Class 12 Ch 13: Probability - Exercise 13.1

Q1

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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 $P(F|E)$.

$P(E|F) = \frac{P(E \cap F)}{P(F)} = \frac{0.2}{0.3} = \frac{2}{3}$.

$P(F|E) = \frac{P(E \cap F)}{P(E)} = \frac{0.2}{0.6} = \frac{1}{3}$.

$\boxed{P(E|F) = \frac{2}{3}, P(F|E) = \frac{1}{3}}$

Q2

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Compute $P(A|B)$ if $P(B) = 0.5$ and $P(A \cap B) = 0.32$.

$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.32}{0.5} = \frac{0.32}{0.50} = \frac{32}{50} = 0.64$.

$\boxed{0.64}$

Q3

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If $P(A) = 0.8, P(B) = 0.5$ and $P(B|A) = 0.4$, find (i) $P(A \cap B)$ (ii) $P(A|B)$ (iii) $P(A \cup B)$.

(i) $P(A \cap B) = P(B|A) \cdot P(A) = 0.4 \times 0.8 = 0.32$.

(ii) $P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.32}{0.5} = 0.64$.

(iii) $P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.8 + 0.5 - 0.32 = 1.3 - 0.32 = 0.98$.

$\boxed{0.32, 0.64, 0.98}$

Q4

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Evaluate $P(A \cup B)$ if $2P(A) = P(B) = \frac{5}{13}$ and $P(A|B) = \frac{2}{5}$.

$P(B) = \frac{5}{13}, P(A) = \frac{5}{26}$.

$P(A \cap B) = P(A|B)P(B) = \frac{2}{5} \times \frac{5}{13} = \frac{2}{13}$.

$P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{5}{26} + \frac{5}{13} - \frac{2}{13}$.

$= \frac{5 + 10 - 4}{26} = \frac{11}{26}$.

$\boxed{\frac{11}{26}}$

Q5

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If $P(A) = \frac{6}{11}, P(B) = \frac{5}{11}$ and $P(A \cup B) = \frac{7}{11}$, find (i) $P(A \cap B)$ (ii) $P(A|B)$ (iii) $P(B|A)$.

(i) $P(A \cap B) = P(A) + P(B) - P(A \cup B) = \frac{6}{11} + \frac{5}{11} - \frac{7}{11} = \frac{4}{11}$.

(ii) $P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{4/11}{5/11} = \frac{4}{5}$.

(iii) $P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{4/11}{6/11} = \frac{4}{6} = \frac{2}{3}$.

$\boxed{\frac{4}{11}, \frac{4}{5}, \frac{2}{3}}$

Q6

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A coin is tossed three times. Determine $P(E|F)$ where E: head on third toss, F: heads on first two tosses.

Sample Space $S$ has 8 outcomes.

$E = \{HHH, HTH, THH, TTH\}$, $F = \{HHH, HHT\}$.

$E \cap F = \{HHH\}$.

$P(F) = \frac{2}{8}, P(E \cap F) = \frac{1}{8}$.

$P(E|F) = \frac{1/8}{2/8} = \frac{1}{2}$.

$\boxed{\frac{1}{2}}$

Q7

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A coin is tossed three times. Determine $P(E|F)$ where (i) E: at least two heads, F: at most two heads. (ii) E: at most two tails, F: at least one tail.

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}. n(S) = 8.

(i) E (at least 2 heads) = {HHH, HHT, HTH, THH}. F (at most 2 heads) = S - {HHH}.

$E \cap F = \{HHT, HTH, THH\}$. $n(F)=7, n(E \cap F)=3$.

$P(E|F) = \frac{n(E \cap F)}{n(F)} = \frac{3}{7}$.

(ii) E (at most 2 tails) = S - {TTT}. F (at least 1 tail) = S - {HHH}.

$E \cap F = S - \{HHH, TTT\}$. $n(F)=7, n(E \cap F)=6$.

$P(E|F) = \frac{n(E \cap F)}{n(F)} = \frac{6}{7}$.

$\boxed{\text{(i) } \frac{3}{7}, \text{ (ii) } \frac{6}{7}}$

Q8

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A die is thrown three times. E: 4 appears on the third toss, F: 6 and 5 appear respectively on first two tosses.

F = "6 and 5 on first two tosses". Outcomes are (6,5,1), (6,5,2), ..., (6,5,6). $n(F)=6$.

E = "4 on third toss".

$E \cap F$ = "6 and 5 on first two, 4 on third" = {(6,5,4)}. $n(E \cap F)=1$.

$P(E|F) = \frac{n(E \cap F)}{n(F)} = \frac{1}{6}$.

$\boxed{\frac{1}{6}}$

Q9

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Mother, father and son line up at random for a family picture. E: son on one end, F: father in middle.

Sample space S = {MFS, MSF, FMS, FSM, SMF, SFM}. n(S)=6.

F (father in middle) = {MFS, SFM}. n(F)=2.

E (son on one end) = {SMF, SFM, MFS, FMS}.

$E \cap F$ = {MFS, SFM}. $n(E \cap F)=2$.

$P(E|F) = \frac{n(E \cap F)}{n(F)} = \frac{2}{2} = 1$.

$\boxed{1}$

Q10

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A black and a red die are rolled. (a) Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5.

Let Black die be first, Red be second. Total outcomes = 36.

F (Black is 5) = $\{(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)\}$. $n(F) = 6$.

E (Sum > 9) = $\{(4,6), (5,5), (5,6), (6,4), (6,5), (6,6)\}$.

$E \cap F = \{(5,5), (5,6)\}$. $n(E \cap F) = 2$.

$P(E|F) = \frac{2}{6} = \frac{1}{3}$.

$\boxed{\frac{1}{3}}$

Q11

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A fair die is rolled. Consider events $E = \{1, 3, 5\}, F = \{2, 3\}$ and $G = \{2, 3, 4, 5\}$. Find $P(E|F)$ and $P(F|E)$.

$E \cap F = \{3\}$. $P(E \cap F) = 1/6$.

$P(F) = 2/6, P(E) = 3/6$.

$P(E|F) = \frac{1/6}{2/6} = \frac{1}{2}$.

$P(F|E) = \frac{1/6}{3/6} = \frac{1}{3}$.

$\boxed{\frac{1}{2}, \frac{1}{3}}$

Q12

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Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that (i) the youngest is a girl, (ii) at least one is a girl?

Sample Space $S = \{BB, BG, GB, GG\}$ (Order: Older, Younger).

Event E (Both girls) = $\{GG\}$.

(i) F (Youngest is girl) = $\{BG, GG\}$. $E \cap F = \{GG\}$. $P(E|F) = \frac{1/4}{2/4} = \frac{1}{2}$.

(ii) G (At least one girl) = $\{BG, GB, GG\}$. $E \cap G = \{GG\}$. $P(E|G) = \frac{1/4}{3/4} = \frac{1}{3}$.

$\boxed{\frac{1}{2}, \frac{1}{3}}$

Q13

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An instructor has a question bank consisting of 300 easy True/False questions, 200 difficult True/False questions, 500 easy multiple choice questions and 400 difficult multiple choice questions. If a question is selected at random from the question bank, what is the probability that it will be an easy question given that it is a multiple choice question?

Total MCQ = $500 (\text{Easy}) + 400 (\text{Diff}) = 900$.

Easy MCQ = 500.

Let E = Easy, F = MCQ. We need $P(E|F)$.

$P(E|F) = \frac{n(E \cap F)}{n(F)} = \frac{500}{900} = \frac{5}{9}$.

$\boxed{\frac{5}{9}}$

Q14

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Given that the two numbers appearing on throwing two dice are different. Find the probability of the event 'the sum of numbers on the dice is 4'.

Let F be the event that numbers are different. Total outcomes = 36. Same numbers = 6. So $n(F) = 36-6=30$.

Let E be the event that sum is 4. E = {(1,3), (2,2), (3,1)}.

$E \cap F$ = sum is 4 and numbers are different = {(1,3), (3,1)}. $n(E \cap F) = 2$.

$P(E|F) = \frac{n(E \cap F)}{n(F)} = \frac{2}{30} = \frac{1}{15}$.

$\boxed{\frac{1}{15}}$

Q15

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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’.

Let E be 'the coin shows a tail' and F be 'at least one die shows a 3'.

For event F to occur, the first roll must be a 3 or a 6.

Case 1: First roll is 3. We roll the die again. A coin is not tossed.

Case 2: First roll is 6. We roll the die again. For F to occur, the second roll must be 3. A coin is not tossed.

In every outcome of event F, a coin is never tossed. Thus, event E (coin shows a tail) cannot happen if F has happened.

Therefore, $E \cap F = \emptyset$, which means $P(E \cap F) = 0$.

$P(E|F) = \frac{P(E \cap F)}{P(F)} = 0$.

$\boxed{0}$

Q16

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If $P(A) = \frac{1}{2}, P(B) = 0$, then $P(A|B)$ is?

$P(A|B) = \frac{P(A \cap B)}{P(B)}$.

Since $P(B) = 0$, division by zero is undefined.

$\boxed{\text{Not defined}}$

Q17

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If A and B are two events such that $P(A|B) = P(B|A)$, then which is correct? (A) A âŠ‚ B (B) A = B (C) A âˆ© B = âˆ… (D) P(A) = P(B)

We are given $P(A|B) = P(B|A)$, assuming $P(A)>0, P(B)>0$.

By definition, $\frac{P(A \cap B)}{P(B)} = \frac{P(B \cap A)}{P(A)}$.

Since $P(A \cap B) = P(B \cap A)$, the equation is $\frac{P(A \cap B)}{P(B)} = \frac{P(A \cap B)}{P(A)}$.

If $P(A \cap B) \neq 0$, we can cancel the term, which gives $\frac{1}{P(B)} = \frac{1}{P(A)}$.

This implies $P(A) = P(B)$.

If $P(A \cap B) = 0$, the equation becomes $0=0$, which is true for any $P(A)$ and $P(B)$. However, option (D) is the most general conclusion derived from the given condition, assuming the events are not mutually exclusive.

$\boxed{\text{(D) } P(A) = P(B)}$

Exercise 13.1 Practice