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NCERT Solutions · Class 12 Maths · NCERT Mathematics Part I and II Part II · Chapter 13

Chapter 13: Probability (Probability)

Step-by-step solutions to all exercises of Chapter 13, Probability (NCERT Class 12 Maths Part II, 2026-27): Exercise 13.1 (conditional probability), 13.2 (multiplication theorem and independent events), 13.3 (total probability and Bayes' theorem) and the Miscellaneous Exercise. All 62 questions are answered, with the key answer highlighted.

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Free NCERT solutions by Notes Bazar · www.notesbazar.in/ncert-solutions/class-12-maths/chapter-13-probability

$P (E ∣ F) = \frac{P ( E \cap F )}{P ( F )}$ , $P (F) \neq = 0$ . Multiplication rule: $P (E \cap F) = P (E) P (F ∣ E)$ . E, F independent ⇔ $P (E \cap F) = P (E) P (F)$ . Total probability: $P (A) = \sum P (E_{i}) P (A ∣ E_{i})$ . Bayes' theorem: $P (E_{i} ∣ A) = \frac{P ( E _{i} ) P ( A ∣ E _{i} )}{\sum _{j} P ( E _{j} ) P ( A ∣ E _{j} )}$ .

Exercise 13.1

Given that E and F are events such that P(E) = 0.6, P(F) = 0.3 and P(E ∩ F) = 0.2, find P(E|F) and P(F|E).

Solution

$P (E ∣ F) = \frac{P ( E \cap F )}{P ( F )} = \frac{0.2}{0.3}$ , $P (F ∣ E) = \frac{0.2}{0.6}$ .

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

Compute P(A|B), if P(B) = 0.5 and P(A ∩ B) = 0.32.

Solution

$P (A ∣ B) = \frac{0.32}{0.5} = 0.64$

$\frac{16}{25}$

If P(A) = 0.8, P(B) = 0.5 and P(B|A) = 0.4, find (i) P(A ∩ B) (ii) P(A|B) (iii) P(A ∪ B)

Solution

(i) $P (A \cap B) = P (A) P (B ∣ A) = 0.8 \times 0.4 = 0.32$

(ii) $P (A ∣ B) = \frac{0.32}{0.5} = 0.64$

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

(i) 0.32 (ii) 0.64 (iii) 0.98

Evaluate P(A ∪ B), if 2P(A) = P(B) = 5/13 and P(A|B) = 2/5.

Solution

$P (B) = \frac{5}{13}$ , $P (A) = \frac{5}{26}$ , $P (A \cap B) = P (B) P (A ∣ B) = \frac{5}{13} \cdot \frac{2}{5} = \frac{2}{13}$ .

$P (A \cup B) = \frac{5}{26} + \frac{10}{26} - \frac{4}{26}$

$\frac{11}{26}$

If P(A) = 6/11, P(B) = 5/11 and P(A ∪ B) = 7/11, find (i) P(A ∩ B) (ii) P(A|B) (iii) P(B|A)

Solution

(i) $P (A \cap B) = \frac{6 + 5 - 7}{11} = \frac{4}{11}$ (ii) $\frac{4/11}{5/11} = \frac{4}{5}$ (iii) $\frac{4/11}{6/11} = \frac{2}{3}$

(i) 4/11 (ii) 4/5 (iii) 2/3

A coin is tossed three times. Determine P(E|F): (i) E: head on third toss, F: heads on first two tosses (ii) E: at least two heads, F: at most two heads (iii) E: at most two tails, F: at least one tail

Solution

Sample space: 8 equally likely outcomes.

(i) F = {HHH, HHT}, E ∩ F = {HHH}: $P (E ∣ F) = \frac{1}{2}$ .

(ii) F = all except HHH (7 outcomes); E ∩ F = {HHT, HTH, THH}: $\frac{3}{7}$ .

(iii) F = all except HHH (7); E = all except TTT, so E ∩ F = F − {TTT} (6 outcomes): $\frac{6}{7}$ .

(i) 1/2 (ii) 3/7 (iii) 6/7

Two coins are tossed once. Determine P(E|F): (i) E: tail appears on one coin, F: one coin shows head (ii) E: no tail appears, F: no head appears

Solution

(i) E = {HT, TH} = F, so $P (E ∣ F) = 1$ .

(ii) E = {HH}, F = {TT}, E ∩ F = ∅: $P (E ∣ F) = 0$ .

(i) 1 (ii) 0

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

Solution

F = {(6, 5, k) : k = 1, …, 6}, 6 outcomes; E ∩ F = {(6, 5, 4)}.

$\frac{1}{6}$

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

Solution

F = {MFS, SFM}. In both, the son is at an end, so E ∩ F = F.

A black and a red dice are rolled. (a) Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5. (b) Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.

Solution

(a) Black = 5: 6 outcomes. Sum > 9 needs red = 5 or 6: 2 outcomes. $\frac{2}{6} = \frac{1}{3}$ .

(b) Red ∈ {1, 2, 3}: 18 outcomes. Sum 8: (red 2, black 6), (red 3, black 5): 2 outcomes. $\frac{2}{18} = \frac{1}{9}$ .

(a) 1/3 (b) 1/9

A fair die is rolled. E = {1, 3, 5}, F = {2, 3}, G = {2, 3, 4, 5}. Find (i) P(E|F) and P(F|E) (ii) P(E|G) and P(G|E) (iii) P((E ∪ F)|G) and P((E ∩ F)|G)

Solution

(i) E ∩ F = {3}: $P (E ∣ F) = \frac{1/6}{2/6} = \frac{1}{2}$ , $P (F ∣ E) = \frac{1/6}{3/6} = \frac{1}{3}$ .

(ii) E ∩ G = {3, 5}: $P (E ∣ G) = \frac{2}{4} = \frac{1}{2}$ , $P (G ∣ E) = \frac{2}{3}$ .

(iii) (E ∪ F) ∩ G = {2, 3, 5}: $\frac{3}{4}$ . (E ∩ F) ∩ G = {3}: $\frac{1}{4}$ .

(i) 1/2, 1/3 (ii) 1/2, 2/3 (iii) 3/4, 1/4

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?

Solution

Writing elder first: S = {BB, BG, GB, GG}.

(i) Youngest a girl: {BG, GG}; both girls: {GG}: $\frac{1}{2}$ .

(ii) At least one girl: {BG, GB, GG}: $\frac{1}{3}$ .

(i) 1/2 (ii) 1/3

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?

Solution

Multiple choice questions: 500 + 400 = 900, of which 500 are easy.

$\frac{500}{900} = \frac{5}{9}$

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'.

Solution

Different numbers: $36 - 6 = 30$ outcomes. Sum 4 with different numbers: (1, 3), (3, 1).

$\frac{2}{30} = \frac{1}{15}$

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'.

Solution

If the first die shows 3 (a multiple of 3), the die is thrown again and no coin is tossed. If the first die is not a multiple of 3, it cannot show 3, and the coin is tossed with only one die thrown. So "coin shows a tail" and "at least one die shows 3" never happen together.

If P(A) = 1/2, P(B) = 0, then P(A|B) is (A) 0 (B) 1/2 (C) not defined (D) 1

Solution

$P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}$ needs $P (B) \neq = 0$ .

(C)

If A and B are events such that P(A|B) = P(B|A), then (A) A ⊂ B but A ≠ B (B) A = B (C) A ∩ B = φ (D) P(A) = P(B)

Solution

$\frac{P ( A \cap B )}{P ( B )} = \frac{P ( A \cap B )}{P ( A )} \Rightarrow P (A) = P (B)$ (when $P (A \cap B) \neq = 0$ ).

(D)

Exercise 13.2

If P(A) = 3/5 and P(B) = 1/5, find P(A ∩ B) if A and B are independent events.

Solution

$P (A \cap B) = P (A) P (B) = \frac{3}{5} \cdot \frac{1}{5}$

$\frac{3}{25}$

Two cards are drawn at random and without replacement from a pack of 52 playing cards. Find the probability that both the cards are black.

Solution

$\frac{26}{52} \cdot \frac{25}{51} = \frac{25}{102}$

$\frac{25}{102}$

A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all the three oranges are good, the box is approved for sale, otherwise, it is rejected. Find the probability that a box containing 15 oranges out of which 12 are good and 3 are bad ones will be approved for sale.

Solution

$\frac{12}{15} \cdot \frac{11}{14} \cdot \frac{10}{13} = \frac{1320}{2730}$

$\frac{44}{91}$

A fair coin and an unbiased die are tossed. Let A be the event 'head appears on the coin' and B be the event '3 on the die'. Check whether A and B are independent events or not.

Solution

12 outcomes. $P (A) = \frac{1}{2}$ , $P (B) = \frac{1}{6}$ , $P (A \cap B) = P ({(H, 3)}) = \frac{1}{12} = P (A) P (B)$ .

A and B are independent.

A die marked 1, 2, 3 in red and 4, 5, 6 in green is tossed. Let A be the event, 'the number is even,' and B be the event, 'the number is red'. Are A and B independent?

Solution

A = {2, 4, 6}, B = {1, 2, 3}, A ∩ B = {2}. $P (A \cap B) = \frac{1}{6}$ , $P (A) P (B) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$ .

Not independent.

Let E and F be events with P(E) = 3/5, P(F) = 3/10 and P(E ∩ F) = 1/5. Are E and F independent?

Solution

$P (E) P (F) = \frac{9}{50} \neq = \frac{1}{5} = \frac{10}{50}$ .

Not independent.

Given that the events A and B are such that P(A) = 1/2, P(A ∪ B) = 3/5 and P(B) = p. Find p if they are (i) mutually exclusive (ii) independent.

Solution

(i) $P (A \cup B) = P (A) + P (B)$ : $\frac{1}{2} + p = \frac{3}{5} \Rightarrow p = \frac{1}{10}$ .

(ii) $P (A \cup B) = P (A) + P (B) - P (A) P (B)$ : $\frac{1}{2} + p - \frac{p}{2} = \frac{3}{5} \Rightarrow \frac{p}{2} = \frac{1}{10} \Rightarrow p = \frac{1}{5}$ .

(i) p = 1/10 (ii) p = 1/5

Let A and B be independent events with P(A) = 0.3 and P(B) = 0.4. Find (i) P(A ∩ B) (ii) P(A ∪ B) (iii) P(A|B) (iv) P(B|A)

Solution

(i) $0.3 \times 0.4 = 0.12$ (ii) $0.3 + 0.4 - 0.12 = 0.58$ (iii) $P (A ∣ B) = P (A) = 0.3$ (iv) $P (B ∣ A) = P (B) = 0.4$ (independent events).

(i) 0.12 (ii) 0.58 (iii) 0.3 (iv) 0.4

If A and B are two events such that P(A) = 1/4, P(B) = 1/2 and P(A ∩ B) = 1/8, find P(not A and not B).

Solution

$P (A \cup B) = \frac{1}{4} + \frac{1}{2} - \frac{1}{8} = \frac{5}{8}$ . $P (A^{'} \cap B^{'}) = 1 - P (A \cup B)$ .

$\frac{3}{8}$

Events A and B are such that P(A) = 1/2, P(B) = 7/12 and P(not A or not B) = 1/4. State whether A and B are independent?

Solution

$P (A^{'} \cup B^{'}) = P ((A \cap B)^{'}) = \frac{1}{4} \Rightarrow P (A \cap B) = \frac{3}{4}$ . But $P (A) P (B) = \frac{1}{2} \cdot \frac{7}{12} = \frac{7}{24} \neq = \frac{3}{4}$ .

A and B are not independent.

Given two independent events A and B such that P(A) = 0.3, P(B) = 0.6. Find (i) P(A and B) (ii) P(A and not B) (iii) P(A or B) (iv) P(neither A nor B)

Solution

(i) $0.3 \times 0.6 = 0.18$ (ii) $P (A) P (B^{'}) = 0.3 \times 0.4 = 0.12$ (iii) $0.3 + 0.6 - 0.18 = 0.72$ (iv) $P (A^{'}) P (B^{'}) = 0.7 \times 0.4 = 0.28$

(i) 0.18 (ii) 0.12 (iii) 0.72 (iv) 0.28

A die is tossed thrice. Find the probability of getting an odd number at least once.

Solution

P(no odd number in three tosses) $= (\frac{1}{2})^{3} = \frac{1}{8}$ .

$1 - \frac{1}{8} = \frac{7}{8}$

Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that (i) both balls are red. (ii) first ball is black and second is red. (iii) one of them is black and other is red.

Solution

With replacement the draws are independent; P(B) $= \frac{10}{18}$ , P(R) $= \frac{8}{18}$ .

(i) $(\frac{8}{18})^{2} = \frac{16}{81}$ (ii) $\frac{10}{18} \cdot \frac{8}{18} = \frac{20}{81}$ (iii) BR or RB: $2 \cdot \frac{20}{81} = \frac{40}{81}$

(i) 16/81 (ii) 20/81 (iii) 40/81

Probability of solving specific problem independently by A and B are 1/2 and 1/3 respectively. If both try to solve the problem independently, find the probability that (i) the problem is solved (ii) exactly one of them solves the problem.

Solution

(i) $1 - P (A^{'}) P (B^{'}) = 1 - \frac{1}{2} \cdot \frac{2}{3} = \frac{2}{3}$

(ii) $P (A) P (B^{'}) + P (A^{'}) P (B) = \frac{1}{2} \cdot \frac{2}{3} + \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{3} + \frac{1}{6}$

(i) 2/3 (ii) 1/2

One card is drawn at random from a well shuffled deck of 52 cards. In which of the following cases are the events E and F independent? (i) E: 'the card drawn is a spade', F: 'the card drawn is an ace' (ii) E: 'the card drawn is black', F: 'the card drawn is a king' (iii) E: 'the card drawn is a king or queen', F: 'the card drawn is a queen or jack'.

Solution

(i) $P (E) = \frac{1}{4}$ , $P (F) = \frac{1}{13}$ , $P (E \cap F) = \frac{1}{52} = P (E) P (F)$ : independent.

(ii) $P (E) = \frac{1}{2}$ , $P (F) = \frac{1}{13}$ , $P (E \cap F) = \frac{2}{52} = \frac{1}{26} = P (E) P (F)$ : independent.

(iii) $P (E) = P (F) = \frac{8}{52} = \frac{2}{13}$ , $E \cap F$ = queens, $P = \frac{4}{52} = \frac{1}{13} \neq = \frac{4}{169}$ : not independent.

Independent in (i) and (ii); not in (iii).

In a hostel, 60% of the students read Hindi newspaper, 40% read English newspaper and 20% read both Hindi and English newspapers. A student is selected at random. (a) Find the probability that she reads neither Hindi nor English newspapers. (b) If she reads Hindi newspaper, find the probability that she reads English newspaper. (c) If she reads English newspaper, find the probability that she reads Hindi newspaper.

Solution

(a) $P (H \cup E) = 0.6 + 0.4 - 0.2 = 0.8$ ; neither $= 0.2$ .

(b) $P (E ∣ H) = \frac{0.2}{0.6} = \frac{1}{3}$ . (c) $P (H ∣ E) = \frac{0.2}{0.4} = \frac{1}{2}$ .

(a) 1/5 (b) 1/3 (c) 1/2

The probability of obtaining an even prime number on each die, when a pair of dice is rolled is (A) 0 (B) 1/3 (C) 1/12 (D) 1/36

Solution

The only even prime is 2, so the only favourable outcome is (2, 2).

(D)

Two events A and B will be independent, if (A) A and B are mutually exclusive (B) P(A′B′) = [1 – P(A)] [1 – P(B)] (C) P(A) = P(B) (D) P(A) + P(B) = 1

Solution

(B) says $P (A^{'} \cap B^{'}) = P (A^{'}) P (B^{'})$ , which holds exactly when A and B are independent.

(B)

Exercise 13.3

An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is noted and is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random. What is the probability that the second ball is red?

Solution

If the first is red (prob $\frac{1}{2}$ ), the urn has 7 R, 5 B; if black ( $\frac{1}{2}$ ), it has 5 R, 7 B.

$P (2nd red) = \frac{1}{2} \cdot \frac{7}{12} + \frac{1}{2} \cdot \frac{5}{12} = \frac{12}{24}$

$\frac{1}{2}$

A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag.

Solution

$P (E_{1}) = P (E_{2}) = \frac{1}{2}$ ; $P (R ∣ E_{1}) = \frac{4}{8} = \frac{1}{2}$ , $P (R ∣ E_{2}) = \frac{2}{8} = \frac{1}{4}$ .

$P (E_{1} ∣ R) = \frac{\frac{1}{2} \cdot \frac{1}{2}}{\frac{1}{2} \cdot \frac{1}{2} + \frac{1}{2} \cdot \frac{1}{4}} = \frac{\frac{1}{4}}{\frac{3}{8}}$

$\frac{2}{3}$

Of the students in a college, it is known that 60% reside in hostel and 40% are day scholars (not residing in hostel). Previous year results report that 30% of all students who reside in hostel attain A grade and 20% of day scholars attain A grade in their annual examination. At the end of the year, one student is chosen at random from the college and he has an A grade, what is the probability that the student is a hostlier?

Solution

$P (H ∣ A) = \frac{0.6 \times 0.3}{0.6 \times 0.3 + 0.4 \times 0.2} = \frac{0.18}{0.26}$

$\frac{9}{13}$

In answering a question on a multiple choice test, a student either knows the answer or guesses. Let 3/4 be the probability that he knows the answer and 1/4 be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability 1/4. What is the probability that the student knows the answer given that he answered it correctly?

Solution

$P (K ∣ C) = \frac{\frac{3}{4} \cdot 1}{\frac{3}{4} \cdot 1 + \frac{1}{4} \cdot \frac{1}{4}} = \frac{\frac{12}{16}}{\frac{13}{16}}$

$\frac{12}{13}$

A laboratory blood test is 99% effective in detecting a certain disease when it is in fact, present. However, the test also yields a false positive result for 0.5% of the healthy person tested. If 0.1 percent of the population actually has the disease, what is the probability that a person has the disease given that his test result is positive?

Solution

$P (D) = 0.001$ , $P (+ ∣ D) = 0.99$ , $P (+ ∣ D^{'}) = 0.005$ .

$P (D ∣ +) = \frac{0.001 \times 0.99}{0.001 \times 0.99 + 0.999 \times 0.005} = \frac{0.00099}{0.005985} = \frac{990}{5985}$

$\frac{22}{133}$

There are three coins. One is a two headed coin, another is a biased coin that comes up heads 75% of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two headed coin?

Solution

$P (E_{1} ∣ H) = \frac{\frac{1}{3} \cdot 1}{\frac{1}{3} ( 1 + \frac{3}{4} + \frac{1}{2} )} = \frac{1}{\frac{9}{4}}$

$\frac{4}{9}$

An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of an accidents are 0.01, 0.03 and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?

Solution

Priors: $\frac{2000}{12000} = \frac{1}{6}$ , $\frac{1}{3}$ , $\frac{1}{2}$ .

$P (S ∣ A) = \frac{\frac{1}{6} ( 0.01 )}{\frac{1}{6} ( 0.01 ) + \frac{1}{3} ( 0.03 ) + \frac{1}{2} ( 0.15 )} = \frac{0.01}{0.01 + 0.06 + 0.45} = \frac{0.01}{0.52}$

$\frac{1}{52}$

A factory has two machines A and B. Past record shows that machine A produced 60% of the items of output and machine B produced 40% of the items. Further, 2% of the items produced by machine A and 1% produced by machine B were defective. All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective. What is the probability that it was produced by machine B?

Solution

$P (B ∣ D) = \frac{0.4 \times 0.01}{0.6 \times 0.02 + 0.4 \times 0.01} = \frac{0.004}{0.016}$

$\frac{1}{4}$

Two groups are competing for the position on the Board of directors of a corporation. The probabilities that the first and the second groups will win are 0.6 and 0.4 respectively. Further, if the first group wins, the probability of introducing a new product is 0.7 and the corresponding probability is 0.3 if the second group wins. Find the probability that the new product introduced was by the second group.

Solution

$P (G_{2} ∣ N) = \frac{0.4 \times 0.3}{0.6 \times 0.7 + 0.4 \times 0.3} = \frac{0.12}{0.54}$

$\frac{2}{9}$

Suppose a girl throws a die. If she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she gets 1, 2, 3 or 4, she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw 1, 2, 3 or 4 with the die?

Solution

$P (E_{1}) = \frac{2}{6} = \frac{1}{3}$ (5 or 6), $P (E_{2}) = \frac{2}{3}$ (1–4). P(exactly one head | three tosses) $= \frac{3}{8}$ ; P(one head | one toss) $= \frac{1}{2}$ .

$P (E_{2} ∣ A) = \frac{\frac{2}{3} \cdot \frac{1}{2}}{\frac{1}{3} \cdot \frac{3}{8} + \frac{2}{3} \cdot \frac{1}{2}} = \frac{\frac{1}{3}}{\frac{1}{8} + \frac{1}{3}} = \frac{\frac{1}{3}}{\frac{11}{24}}$

$\frac{8}{11}$

A manufacturer has three machine operators A, B and C. The first operator A produces 1% defective items, where as the other two operators B and C produce 5% and 7% defective items respectively. A is on the job for 50% of the time, B is on the job for 30% of the time and C is on the job for 20% of the time. A defective item is produced, what is the probability that it was produced by A?

Solution

$P (A ∣ D) = \frac{0.5 \times 0.01}{0.5 \times 0.01 + 0.3 \times 0.05 + 0.2 \times 0.07} = \frac{0.005}{0.005 + 0.015 + 0.014} = \frac{0.005}{0.034}$

$\frac{5}{34}$

A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond.

Solution

$P (E_{1}) = \frac{13}{52} = \frac{1}{4}$ (lost card a diamond), $P (E_{2}) = \frac{3}{4}$ .

$P (A ∣ E_{1}) = \frac{^{12} C _{2}}{^{51} C _{2}}$ , $P (A ∣ E_{2}) = \frac{^{13} C _{2}}{^{51} C _{2}}$ ; $^{12} C_{2} = 66$ , $^{13} C_{2} = 78$ .

$P (E_{1} ∣ A) = \frac{\frac{1}{4} \cdot 66}{\frac{1}{4} \cdot 66 + \frac{3}{4} \cdot 78} = \frac{66}{66 + 234} = \frac{66}{300}$

$\frac{11}{50}$

Probability that A speaks truth is 4/5. A coin is tossed. A reports that a head appears. The probability that actually there was head is (A) 4/5 (B) 1/2 (C) 1/5 (D) 2/5

Solution

$P (H ∣ reports H) = \frac{\frac{1}{2} \cdot \frac{4}{5}}{\frac{1}{2} \cdot \frac{4}{5} + \frac{1}{2} \cdot \frac{1}{5}} = \frac{4}{5}$

(A)

If A and B are two events such that A ⊂ B and P(B) ≠ 0, then which of the following is correct? (A) P(A|B) = P(B)/P(A) (B) P(A|B) < P(A) (C) P(A|B) ≥ P(A) (D) None of these

Solution

$A \subset B \Rightarrow A \cap B = A$ , so $P (A ∣ B) = \frac{P ( A )}{P ( B )} \geq P (A)$ as $P (B) \leq 1$ .

(C)

Miscellaneous Exercise on Chapter 13

A and B are two events such that P (A) ≠ 0. Find P(B|A), if (i) A is a subset of B (ii) A ∩ B = φ

Solution

(i) $A \cap B = A$ , so $P (B ∣ A) = \frac{P ( A )}{P ( A )} = 1$ . (ii) $P (A \cap B) = 0$ , so $P (B ∣ A) = 0$ .

(i) 1 (ii) 0

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.

Solution

S = {MM, MF, FM, FF} (elder first).

(i) At least one male: {MM, MF, FM}; both male: {MM}: $\frac{1}{3}$ .

(ii) Elder female: {FM, FF}; both female: {FF}: $\frac{1}{2}$ .

(i) 1/3 (ii) 1/2

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.

Solution

$P (M ∣ G) = \frac{\frac{1}{2} ( 0.05 )}{\frac{1}{2} ( 0.05 ) + \frac{1}{2} ( 0.0025 )} = \frac{0.05}{0.0525} = \frac{500}{525}$

$\frac{20}{21}$

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?

Solution

Each person is right-handed with $p = 0.9$ , independently. Let X = number of right-handed people among 10: $P (X = r) =^{10} C_{r} (0.9)^{r} (0.1)^{10 - r}$ .

$P (X \leq 6) = 1 - P (X \geq 7) = 1 - \sum_{r = 7}^{10}^{10} C_{r} (0.9)^{r} (0.1)^{10 - r}$

$= 1 - (0.0574 + 0.1937 + 0.3874 + 0.3487) = 1 - 0.9872$

$1 - \sum_{r = 7}^{10}^{10} C_{r} (0.9)^{r} (0.1)^{10 - r} \approx 0.0128$

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

Solution

366 days = 52 weeks + 2 days. The 2 extra days are one of the 7 pairs (Sun–Mon, Mon–Tue, Tue–Wed, Wed–Thu, Thu–Fri, Fri–Sat, Sat–Sun). Two of them contain a Tuesday.

$\frac{2}{7}$

Four boxes contain marbles (Red, White, Black): A (1, 6, 3), B (6, 2, 2), C (8, 1, 1), D (0, 6, 4). 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?

Solution

Each box has 10 marbles and prior $\frac{1}{4}$ . $P (R ∣ A) = \frac{1}{10}$ , $P (R ∣ B) = \frac{6}{10}$ , $P (R ∣ C) = \frac{8}{10}$ , $P (R ∣ D) = 0$ . Sum of likelihoods $= \frac{15}{10}$ .

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

1/15, 2/5, 8/15

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 chances 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?

Solution

$P (H ∣ Y) = 0.4 \times 0.7 = 0.28$ , $P (H ∣ D) = 0.4 \times 0.75 = 0.30$ , $P (Y) = P (D) = \frac{1}{2}$ .

$P (Y ∣ H) = \frac{\frac{1}{2} ( 0.28 )}{\frac{1}{2} ( 0.28 ) + \frac{1}{2} ( 0.30 )} = \frac{28}{58}$

$\frac{14}{29}$

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? (Assume that the individual entries of the determinant are chosen independently, each value being assumed with probability 1/2).

Solution

$a c b d = a d - b c > 0$ needs $a d = 1$ and $b c = 0$ . There are $2^{4} = 16$ equally likely matrices. $a = d = 1$ and $(b, c) \in {(0, 0), (0, 1), (1, 0)}$ : 3 matrices.

$\frac{3}{16}$

An electronic assembly consists of two subsystems, say, A and B. P(A fails) = 0.2, P(B fails alone) = 0.15, P(A and B fail) = 0.15. Evaluate (i) P(A fails|B has failed) (ii) P(A fails alone)

Solution

$P (B fails) = P (B alone) + P (A and B) = 0.15 + 0.15 = 0.3$ .

(i) $\frac{0.15}{0.3} = 0.5$ (ii) $P (A) - P (A \cap B) = 0.2 - 0.15 = 0.05$

(i) 0.5 (ii) 0.05

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.

Solution

Transferred red ( $\frac{3}{7}$ ): Bag II has 5 R, 5 B, $P (R) = \frac{5}{10}$ . Transferred black ( $\frac{4}{7}$ ): 4 R, 6 B, $P (R) = \frac{4}{10}$ .

$P (black ∣ R) = \frac{\frac{4}{7} \cdot \frac{4}{10}}{\frac{3}{7} \cdot \frac{5}{10} + \frac{4}{7} \cdot \frac{4}{10}} = \frac{16}{15 + 16}$

$\frac{16}{31}$

If A and B are two events such that P(A) ≠ 0 and P(B | A) = 1, then (A) A ⊂ B (B) B ⊂ A (C) B = φ (D) A = φ

Solution

$P (A \cap B) = P (A)$ , which holds when $A \subset B$ .

(A)

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

Solution

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

(C)

If A and B are any two events such that P(A) + P(B) – P(A 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

Solution

$P (B) = P (A \cap B)$ , so $P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )} = 1$ .

(B)

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