Class 12 Ch 13: Probability - Miscellaneous Exercise

Q1

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A and B are two events such that $P(A) \neq 0$. Find $P(B|A)$ if (i) A is a subset of B (ii) $A \cap B = \phi$.

(i) If $A \subset B$, then $A \cap B = A$.

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{P(A)}{P(A)} = 1$.

(ii) If $A \cap B = \phi$, then $P(A \cap B) = 0$.

$P(B|A) = \frac{0}{P(A)} = 0$.

$\boxed{\text{(i) } 1, \text{ (ii) } 0}$

Q2

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A couple has two children. (i) Find the probability that both children are males, if it is known that at least one of the children is male. (ii) Find the probability that both children are females, if it is known that the elder child is a female.

Sample Space $S = \{BB, BG, GB, GG\}$.

(i) Let $E$: Both males $\{BB\}$, $F$: At least one male $\{BB, BG, GB\}$.

$P(E|F) = \frac{P(E \cap F)}{P(F)} = \frac{1/4}{3/4} = \frac{1}{3}$.

(ii) Let $A$: Both females $\{GG\}$, $B$: Elder is female $\{GB, GG\}$.

$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{1/4}{2/4} = \frac{1}{2}$.

$\boxed{\text{(i) } \frac{1}{3}, \text{ (ii) } \frac{1}{2}}$

Q3

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Suppose that 5% of men and 0.25% of women have grey hair. A grey haired person is selected at random. What is the probability of this person being male? Assume that there are equal number of males and females.

$P(M) = P(W) = 0.5$. $G$: Grey hair.

$P(G|M) = 0.05$, $P(G|W) = 0.0025$.

$P(M|G) = \frac{0.5 \times 0.05}{0.5 \times 0.05 + 0.5 \times 0.0025} = \frac{0.05}{0.0525} = \frac{500}{525} = \frac{20}{21}$.

$\boxed{\frac{20}{21}}$

Q4

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Suppose that 90% of people are right-handed. What is the probability that at most 6 of a random sample of 10 people are right-handed?

$p = 0.9, q = 0.1, n = 10$.

$P(X \le 6) = 1 - P(X > 6) = 1 - [P(7) + P(8) + P(9) + P(10)]$.

$= 1 - \sum_{r=7}^{10} {^{10}C_r} (0.9)^r (0.1)^{10-r}$.

$\boxed{\sum_{r=0}^{6} {^{10}C_r} (0.9)^r (0.1)^{10-r}}$

Q5

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An urn contains 25 balls of which 10 balls bear a mark 'X' and the remaining 15 bear a mark 'Y'. A ball is drawn at random from the urn, its mark is noted down and it is replaced. If 6 balls are drawn in this way, find the probability that (i) all will bear 'X' mark. (ii) not more than 2 will bear 'Y' mark. (iii) at least one ball will bear 'Y' mark. (iv) the number of balls with 'X' mark and 'Y' mark will be equal.

$P(X) = \frac{10}{25} = \frac{2}{5}, P(Y) = \frac{15}{25} = \frac{3}{5}$. $n=6$.

(i) All X: $P(6X) = (\frac{2}{5})^6$.

(ii) Not more than 2 Y: $P(Y=0) + P(Y=1) + P(Y=2)$.
$= (\frac{2}{5})^6 + 6(\frac{2}{5})^5(\frac{3}{5}) + 15(\frac{2}{5})^4(\frac{3}{5})^2$.

(iii) At least one Y: $1 - P(\text{No Y}) = 1 - (\frac{2}{5})^6$.

(iv) Equal X and Y (3 each): $^{6}C_3 (\frac{2}{5})^3 (\frac{3}{5})^3 = 20 (\frac{6}{25})^3$.

$\boxed{\text{See steps}}$

Q6

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In a hurdle race, a player has to clear 10 hurdles. The probability that he will clear each hurdle is 5/6. What is the probability that he will knock down fewer than 2 hurdles?

$P(\text{Clear}) = 5/6, P(\text{Knock}) = 1/6$. $n=10$.

Fewer than 2 knocks $\Rightarrow$ 0 or 1 knock.

$P(K=0) + P(K=1) = (\frac{5}{6})^{10} + ^{10}C_1 (\frac{1}{6})(\frac{5}{6})^9$.

$= (\frac{5}{6})^9 [\frac{5}{6} + \frac{10}{6}] = (\frac{5}{6})^9 (\frac{15}{6}) = \frac{5}{2} (\frac{5}{6})^9$.

$\boxed{\frac{5}{2} (\frac{5}{6})^9}$

Q7

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A die is tossed again and again until three sixes are obtained. Find the probability of obtaining the third six in the sixth throw.

We need exactly 2 sixes in the first 5 throws, and a six in the 6th throw.

$P(2 \text{ sixes in 5}) = ^{5}C_2 (\frac{1}{6})^2 (\frac{5}{6})^3 = 10 \times \frac{1}{36} \times \frac{125}{216}$.

Total Prob = $10 \times \frac{125}{7776} \times \frac{1}{6} = \frac{1250}{46656} = \frac{625}{23328}$.

$\boxed{\frac{625}{23328}}$

Q8

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If a leap year is selected at random, what is the chance that it will contain 53 Tuesdays?

Leap year = 366 days = 52 weeks + 2 days.

The 2 extra days can be (Mon,Tue), (Tue,Wed), (Wed,Thu), (Thu,Fri), (Fri,Sat), (Sat,Sun), (Sun,Mon).

Favorable cases for Tuesday: (Mon,Tue), (Tue,Wed). Total = 2.

$\boxed{\frac{2}{7}}$

Q9

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An experiment succeeds twice as often as it fails. Find the probability that in the next six trials, there will be at least 4 successes.

$p = 2q \Rightarrow p = 2/3, q = 1/3$. $n=6$.

$P(X \ge 4) = P(4) + P(5) + P(6)$.

$= ^{6}C_4 (\frac{2}{3})^4 (\frac{1}{3})^2 + ^{6}C_5 (\frac{2}{3})^5 (\frac{1}{3}) + ^{6}C_6 (\frac{2}{3})^6$.

$= \frac{1}{729} [15(16) + 6(32) + 64] = \frac{240 + 192 + 64}{729} = \frac{496}{729}$.

$\boxed{\frac{496}{729}}$

Q10

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How many times must a man toss a fair coin so that the probability of having at least one head is more than 90%?

$1 - P(\text{No Head}) > 0.9 \Rightarrow 1 - (1/2)^n > 0.9$.

$(1/2)^n < 0.1 \Rightarrow 2^n > 10$.

For $n=3, 8 < 10$. For $n=4, 16 > 10$.

$\boxed{4}$

Q11

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In a game, a man wins a rupee for a six and loses a rupee for any other number when a fair die is thrown. The man decided to throw a die thrice but to quit as and when he gets a six. Find the expected value of the amount he wins / loses.

Outcomes: 6 (Win 1), Not-6,6 (Lose 1, Win 1 = 0), Not-6,Not-6,6 (Lose 2, Win 1 = -1), Not-6,Not-6,Not-6 (Lose 3).

$P(1) = 1/6$. $P(0) = 5/6 \times 1/6 = 5/36$. $P(-1) = (5/6)^2 \times 1/6 = 25/216$. $P(-3) = (5/6)^3 = 125/216$.

$E(X) = 1(1/6) + 0(5/36) - 1(25/216) - 3(125/216) = \frac{36 - 25 - 375}{216} = -\frac{364}{216} = -\frac{91}{54}$.

$\boxed{-\frac{91}{54}}$

Q12

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Suppose we have four boxes A, B, C and D containing coloured marbles. Box A: 1R, 6W, 3B. Box B: 6R, 2W, 2B. Box C: 8R, 1W, 1B. Box D: 0R, 6W, 4B. One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from box A?, box B?, box C?

$P(A)=P(B)=P(C)=P(D)=1/4$.

$P(R|A)=1/10, P(R|B)=6/10, P(R|C)=8/10, P(R|D)=0$.

Total $P(R) = \frac{1}{4} (\frac{1}{10} + \frac{6}{10} + \frac{8}{10} + 0) = \frac{15}{40} = \frac{3}{8}$.

$P(A|R) = \frac{1/40}{15/40} = \frac{1}{15}$. $P(B|R) = \frac{6/40}{15/40} = \frac{6}{15} = \frac{2}{5}$. $P(C|R) = \frac{8/40}{15/40} = \frac{8}{15}$.

$\boxed{\frac{1}{15}, \frac{2}{5}, \frac{8}{15}}$

Q13

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Assume that the chances of a patient having a heart attack is 40%. It is also assumed that a meditation and yoga course reduce the risk of heart attack by 30% and prescription of certain drug reduces its chance by 25%. At a time a patient can choose any one of the two options with equal probabilities. It is given that after going through one of the two options the patient selected at random suffers a heart attack. Find the probability that the patient followed a course of meditation and yoga?

Base risk = 40%.

Risk with Yoga ($E_1$): $40\% \times 0.7 = 28\%$. Risk with Drug ($E_2$): $40\% \times 0.75 = 30\%$.

$P(E_1)=P(E_2)=0.5$. $A$: Heart Attack.

$P(E_1|A) = \frac{0.5 \times 0.28}{0.5 \times 0.28 + 0.5 \times 0.30} = \frac{0.28}{0.58} = \frac{14}{29}$.

$\boxed{\frac{14}{29}}$

Q14

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If each element of a second order determinant is either zero or one, what is the probability that the value of the determinant is positive?

Total matrices = $2^4 = 16$. Determinant $\Delta = ad - bc$.

For $\Delta > 0$, we need $ad=1$ and $bc=0$.

$ad=1 \Rightarrow a=1, d=1$. $bc=0 \Rightarrow (b,c) \in \{(0,0), (0,1), (1,0)\}$.

Total favorable = 3. Probability = $3/16$.

$\boxed{\frac{3}{16}}$

Q15

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An electronic assembly consists of two subsystems, say, A and B. From previous testing procedures, the following probabilities are assumed to be known: P(A fails) = 0.2, P(B fails alone) = 0.15, P(A and B fail) = 0.15. Evaluate the following probabilities: (i) P(A fails | B has failed) (ii) P(A fails alone).

$P(A)=0.2, P(B \cap A')=0.15, P(A \cap B)=0.15$.

$P(B) = P(B \cap A') + P(A \cap B) = 0.15 + 0.15 = 0.3$.

(i) $P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.15}{0.3} = 0.5$.

(ii) $P(A \text{ alone}) = P(A \cap B') = P(A) - P(A \cap B) = 0.2 - 0.15 = 0.05$.

$\boxed{0.5, 0.05}$

Q16

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Bag I contains 3 red and 4 black balls and Bag II contains 4 red and 5 black balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black.

$E_1$: Transferred Red ($3/7$), $E_2$: Transferred Black ($4/7$).

If $E_1$, Bag II has 5R, 5B. $P(R|E_1) = 5/10 = 1/2$.

If $E_2$, Bag II has 4R, 6B. $P(R|E_2) = 4/10 = 2/5$.

$P(E_2|R) = \frac{4/7 \times 2/5}{3/7 \times 1/2 + 4/7 \times 2/5} = \frac{8/35}{3/14 + 8/35} = \frac{16/70}{15/70 + 16/70} = \frac{16}{31}$.

$\boxed{\frac{16}{31}}$

Q17

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If A and B are two events such that $P(A) \neq 0$ and $P(B|A) = 1$, then (A) $A \subset B$ (B) $B \subset A$ (C) $B = \phi$ (D) $A = \phi$.

$P(B|A) = 1 \Rightarrow P(A \cap B) = P(A)$.

This implies all outcomes in A are also in B. So $A \subset B$.

$\boxed{\text{(A) } A \subset B}$

Q18

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If $P(A|B) > P(A)$, then which of the following is correct: (A) $P(B|A) < P(B)$ (B) $P(A \cap B) < P(A)P(B)$ (C) $P(B|A) > P(B)$ (D) $P(B|A) = P(B)$.

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

Dividing by $P(A)$, we get $\frac{P(A \cap B)}{P(A)} > P(B) \Rightarrow P(B|A) > P(B)$.

$\boxed{\text{(C) } P(B|A) > P(B)}$

Q19

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If A and B are any two events such that $P(A) + P(B) - P(A \text{ and } B) = P(A)$, then (A) $P(B|A) = 1$ (B) $P(A|B) = 1$ (C) $P(B|A) = 0$ (D) $P(A|B) = 0$.

$P(A) + P(B) - P(A \cap B) = P(A) \Rightarrow P(B) - P(A \cap B) = 0 \Rightarrow P(A \cap B) = P(B)$.

This implies $B \subset A$.

Then $P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{P(B)}{P(B)} = 1$.

$\boxed{\text{(B) } P(A|B) = 1}$