).

🎯 Exam Tip: Remember that \( P(E) + P(\overline{E}) = 1 \) for any event E. This formula is fundamental for calculating complementary probabilities.

Question 2. If there are 6 female members and 4 male members in a village panel, and one member for a committee is selected, what is the probability of selecting a female?
Answer: Total number of female members = 6. Total number of male members = 4. The total number of members available for selection is the sum of female and male members, which is \( 6 + 4 = 10 \). We want to find the probability of selecting one female member.
The number of ways to select 1 female member from 6 is \( ^6C_1 = 6 \).
The total number of ways to select 1 member from 10 is \( ^{10}C_1 = 10 \).
The probability of selecting a female is the number of favorable outcomes divided by the total number of outcomes.
\( P(\text{female}) = \frac{\text{Number of female members}}{\text{Total number of members}} = \frac{6}{10} = \frac{3}{5} \)
Thus, the probability of selecting a female is \( \frac{3}{5} \). This basic calculation shows the likelihood of a specific outcome.
In simple words: There are 6 women and 4 men, making 10 people in total. If you pick one person, the chance it's a woman is 6 out of 10, which simplifies to 3 out of 5.

🎯 Exam Tip: When selecting a single item, the probability is simply the count of the desired item divided by the total count. Always simplify fractions to their lowest terms.

Question 3. In throwing a dice find the probability of following events.
(i) Appear a prime number
(ii) appear 1 or less than 1
(iii) appear number less than 6.
Answer: When a fair die is thrown, the possible outcomes form the sample space (S). The total number of outcomes is \( n(S) \).
The sample space is \( S = \{1, 2, 3, 4, 5, 6\} \). So, \( n(S) = 6 \).

(i) Let E be the event of appearing a prime number. Prime numbers are those greater than 1 that have only two divisors: 1 and themselves. The prime numbers in our sample space are 2, 3, and 5.
\( E = \{2, 3, 5\} \)
The number of outcomes in event E is \( n(E) = 3 \).
The probability of event E is \( P(E) = \frac{n(E)}{n(S)} \)
\( P(E) = \frac{3}{6} = \frac{1}{2} \)
Thus, the probability to get a prime number is \( \frac{1}{2} \).

(ii) Let B be the event of getting a number 1 or less than 1. On a standard die, the only number that is 1 or less than 1 is 1 itself.
\( B = \{1\} \)
The number of outcomes in event B is \( n(B) = 1 \).
The probability of event B is \( P(B) = \frac{n(B)}{n(S)} \)
\( P(B) = \frac{1}{6} \)
Thus, the probability to get a number 1 or less than 1 is \( \frac{1}{6} \).

(iii) Let D be the event of appearing a number less than 6. The numbers in our sample space that are less than 6 are 1, 2, 3, 4, and 5.
\( D = \{1, 2, 3, 4, 5\} \)
The number of outcomes in event D is \( n(D) = 5 \).
The probability of event D is \( P(D) = \frac{n(D)}{n(S)} \)
\( P(D) = \frac{5}{6} \)
Thus, the probability to appear a number less than 6 is \( \frac{5}{6} \). Knowing the sample space helps in easily identifying outcomes.
In simple words: When you roll a die, there are 6 possible results. For prime numbers (2, 3, 5), it's 3 out of 6, or \( \frac{1}{2} \). For 1 or less, it's just 1 (the number 1), so 1 out of 6. For numbers less than 6 (1, 2, 3, 4, 5), it's 5 out of 6.

🎯 Exam Tip: Always clearly list the sample space and the favorable outcomes for each event before calculating probabilities. Make sure you understand what "prime number," "less than," or "at least" means for the specific question.

Question 4. A coin is tossed 4 times. Find the probability to get tail at least three times in these throw.
Answer: When a coin is tossed 4 times, each toss can result in either a Head (H) or a Tail (T). The total number of possible outcomes (sample space) is \( 2^4 = 16 \).
The sample space S is:
S = {HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT}
So, \( n(S) = 16 \).
Let A be the event of getting a tail at least three times. This means getting exactly 3 tails or exactly 4 tails.
Outcomes with exactly 3 tails: {HTTT, THTT, TTHT, TTTH}
Outcomes with exactly 4 tails: {TTTT}
So, the event A is:
\( A = \{\text{HTTT, THTT, TTHT, TTTH, TTTT}\} \)
The number of outcomes in event A is \( n(A) = 5 \). This collection covers all instances of three or more tails.
The required probability is \( P(A) = \frac{n(A)}{n(S)} \)
\( P(A) = \frac{5}{16} \)
Thus, the probability of getting a tail at least three times is \( \frac{5}{16} \).
In simple words: If you flip a coin four times, there are 16 total ways it can land. We want to know how many ways have 3 or 4 tails. Counting them, there are 5 such ways. So, the chance is 5 out of 16.

🎯 Exam Tip: For problems involving multiple coin tosses, it's helpful to either list out the sample space systematically or use binomial probability if the number of trials is large. "At least three" means three or more, so include all possibilities that meet that condition.

Question 5. If a coin and a dice are thrown simultaneously, then what is the probability that tail appear on coin and event number appear on dice?
Answer: When a coin and a die are thrown simultaneously, the sample space (S) is formed by combining the outcomes of each. The coin can land on Head (H) or Tail (T), and the die can land on 1, 2, 3, 4, 5, or 6.
The sample space S is:
\( S = \{(H,1), (H,2), (H,3), (H,4), (H,5), (H,6), (T,1), (T,2), (T,3), (T,4), (T,5), (T,6)\} \)
The total number of outcomes is \( n(S) = 2 \times 6 = 12 \).
Let A be the event that a tail appears on the coin AND an even number appears on the die.
The even numbers on a die are {2, 4, 6}.
So, the outcomes for event A are combinations of Tail with an even number:
\( A = \{(T,2), (T,4), (T,6)\} \)
The number of outcomes in event A is \( n(A) = 3 \). This set represents all successful outcomes.
The required probability is \( P(A) = \frac{n(A)}{n(S)} \)
\( P(A) = \frac{3}{12} = \frac{1}{4} \)
Thus, the probability that a tail appears on the coin and an even number appears on the die is \( \frac{1}{4} \).
In simple words: When a coin and a die are tossed together, there are 12 possible results. We are looking for a "Tail" on the coin AND an "even number" on the die. The combinations (T,2), (T,4), (T,6) fit this, so there are 3 such results. The chance is 3 out of 12, or \( \frac{1}{4} \).

🎯 Exam Tip: For combined events, list the full sample space or multiply the number of outcomes for each independent event. Ensure you understand "AND" (intersection) versus "OR" (union) in probability problems.

Question 7. To solve a problem, the probability for A not to solve it is \( \frac{4}{7} \). The probability for B to solve it is \( \frac{5}{12} \). What is the probability that:
(i) Problem will be solved
(ii) Problem will not be solved
(iii) Problem will be solved only by one
Answer: We are given the following probabilities:
Probability for A not to solve the problem: \( P(\overline{A}) = \frac{4}{7} \)
From this, the probability for A to solve the problem is: \( P(A) = 1 - P(\overline{A}) = 1 - \frac{4}{7} = \frac{3}{7} \)
Probability for B to solve the problem: \( P(B) = \frac{5}{12} \)
From this, the probability for B not to solve the problem is: \( P(\overline{B}) = 1 - P(B) = 1 - \frac{5}{12} = \frac{7}{12} \)
Assume A and B solving the problem are independent events.

(i) **Probability that the problem will be solved:**
The problem is solved if A solves it OR B solves it OR both solve it. This is \( P(A \cup B) \).
Using the formula for the union of two independent events:
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
Since A and B are independent, \( P(A \cap B) = P(A) \times P(B) \).
\( P(A \cup B) = P(A) + P(B) - P(A)P(B) \)
\( P(A \cup B) = \frac{3}{7} + \frac{5}{12} - \left(\frac{3}{7} \times \frac{5}{12}\right) \)
\( P(A \cup B) = \frac{3}{7} + \frac{5}{12} - \frac{15}{84} \)
To add and subtract these fractions, find a common denominator, which is 84.
\( P(A \cup B) = \frac{3 \times 12}{7 \times 12} + \frac{5 \times 7}{12 \times 7} - \frac{15}{84} \)
\( P(A \cup B) = \frac{36}{84} + \frac{35}{84} - \frac{15}{84} \)
\( P(A \cup B) = \frac{36 + 35 - 15}{84} \)
\( P(A \cup B) = \frac{71 - 15}{84} \)
\( P(A \cup B) = \frac{56}{84} \)
This fraction can be simplified by dividing both numerator and denominator by 28.
\( P(A \cup B) = \frac{56 \div 28}{84 \div 28} = \frac{2}{3} \)
Alternatively, the problem is solved if it's not the case that *neither* A nor B solves it. So, \( P(A \cup B) = 1 - P(\overline{A} \cap \overline{B}) = 1 - P(\overline{A})P(\overline{B}) \).
\( P(A \cup B) = 1 - \left(\frac{4}{7} \times \frac{7}{12}\right) \)
\( P(A \cup B) = 1 - \frac{28}{84} \)
\( P(A \cup B) = 1 - \frac{1}{3} \)
\( P(A \cup B) = \frac{2}{3} \)
The probability that the problem will be solved is \( \frac{2}{3} \). This calculation uses the concept of complementary events effectively.

(ii) **Probability that the problem will not be solved:**
The problem will not be solved if neither A solves it nor B solves it. This is \( P(\overline{A} \cap \overline{B}) \).
Since A and B solving are independent, their complements are also independent.
\( P(\overline{A} \cap \overline{B}) = P(\overline{A}) \times P(\overline{B}) \)
\( P(\overline{A} \cap \overline{B}) = \frac{4}{7} \times \frac{7}{12} \)
\( P(\overline{A} \cap \overline{B}) = \frac{28}{84} \)
\( P(\overline{A} \cap \overline{B}) = \frac{1}{3} \)
The probability that the problem will not be solved is \( \frac{1}{3} \). This also confirms the result from part (i) as \( 1 - \frac{2}{3} = \frac{1}{3} \).

(iii) **Probability that the problem will be solved only by one:**
This means either A solves it and B does not, OR B solves it and A does not. This can be written as \( P(A \cap \overline{B}) + P(\overline{A} \cap B) \).
Since A and B are independent events, these probabilities can be calculated as:
\( P(A \cap \overline{B}) = P(A) \times P(\overline{B}) = \frac{3}{7} \times \frac{7}{12} = \frac{21}{84} = \frac{1}{4} \)
\( P(\overline{A} \cap B) = P(\overline{A}) \times P(B) = \frac{4}{7} \times \frac{5}{12} = \frac{20}{84} = \frac{5}{21} \)
Now, sum these two probabilities:
\( P(\text{solved only by one}) = \frac{1}{4} + \frac{5}{21} \)
To add these fractions, find a common denominator, which is 84.
\( P(\text{solved only by one}) = \frac{1 \times 21}{4 \times 21} + \frac{5 \times 4}{21 \times 4} \)
\( P(\text{solved only by one}) = \frac{21}{84} + \frac{20}{84} \)
\( P(\text{solved only by one}) = \frac{21 + 20}{84} \)
\( P(\text{solved only by one}) = \frac{41}{84} \)
The probability that the problem will be solved only by one person is \( \frac{41}{84} \). Breaking down the event into mutually exclusive cases simplifies the problem.
In simple words: First, we find the chance of A solving it (\( \frac{3}{7} \)) and B solving it (\( \frac{5}{12} \)).
(i) The problem is solved if at least one person solves it. This chance is \( \frac{2}{3} \).
(ii) The problem is not solved if neither person solves it. This chance is \( \frac{1}{3} \).
(iii) The problem is solved by only one person if A solves it and B does not, OR B solves it and A does not. This combined chance is \( \frac{41}{84} \).

🎯 Exam Tip: When dealing with multiple events, clearly identify what each part of the question (e.g., "solved," "not solved," "only by one") translates to in terms of set operations (union, intersection, complement) and apply the correct probability formulas for independent events.

Question 8. An equipment will work only when its three components A, B and C are in working condition. Probability that A will not work is 0.15, probability that B will not work is 0.05, and probability that C will not work is 0.10. What is the probability that the equipment will not be in working condition before year ending?
Answer: We are given the probabilities that the components will not work (their complements):
\( P(\overline{A}) = 0.15 \)
\( P(\overline{B}) = 0.05 \)
\( P(\overline{C}) = 0.10 \)
From these, we can find the probabilities that each component *will* work:
\( P(A) = 1 - P(\overline{A}) = 1 - 0.15 = 0.85 \)
\( P(B) = 1 - P(\overline{B}) = 1 - 0.05 = 0.95 \)
\( P(C) = 1 - P(\overline{C}) = 1 - 0.10 = 0.90 \)
The equipment works only if all three components A, B, and C are in working condition. Assuming the components work independently, the probability that the equipment *will* work is:
\( P(\text{Equipment works}) = P(A) \times P(B) \times P(C) \)
\( P(\text{Equipment works}) = 0.85 \times 0.95 \times 0.90 \)
\( P(\text{Equipment works}) = 0.8075 \times 0.90 \)
\( P(\text{Equipment works}) = 0.72675 \)
We need to find the probability that the equipment *will not* be in working condition. This is the complement of the equipment working.
\( P(\text{Equipment will not work}) = 1 - P(\text{Equipment works}) \)
\( P(\text{Equipment will not work}) = 1 - 0.72675 \)
\( P(\text{Equipment will not work}) = 0.27325 \)
Thus, the probability that the equipment will not be in working condition before year ending is 0.27325. This uses the principle of independent events for combined probabilities.
In simple words: First, find the chance of each part working (A: 0.85, B: 0.95, C: 0.90). Since the whole machine works only if all parts work, multiply these chances together (0.85 * 0.95 * 0.90 = 0.72675). The chance the machine *won't* work is 1 minus the chance it *will* work (1 - 0.72675 = 0.27325).

🎯 Exam Tip: For systems that require all components to work, multiply the individual probabilities of each component working. For calculating the probability of failure, use the complement rule: \( P(\text{failure}) = 1 - P(\text{success}) \).

Question 9. From a deck of cards, two cards are randomly drawn one by one. If card once drawn is not replaced, what is the probability to get two aces in the first attempt and two kings in the second attempt?
Answer: We have a standard deck of 52 cards. We are drawing cards without replacement.

**First attempt: Getting two aces**
The total number of ways to draw 2 cards from 52 is \( ^{52}C_2 \).
\( ^{52}C_2 = \frac{52 \times 51}{2 \times 1} = 26 \times 51 = 1326 \)
The number of ways to draw 2 aces from the 4 aces in the deck is \( ^4C_2 \).
\( ^4C_2 = \frac{4 \times 3}{2 \times 1} = 6 \)
The probability of getting two aces in the first attempt, \( P(E_1) \), is:
\( P(E_1) = \frac{^4C_2}{^{52}C_2} = \frac{6}{1326} = \frac{1}{221} \)

**Second attempt: Getting two kings after the first draw**
After the first attempt (drawing 2 aces without replacement), there are \( 52 - 2 = 50 \) cards left in the deck. The number of aces is now 2 (if we consider aces still in the deck, but we only care about kings). The number of kings is still 4.
The total number of ways to draw 2 cards from the remaining 50 is \( ^{50}C_2 \).
\( ^{50}C_2 = \frac{50 \times 49}{2 \times 1} = 25 \times 49 = 1225 \)
The number of ways to draw 2 kings from the 4 kings remaining is \( ^4C_2 = 6 \).
The probability of getting two kings in the second attempt, given that two aces were drawn first, \( P(E_2 | E_1) \), is:
\( P(E_2 | E_1) = \frac{^4C_2}{^{50}C_2} = \frac{6}{1225} \)

**Combined probability:**
The probability of getting two aces in the first attempt AND two kings in the second attempt is the product of these probabilities because the events are sequential and dependent (due to no replacement).
\( P(E_1 \text{ and } E_2) = P(E_1) \times P(E_2 | E_1) \)
\( P(E_1 \text{ and } E_2) = \frac{1}{221} \times \frac{6}{1225} \)
\( P(E_1 \text{ and } E_2) = \frac{6}{270725} \)
Thus, the probability of drawing two aces and then two kings without replacement is \( \frac{6}{270725} \). This type of problem highlights the importance of adjusting the sample space for sequential events without replacement.
In simple words: We draw cards two times, two cards each time, without putting them back. First, we find the chance of getting two aces from the 52 cards, which is 6 out of 1326. After taking out two aces, there are 50 cards left. Then, we find the chance of getting two kings from these 50 cards, which is 6 out of 1225. To get both events, we multiply these two chances: \( \frac{6}{270725} \).

🎯 Exam Tip: For "without replacement" problems, remember that the total number of items and the number of specific items decrease after each draw. Use combinations (\( ^nC_r \)) for selecting groups of items.

Question 11. Imagine that the ratio of male children to total children is 1:2. Find the probability that in a family out of 5 children, there are:
(i) all boys
(ii) three boys and two girls.
Answer: The ratio of male children to total children is 1:2. This means the probability of having a male child (boy) is \( P(\text{boy}) = \frac{1}{2} \). Consequently, the probability of having a female child (girl) is also \( P(\text{girl}) = 1 - \frac{1}{2} = \frac{1}{2} \).
This is a binomial probability problem since there are a fixed number of trials (5 children), each trial has two outcomes (boy or girl), and the probability of success (boy) is constant.
Let \( p = P(\text{boy}) = \frac{1}{2} \) and \( q = P(\text{girl}) = \frac{1}{2} \). The number of children is \( n = 5 \).

(i) **Probability of having all boys out of 5 children:**
This means we want \( k = 5 \) boys. Using the binomial probability formula \( P(X=k) = ^nC_k p^k q^{n-k} \):
\( P(\text{all boys}) = ^5C_5 \left(\frac{1}{2}\right)^5 \left(\frac{1}{2}\right)^{5-5} \)
\( P(\text{all boys}) = 1 \times \left(\frac{1}{2}\right)^5 \times \left(\frac{1}{2}\right)^0 \)
\( P(\text{all boys}) = 1 \times \frac{1}{32} \times 1 \)
\( P(\text{all boys}) = \frac{1}{32} \)
The probability of having all boys is \( \frac{1}{32} \). This is a straightforward application of the binomial formula.

(ii) **Probability of having three boys and two girls out of 5 children:**
This means we want \( k = 3 \) boys (and therefore \( 5 - 3 = 2 \) girls).
\( P(\text{3 boys, 2 girls}) = ^5C_3 \left(\frac{1}{2}\right)^3 \left(\frac{1}{2}\right)^{5-3} \)
\( P(\text{3 boys, 2 girls}) = ^5C_3 \left(\frac{1}{2}\right)^3 \left(\frac{1}{2}\right)^2 \)
First, calculate \( ^5C_3 \):
\( ^5C_3 = \frac{5!}{3!(5-3)!} = \frac{5!}{3!2!} = \frac{5 \times 4}{2 \times 1} = 10 \)
Now, substitute this back into the probability formula:
\( P(\text{3 boys, 2 girls}) = 10 \times \frac{1}{8} \times \frac{1}{4} \)
\( P(\text{3 boys, 2 girls}) = 10 \times \frac{1}{32} \)
\( P(\text{3 boys, 2 girls}) = \frac{10}{32} \)
This fraction can be simplified by dividing both numerator and denominator by 2.
\( P(\text{3 boys, 2 girls}) = \frac{5}{16} \)
The probability of having three boys and two girls is \( \frac{5}{16} \). This shows how probabilities for different combinations within a set of trials are calculated.
In simple words: The chance of having a boy is 1 in 2, and a girl is also 1 in 2. If a family has 5 children:
(i) The chance of all 5 being boys is 1 out of 32.
(ii) The chance of having 3 boys and 2 girls is 10 out of 32, which simplifies to 5 out of 16.

🎯 Exam Tip: Recognize binomial distribution problems by "fixed number of trials," "two outcomes," and "constant probability." Remember the formula \( ^nC_k p^k q^{n-k} \) and how to calculate combinations (\( ^nC_k \)).

Question. Three people A, B, and C attempt to hit a target. The probability that A hits the target is \( P(A) = \frac{1}{2} \), B hits is \( P(B) = \frac{1}{2} \), and C hits is \( P(C) = \frac{3}{4} \). What is the probability that at least two people hit the target?
Answer: We are given the probabilities that A, B, and C hit the target:
\( P(A) = \frac{1}{2} \)
\( P(B) = \frac{1}{2} \)
\( P(C) = \frac{3}{4} \)
From these, we can find the probabilities that each person *misses* the target (their complements):
\( P(\overline{A}) = 1 - P(A) = 1 - \frac{1}{2} = \frac{1}{2} \)
\( P(\overline{B}) = 1 - P(B) = 1 - \frac{1}{2} = \frac{1}{2} \)
\( P(\overline{C}) = 1 - P(C) = 1 - \frac{3}{4} = \frac{1}{4} \)
We want to find the probability that *at least two* people hit the target. This means either exactly two people hit the target, or all three people hit the target.
These are mutually exclusive events, so we can sum their probabilities.

**Case 1: Exactly two people hit the target.**
There are three ways for exactly two people to hit the target:
1. A and B hit, C misses: \( P(A \cap B \cap \overline{C}) = P(A)P(B)P(\overline{C}) = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{4} = \frac{1}{16} \)
2. A and C hit, B misses: \( P(A \cap \overline{B} \cap C) = P(A)P(\overline{B})P(C) = \frac{1}{2} \times \frac{1}{2} \times \frac{3}{4} = \frac{3}{16} \)
3. B and C hit, A misses: \( P(\overline{A} \cap B \cap C) = P(\overline{A})P(B)P(C) = \frac{1}{2} \times \frac{1}{2} \times \frac{3}{4} = \frac{3}{16} \)
The probability of exactly two people hitting the target is the sum of these:
\( P(\text{exactly 2 hits}) = \frac{1}{16} + \frac{3}{16} + \frac{3}{16} = \frac{1+3+3}{16} = \frac{7}{16} \)

**Case 2: All three people hit the target.**
This means A, B, and C all hit:
\( P(A \cap B \cap C) = P(A)P(B)P(C) = \frac{1}{2} \times \frac{1}{2} \times \frac{3}{4} = \frac{3}{16} \)

**Total probability for at least two people hitting the target:**
This is the sum of the probabilities from Case 1 and Case 2.
\( P(\text{at least 2 hits}) = P(\text{exactly 2 hits}) + P(\text{all 3 hits}) \)
\( P(\text{at least 2 hits}) = \frac{7}{16} + \frac{3}{16} \)
\( P(\text{at least 2 hits}) = \frac{7+3}{16} = \frac{10}{16} \)
This fraction can be simplified by dividing both numerator and denominator by 2.
\( P(\text{at least 2 hits}) = \frac{5}{8} \)
The probability that at least two people hit the target is \( \frac{5}{8} \). By considering all combinations, we ensure a complete solution.
In simple words: We want to find the chance that 2 or 3 people hit the target. We calculate the chance for exactly 2 people to hit (A and B hit, C misses; or A and C hit, B misses; or B and C hit, A misses). We also calculate the chance for all 3 to hit. We add these chances together to get the final answer.

🎯 Exam Tip: For "at least" type problems, list all the ways the condition can be met. Often, it's easier to calculate the probabilities of the opposite events and subtract from 1, but for small numbers of events, direct calculation can be clearer. Always assume independence unless stated otherwise.