Chapter 13 Probability

Frequently Asked Questions

NCERT Solutions Class 12 Maths Chapter 13 Probability

Chapter 13 Probability from Class 12 Maths is about helping students to understand how to measure uncertainty and predict outcomes using mathematical reasoning. Using the earlier fundamentals on probability, this chapter introduces advanced ideas including conditional probability, multiplication theorem, independent events, and Bayes’ theorem. These topics are important for board exams and are also useful in certain real-life decision-making situations.

ALLEN provides NCERT Solutions for Class 12 Maths Chapter 13 Probability based on the latest CBSE syllabus. The solutions break down probability questions into clear, step-by-step explanations, making topics such as conditional probability, Bayes’ theorem, and other important concepts easier to understand. With the help of these solutions, students may determine the proper approach for various problems, increase their accuracy, and efficiently prepare for CBSE board exams and competitive examinations

1.0Key Concepts of Class 12 Maths Chapter 13 Probability

Class 12 Maths Chapter 13 Probability covers important methods for finding the probability of events and solving questions based on different conditions. The main topics include:

Conditional Probability: Finding the probability of an event when the occurrence of another event is already known.
Multiplication Theorem of Probability: Calculating the probability of two or more events occurring together.
Independent Events: Understanding events where the occurrence of one event does not change the probability of another event.
Bayes’ Theorem: Finding the probability of an event when related information or conditions are given.
Total Probability Theorem: Calculating the probability of an event by considering different mutually exclusive cases.
Applications of Probability: Solving questions based on probability in different situations and practical examples.

2.0NCERT Class 12 Maths Chapter 13 Probability : Detailed Solutions

EXERCISE - 13.1

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 $P (F ∣ E)$ . Sol. It is given that $P (E) = 0.6, P (F) = 0.3$ , and $P (E \cap F) = 0.2$ $\Rightarrow P (E ∣ F) = \frac{P ( E \cap F )}{P ( F )} = \frac{0.2}{0.3} = \frac{2}{3}$ $\Rightarrow P (F ∣ E) = \frac{P ( F \cap E )}{P ( E )} = \frac{0.2}{0.6} = \frac{1}{3}$
Compute $P (A ∣ B)$ , if $P (B) = 0.5$ and $P (A \cap B) = 0.32$ Sol. It is given that $P (B) = 0.5$ and $P (A \cap B) = 0.32$ $\Rightarrow P (\frac{A}{B}) = \frac{P ( A \cap B )}{P ( B )} = \frac{0.32}{0.5} = \frac{16}{25}$
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) $P (A \cup B)$ Sol. It is given that $P (A) = 0.8, P (B) = 0.5$ , and $P (B ∣ A) = 0.4$ (i) $P (B ∣ A) = 0.4$

$∴ \Rightarrow \Rightarrow \frac{P ( B \cap A )}{P ( A )} = 0.4 \frac{P ( A \cap B )}{0.8} = 0.4 P (A \cap B) = 0.32 [∵ P (\frac{B}{A}) = \frac{P ( B \cap A )}{P ( A )}] [∵ B \cap A = A \cap B]$

(ii) $P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}$

$[∵ P (\frac{A}{B}) = \frac{P ( A \cap B )}{P ( B )}]$

$\Rightarrow P (A ∣ B) = \frac{0.32}{0.5} = 0.64$ (iii) $P (A \cup B) = P (A) + P (B) - P (A \cap B)$ $\Rightarrow P (A \cup B) = 0.8 + 0.5 - 0.32 = 0.98$

Evaluate $P (A \cup B)$ , if 2P(A) = P(B) = $\frac{5}{13}$ and $P (A ∣ B) = \frac{2}{5}$ Sol. It is given that, $2 P (A) = P (B) = \frac{5}{13}$ $\Rightarrow P (A) = \frac{5}{26}$ and $P (B) = \frac{5}{13}$ Now, $P (A ∣ B) = \frac{2}{5} \Rightarrow \frac{P ( A \cap B )}{P ( B )} = \frac{2}{5}$

$[∵ P (\frac{A}{B}) = \frac{P ( A \cap B )}{P ( B )}]$

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

It is known that,

$P (A \cup B) = P (A) + P (B) - P (A \cap B)$

$\Rightarrow P (A \cup B) = \frac{5}{26} + \frac{5}{13} - \frac{2}{13}$ $\Rightarrow P (A \cup B) = \frac{5 + 10 - 4}{26}$ $\Rightarrow P (A \cup B) = \frac{11}{26}$

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) $P (B ∣ A)$ Sol. It is given that $P (A) = \frac{6}{11}, P (B) = \frac{5}{11}$ and

$P (A \cup B) = \frac{7}{11}$

(i) $P (A \cup B) = \frac{7}{11}$

$[∵ P (A \cup B) = P (A) + P (B) - P (A \cap B)]$

$∴ P (A) + P (B) - P (A \cap B) = \frac{7}{11}$ $\Rightarrow \frac{6}{11} + \frac{5}{11} - P (A \cap B) = \frac{7}{11}$ $\Rightarrow P (A \cap B) = \frac{11}{11} - \frac{7}{11} = \frac{4}{11}$

(ii) It is known that, $P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}$ $\Rightarrow P (A ∣ B) = \frac{\frac{4}{11}}{\frac{5}{11}} = \frac{4}{5}$ (iii) It is known that, $P (B ∣ A) = \frac{P ( B \cap A )}{P ( A )}$ $\Rightarrow P (B ∣ A) = \frac{\frac{4}{11}}{\frac{6}{11}} = \frac{4}{6} = \frac{2}{3}$

Determine $P (E ∣ F)$ in Q. 6 to 9.

A coin is tossed three times, where

(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

Sol. If a coin is tossed three times, then the sample space S is $S = {HHH, HHT, HTH, HTT, THH, THT$ , TTH, TTT}

$\Rightarrow n (S) = 8$

(i) $E = {HHH, HTH, THH, TTH}$ $F = {HHH, HHT}$ $∴ E \cap F = {HHH}$ $∴ P (F) = \frac{2}{8} = \frac{1}{4} and P (E \cap F) = \frac{1}{8}$ $∴ P (E ∣ F) = \frac{P ( E \cap F )}{P ( F )} = \frac{\frac{1}{8}}{\frac{1}{4}} = \frac{4}{8} = \frac{1}{2}$ (ii) $E = {HHH, HHT, HTH, THH}$ F = {HHT, HTH, HTT, THH, THT, TTH, TTT} $∴ E \cap F = {HHT, HTH, THH}$ Clearly, $P (E \cap F) = \frac{3}{8}$ and $P (F) = \frac{7}{8}$

$P (E ∣ F) = \frac{P ( E \cap F )}{P ( F )} = \frac{\frac{3}{8}}{\frac{7}{8}} = \frac{3}{7}$

(iii) $E = {HHH, HHT, HTT, HTH, THH, THT, TTH}$ F = {HHT, HTT, HTH, THH, THT, TTH, TTT} $∴ E \cap F = {HHT, HTT, HTH, THH, THT, TTH}$ $P (F) = \frac{7}{8}$ and $P (E \cap F) = \frac{6}{8}$ Therefore, $P (E ∣ F) = \frac{P ( E \cap F )}{P ( F )} = \frac{\frac{6}{8}}{\frac{7}{8}} = \frac{6}{7}$

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

Sol. If two coins are tossed once, then the sample space S is $S = {HH, HT, TH, TT}$

(i) $E = {HT, TH}, F = {HT, TH}$ $∴ E \cap F = {HT, TH}$

$P (F) = \frac{2}{4} = \frac{1}{2} P (E \cap F) = \frac{2}{4} = \frac{1}{2}$

$\Rightarrow P (E ∣ F) = \frac{P ( E \cap F )}{P ( F )} = \frac{\frac{1}{2}}{\frac{1}{2}} = 1$ (ii) $E = {HH}$ [Set of events having no tail] $F = {TT}$ [Set of events having no head] $∴ E \cap F = ϕ$ $P (E) = \frac{1}{4}, P (F) = \frac{1}{4}$ and $P (E \cap F) = 0$ $∴ P (E ∣ F) = \frac{P ( E \cap F )}{P ( F )} = \frac{0}{1/4} = 0$

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

Sol. If a die is thrown three times, then the number of elements in the sample space will be

$6 \times 6 \times 6 = 216$

The sample space is

$S = E = {(x, y, z) : x, y, z \in 1, 2, 3, 4, 5, 6} ⎩ ⎨ ⎧ (1, 1, 4), (1, 2, 4), \dots .. (1, 6, 4) (2, 1, 4), (2, 2, 4), \dots ... (2, 6, 4) (3, 1, 4), (3, 2, 4), \dots .. (3, 6, 4) (4, 1, 4), (4, 2, 4), \dots ... (4, 6, 4) (5, 1, 4), (5, 2, 4), \dots ... (5, 6, 4) (6, 1, 4), (6, 2, 4), \dots .. (6, 6, 4) ⎭ ⎬ ⎫$

$\Rightarrow n (E) = 36 F = {(6, 5, 1), (6, 5, 2), (6, 5, 3), (6, 5, 4), (6, 5, 5), (6, 5, 6)}$

$\Rightarrow P (E ∣ F) = \frac{P ( E \cap F )}{P ( F )} = \frac{\frac{1}{216}}{\frac{6}{216}} = \frac{1}{6}$

Mother, father and son line up at random for a family picture, E : son on one end, F: father in middle. Sol. If mother (M), father (F), and son (S) line up for the family picture, then the sample space will be $S = {MFS, MSF, FMS, FSM, SMF, SFM}$

$\Rightarrow E F = {MFS, FMS, SMF, SFM}, = {MFS, SFM}$

$\Rightarrow E \cap F = {MFS, SFM} = 2 P (E \cap F) = \frac{2}{6} = \frac{1}{3} and P (F) = \frac{2}{6} = \frac{1}{3}$

$\Rightarrow P (E ∣ F) = \frac{P ( E \cap F )}{P ( F )} = \frac{\frac{1}{3}}{\frac{1}{3}} = 1$

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. Sol. Let the first observation be from the black die and second from the red die. When two dice (one black and another red) are rolled, the sample space $S = 6 \times 6 = 36$ number of elements. (a) Let A : Obtaining a sum greater than 9 $= {(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)}$ B : Black die results in a 5 $= {(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}$

$∴ A \cap B = {(5, 5), (5, 6)}$

The conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5, is given by $P (A ∣ B)$ .

$∴ P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )} = \frac{\frac{2}{36}}{\frac{6}{36}} = \frac{2}{6} = \frac{1}{3}$

(b) E : Sum of the observations is $8 = {(2, 6), (3, 5)$ , (4, 4), (5, 3), (6, 2)} F : Red die resulted in a number less than 4

$= ⎩ ⎨ ⎧ (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (5, 3), (6, 1), (6, 2), (6, 3) ⎭ ⎬ ⎫$

$∴ E \cap F = {(5, 3), (6, 2)}$ $P (F) = \frac{18}{36}$ and $P (E \cap F) = \frac{2}{36}$ The conditional probability of obtaining the sum equal to 8, given that the red die resulted in a number less than 4, is given by P (E | F). Therefore, $P (E ∣ F) = \frac{P ( E \cap F )}{P ( F )} = \frac{\frac{2}{36}}{\frac{18}{36}} = \frac{2}{18} = \frac{1}{9}$

A fair die is rolled. Consider events $E = {1, 3, 5}, F = {2, 3}$ and $G = {2, 3, 4, 5}$ . Find (i) $P (E ∣ F)$ and $P (F ∣ E)$ (ii) $P (E ∣ G)$ and $P (G ∣ E)$ (iii) $P ((E \cup F) ∣ G)$ and $P ((E \cap F) ∣ G)$

Sol. When a fair die is rolled, the sample space S will be $S = {1, 2, 3, 4, 5, 6}$ It is given that

$E = {1, 3, 5}, F = {2, 3} and G = {2, 3, 4, 5}$

$∴ P (E) = \frac{3}{6} = \frac{1}{2}, P (F) = \frac{2}{6} = \frac{1}{3}, P (G) = \frac{4}{6} = \frac{2}{3}$ (i) $E \cap F = {3} \Rightarrow P (E \cap F) = \frac{1}{6}$ $∴ P (E ∣ F) = \frac{P ( E \cap F )}{P ( F )} = \frac{\frac{1}{6}}{\frac{1}{3}} = \frac{1}{2};$

$P (F ∣ E) = \frac{P ( F \cap E )}{P ( E )} = \frac{\frac{1}{6}}{\frac{1}{2}} = \frac{1}{3}$

(ii) $E \cap G = {3, 5} ∴ P (E \cap G) = \frac{2}{6} = \frac{1}{3}$ $∴ P (E ∣ G) = \frac{P ( E \cap G )}{P ( G )} = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2};$

$P (G ∣ E) = \frac{P ( G \cap E )}{P ( E )} = \frac{\frac{1}{3}}{\frac{1}{2}} = \frac{2}{3}$

(iii) $E \cup F = {1, 2, 3, 5}$

$(E \cup F) \cap G = {1, 2, 3, 5} \cap {2, 3, 4, 5} = {2, 3, 5}$

$E \cap F = {3}$

$(E \cap F) \cap G = {3} \cap {2, 3, 4, 5} = {3}$

$∴ P ((E \cup F) \cap G) P ((E \cap F) \cap G) P ((E \cap F) ∣ G) = \frac{3}{6} = \frac{1}{2}, = \frac{1}{6} = \frac{P (( E \cap F ) \cap G )}{P ( G )} = \frac{\frac{1}{6}}{\frac{2}{3}} = \frac{1}{6} \times \frac{3}{2} = \frac{1}{4}$

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? Sol. Let b and g represent the boy and the girl child respectively. If a family has two children, the sample space will be

$S = {(b, b), (b, g), (g, b), (g, g)}$

Let A be the event that both children are girls. $∴ A = {(g, g)}$ (i) Let B be the event that the youngest child is a girl.

$∴ B = [(b, g), (g, g)]$

$\Rightarrow A \cap B = {(g, g)}$

$∴ P (B) = \frac{2}{4} = \frac{1}{2}, P (A \cap B) = \frac{1}{4}$

The conditional probability that both are girls, given that the youngest child is a girl, is given by $P (A ∣ B)$ .

$P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )} = \frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{2}$

Therefore, the required probability is $\frac{1}{2}$ .

(ii) Let C be the event that at least one child is a girl. $∴ C = {(b, g), (g, b), (g, g)}$ $\Rightarrow A \cap C = {g, g}$ $\Rightarrow P (C) = \frac{3}{4} \Rightarrow P (A \cap C) = \frac{1}{4}$ The conditional probability that both are girls, given that at least one child is a girl, is given by $P (A ∣ C)$ . Therefore, $P (A ∣ C) = \frac{P ( A \cap C )}{P ( C )} = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3}$

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? Sol. The given data can be tabulated as

	True/False	Multiple Choice	Total
Easy	300	500	800
Difficult	200	400	600
Total	500	900	1400

Let us denote $E =$ easy questions, $M =$ multiple choice questions, D = difficult questions, and T = True/False questions Total number of questions $= 1400$ Total number of multiple choice questions = 900 Therefore, probability of selecting an easy multiple choice question is

$P (E \cap M) = \frac{500}{1400} = \frac{5}{14}$

Probability of selecting a multiple choice question,

$P (M) = \frac{900}{1400} = \frac{9}{14}$

$P (E ∣ M)$ represents the probability that a randomly selected question will be an easy question, given that it is a multiple choice question.

$∴ P (E ∣ M) = \frac{P ( E \cap M )}{P ( M )} = \frac{\frac{5}{14}}{\frac{9}{14}} = \frac{5}{9}$ Therefore, the required probability is $\frac{5}{9}$

Given that the two numbers appearing on throwing the two dice are different. Find the probability of the event 'the sum of numbers on the dice is 4'. Sol. When two dice are thrown, number of observations in the sample space $= 6 \times 6 = 36$ Let A be the event that the sum of the numbers on both dice is 4 and B be the event that the two numbers appearing on throwing the two dice are different. $∴ A = {(1, 3), (2, 2), (3, 1)}$ $\Rightarrow n (A) = 3$ $B = ⎩ ⎨ ⎧ (1, 2), (1, 3), (1, 4), (1, 5), (1, 6) (2, 1), (2, 3), (2, 4), (2, 5), (2, 6) (3, 1), (3, 2), (3, 4), (3, 5), (3, 6) (4, 1), (4, 2), (4, 3), (4, 5), (4, 6) (5, 1), (5, 2), (5, 3), (5, 4), (5, 6) (6, 1), (6, 2), (6, 3), (6, 4), (6, 5) ⎭ ⎬ ⎫ = n (B) = 30$ $A \cap B = {(1, 3), (3, 1)}$ $∴ P (B) = \frac{30}{36} = \frac{5}{6}$ and $P (A \cap B) = \frac{2}{36} = \frac{1}{18}$

Let $P (A ∣ B)$ represent the probability that the sum of the numbers on both dice is 4, given that the two numbers appearing on throwing the two dice are different.

$∴ P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )} = \frac{\frac{1}{18}}{\frac{5}{6}} = \frac{1}{15}$

Therefore, the required probability is $\frac{1}{15}$ .

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'. Sol. The sample space of the experiment is,

$S = ⎩ ⎨ ⎧ (1, H), (1, T), (2, H), (2, T), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6) (4, H), (4, T), (5, H), (5, T), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) ⎭ ⎬ ⎫$

Let A be the event that the coin shows a tail and B be the event that at least one die shows 3.

$∴ A = {(1, T), (2, T), (4, T), (5, T)} B = {(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (6, 3)}$

because here are no common elements. Then,

$P (B) = P ({3, 1}) + P ({3, 2}) + P ({3, 3}) + P ({3, 4}) + P ({3, 5}) + P ({3, 6}) + P ({6, 3}) = \frac{1}{36} + \frac{1}{36} + \frac{1}{36} + \frac{1}{36} + \frac{1}{36} + \frac{1}{36} + \frac{1}{36} = \frac{7}{36}$

Probability of the event that the coin shows a tail, given that at least one die shows 3, is given by $P (A ∣ B)$ . Therefore, $P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )} = \frac{0}{\frac{7}{36}} = 0$ Choose the correct answer in Q. 16 and 17.

If $P (A) = \frac{1}{2}, P (B) = 0$ , then $P (A ∣ B)$ is (A) 0 (B) $\frac{1}{2}$ (C) not defined (D) 1 Sol. (C) It is given that $P (A) = \frac{1}{2}$ and $P (B) = 0$

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

Therefore, $P (A ∣ B)$ is not defined. Thus, the correct Answer is C.

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) $A \cap B = ϕ$ (D) $P (A) = P (B)$ Sol. (D) It is given that, $P (A ∣ B) = P (B ∣ A)$

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

Thus, the correct Answer is D.

EXERCISE 13.2

If $P (A) = \frac{3}{5}$ and $P (B) = \frac{1}{5}$ , find $P (A \cap B)$ if A and B are independent events. Sol. It is given that $P (A) = \frac{3}{5}$ and $P (B) = \frac{1}{5}$ A and B are independent events. Therefore,

$P (A \cap B) = P (A) \cdot P (B) = \frac{3}{5} \cdot \frac{1}{5} = \frac{3}{25}$

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. Sol. There are 26 black cards in a deck of 52 cards. Let A : event that $1^{st}$ card is black, B : event that $2^{nd}$ card is black. P(getting a black card in the first draw)

$= P (A) = \frac{26}{52} = \frac{1}{2}$

P (getting a black card on the second draw)

$= P (B / A) = \frac{25}{51} (∵ card is not replaced)$

Thus, P(getting both the cards black)

$= P (A \cap B) = P (A) \cdot P (B / A) = \frac{1}{2} \times \frac{25}{51} = \frac{25}{102}$

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. Sol. Let A, B, and C be the events respectively that the first, second, and third drawn orange is good. Therefore, probability that first drawn orange is good, $P (A) = \frac{12}{15}$

Since second orange is drawn without replacement (now total number of good oranges will be 11 and total oranges will be 14

∴ Conditional probability of B , given that A has already occurred is

$P (B / A) \Rightarrow P (B / A) = \frac{11}{14}$

Again, the third orange is drawn without replacement (now total number of good oranges will be 10 and total number of oranges is 13 ). ∴ Conditional probability of C, given that A & B have already occurred is

$P (C / AB) \Rightarrow P (C / AB) = \frac{10}{13}$

The box is approved for sale, if all the three oranges are good. Thus, probability of getting all the oranges good

$= P (A \cap B \cap C) = P (A) \cdot P (B / A) \cdot P (C / AB) = \frac{12}{15} \times \frac{11}{14} \times \frac{10}{13} = \frac{44}{91}$

Therefore, the probability that the box is approved for sale is $\frac{44}{91}$

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. Sol. If a fair coin and an unbiased die are tossed, then the sample space S is given by,

$S = {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6) (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)}$

$\Rightarrow n (S) = 12$

Let

$A: Head appears on the coin A = {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6)}$

$\Rightarrow P (A) = \frac{6}{12} = \frac{1}{2} B : 3 on die = {(H, 3), (T, 3)} P (B) = \frac{2}{12} = \frac{1}{6}$

$∴ A \cap B = {(H, 3)}$

$P (A \cap B) = \frac{1}{12}$

Also, $P (A) \cdot P (B) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12} = P (A \cap B)$ Therefore, A and B are independent events.

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? Sol. When a die is thrown, the sample space (S) is $S = {1, 2, 3, 4, 5, 6}$ Let A: the number is even $= {2, 4, 6}$

$\Rightarrow P (A) = \frac{3}{6} = \frac{1}{2}$

B: the number is red $= {1, 2, 3}$

$\Rightarrow P (B) = \frac{3}{6} = \frac{1}{2}$

$∴ A \cap B = {2}$

$\Rightarrow P (A \cap B) = \frac{1}{6}$

Also, $P (A) \cdot P (B) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \neq = \frac{1}{6} \neq = P (A \cap B)$

$\Rightarrow P (A) \cdot P (B) \neq = P (A \cap B)$

Therefore, A and B are not independent.

Let E and F be events with $P (E) = \frac{3}{5}, P (F) = \frac{3}{10}$ and $P (E \cap F) = \frac{1}{5}$ . Are E and F independent? Sol. It is given that $P (E) = \frac{3}{5}, P (F) = \frac{3}{10}$ and

$P (E \cap F) = \frac{1}{5}$

Now, $P (E) \cdot P (F) = \frac{3}{5} \cdot \frac{3}{10} = \frac{9}{50} \neq = \frac{1}{5}$

$\Rightarrow P (E \cap F)$

$\Rightarrow P (E) \cdot P (F) \neq = P (E \cap F)$ Therefore, E and F are not independent.

Given that the events A and B are such that $P (A) = \frac{1}{2}, P (A \cup B) = \frac{3}{5}$ and $P (B) = p$ . Find p if they are (i) mutually exclusive (ii) independent.

Sol. It is given that $P (A) = \frac{1}{2}, P (A \cup B) = \frac{3}{5}$ and $P (B) = p$ (i) When A and B are mutually exclusive, $A \cap B = ϕ$

$∴ P (A \cap B) = 0$

It is known that,

$P (A \cup B) = P (A) + P (B) - P (A \cap B)$

$\Rightarrow \frac{3}{5} = \frac{1}{2} + p - 0 \Rightarrow p = \frac{3}{5} - \frac{1}{2} = \frac{1}{10}$

(ii) When A and B are independent, events then,

$P (A \cap B) = P (A) \cdot P (B) = \frac{1}{2} p$

It is known that,

$P (A \cup B) = P (A) + P (B) - P (A \cap B)$

$\Rightarrow \frac{3}{5} = \frac{1}{2} + p - \frac{1}{2} p \Rightarrow \frac{3}{5} = \frac{1}{2} + \frac{p}{2}$

$\Rightarrow \frac{p}{2} = \frac{3}{5} - \frac{1}{2} = \frac{1}{10} \Rightarrow p = \frac{2}{10} = \frac{1}{5}$

Let A and B be independent events with $P (A) = 0.3$ and $P (B) = 0.4$ . Find (i) $P (A \cap B)$ , (ii) $P (A \cup B)$ ,

$(iii) P (A ∣ B), (iv) P (B ∣ A)$

Sol. It is given that $P (A) = 0.3$ and $P (B) = 0.4$

(i) If A and B are independent events, then

$P (A \cap B) = P (A) \cdot P (B) = 0.3 \times 0.4 = 0.12$

(ii) $P (A \cup B) = P (A) + P (B) - P (A \cap B)$

$\Rightarrow P (A \cup B) = 0.3 + 0.4 - 0.12 = 0.58$

(iii) It is known that,

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

$\Rightarrow P (A ∣ B) = \frac{0.12}{0.4} = 0.3$

(iv) It is known that,

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

$\Rightarrow P (B ∣ A) = \frac{0.12}{0.3} = 0.4$

If A and B are two events such that $P (A) = \frac{1}{4}$ ,

$P (B) = \frac{1}{2} and P (A \cap B) = \frac{1}{8}, find P (not A and$

not B). Sol. It is given that $P (A) = \frac{1}{4}, P (B) = \frac{1}{2}$ and

$P (A \cap B) = \frac{1}{8}$

$P ($ not A and not B $) = P (A^{'} \cap B^{'})$

$P (not A and not B) = P ((A \cup B)^{'}) [∵ A^{'} \cap B^{'} = (A \cup B)^{'}] = 1 - P (A \cup B) = 1 - [P (A) + P (B) - P (A \cap B)] = 1 - [\frac{1}{4} + \frac{1}{2} - \frac{1}{8}] = 1 - \frac{5}{8} = \frac{3}{8}$

Events A and B are such that $P (A) = \frac{1}{2}$ ,

$P (B) = \frac{7}{12} and P (not A or not B) = \frac{1}{4} . State$

whether A and B are independent? Sol. It is given that $P (A) = \frac{1}{2}, P (B) = \frac{7}{12}$

$and P (not A or not B) = \frac{1}{4} .$

$\Rightarrow P (A^{'} \cup B^{'}) = \frac{1}{4}$

$\Rightarrow P ((A \cap B)^{'}) = \frac{1}{4} [∵ A^{'} \cup B^{'} = (A \cap B)^{'}]$

$\Rightarrow 1 - P (A \cap B) = \frac{1}{4}$

$\Rightarrow P (A \cap B) = \frac{3}{4}$

$However, P (A) \cdot P (B) = \frac{1}{2} \cdot \frac{7}{12} = \frac{7}{24}$

Here, $\frac{3}{4} \neq = \frac{7}{24}$

$∴ P (A \cap B) \neq = P (A) \cdot P (B)$

Therefore, A and B are not independent events.

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) Sol. It is given that $P (A) = 0.3$ and $P (B) = 0.6$ Also, A and B are independent events. (i) $P (A$ and B $) = P (A) \cdot P (B)$ $\Rightarrow P (A \cap B) = 0.3 \times 0.6 = 0.18$ (ii) $P (A$ and not B $)$

$= P (A \cap B^{'}) = P (A) - P (A \cap B) = 0.3 - 0.18 = 0.12$

(iii) $P (A$ or B $) = P (A \cup B)$

$= P (A) + P (B) - P (A \cap B) = 0.3 + 0.6 - 0.18 = 0.72$

(iv) P (neither A nor B )

$= P (A^{'} \cap B^{'}) = P ((A \cup B)^{'}) = 1 - P (A \cup B) = 1 - 0.72 = 0.28$

A die is tossed thrice. Find the probability of getting an odd number at least once. Sol. Probability of getting an odd number in a single throw of a die $= \frac{3}{6} = \frac{1}{2}$ Similarly, probability of getting an even number $= \frac{3}{6} = \frac{1}{2}$ Probability of getting an even number three times $= \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$ Therefore, probability of getting an odd number at least once $= 1$ - Probability of getting an odd number in none of the throws = 1 - Probability of getting an even number thrice

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

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. Sol. Let R : event that drawn ball is red B : event that drawn ball is black $∴ P (R) = \frac{8}{18} = \frac{4}{9}$ and $P (B) = \frac{10}{18} = \frac{5}{9}$ (i) P(Both balls are red)

$= P (R \cap R) = P (R) \cdot P (R) (∵ Ball is replaced) = \frac{4}{9} \cdot \frac{4}{9} = \frac{16}{81}$

(ii) P(first ball is black and second is red)

$= P (B \cap R) = P (B) \cdot P (R) (∵ Ball is replaced) = \frac{5}{9} \cdot \frac{4}{9} = \frac{20}{81}$

(iii) P (one of them is black and other red)

$= P [(R \cap B) \cup (B \cap R)] = P (R \cap B) + P (B \cap R) = P (R) \cdot P (B) + P (B) \cdot P (R) (∵ Ball is replaced) = \frac{4}{9} \cdot \frac{5}{9} + \frac{5}{9} \cdot \frac{4}{9} = \frac{20}{81} + \frac{20}{81} = \frac{40}{81}$

Probability of solving specific problem independently by A and B are $\frac{1}{2}$ and $\frac{1}{3}$ 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. Sol. Let $E_{1}$ : event that A solves the problem $E_{2}$ : event that B solves the problem Then $P (E_{1}) = \frac{1}{2}$ and $P (E_{2}) = \frac{1}{3}$ $\Rightarrow P (\overline{E}_{1}) = 1 - \frac{1}{2} = \frac{1}{2}$ and $P (\overline{E}_{2}) = 1 - \frac{1}{3} = \frac{2}{3}$

Clearly, $E_{1}$ and $E_{2}$ are independent events :

(i) P(the problem is solved) $= P$ (atleast one of A and B solves the problem)

$= P (E_{1} \cup E_{2}) = P (E_{1}) + P (E_{2}) - P (E_{1} \cap E_{2}) = P (E_{1}) + P (E_{2}) - P (E_{1}) \cdot P (E_{2}) = \frac{1}{2} + \frac{1}{3} - \frac{1}{2} \cdot \frac{1}{3} = \frac{2}{3}$

(ii) P (exactly one of them solves the problem)

$= P [(E_{1} \cap \overset{ˉ}{E}_{2}) \cup (\overset{ˉ}{E}_{1} \cap E_{2})] = P (E_{1} \cap \overset{ˉ}{E}_{2}) + P (\overset{ˉ}{E}_{1} \cap E_{2}) = P (E_{1}) \cdot P (\overset{ˉ}{E}_{2}) + P (\overset{ˉ}{E}_{1}) \cdot P (E_{2}) = \frac{1}{2} \times \frac{2}{3} + \frac{1}{3} \times \frac{1}{2} = \frac{1}{2}$

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'

Sol.

(i) In a deck of 52 cards, 13 cards are spades and 4 cards are aces. $∴ P (E) = P ($ the card drawn is a spade $) = \frac{13}{52} = \frac{1}{4}$ $∴ P (F) = P (the card drawn is an ace) = \frac{4}{52} = \frac{1}{13}$

In the deck of cards, only 1 card is an ace of spades. P(E $\cap F)$ $= P ($ the card drawn is spade and an ace $) = \frac{1}{52}$ $P (E) \cdot P (F) = \frac{1}{4} \cdot \frac{1}{13} = \frac{1}{52} = P (E \cap F)$ $\Rightarrow P (E) \cdot P (F) = P (E \cap F)$ Therefore, the events E and F are independent.

(ii) In a deck of 52 cards, 26 cards are black and 4 cards are kings. $∴ P (E) = P (the card drawn is black) = \frac{26}{52} = \frac{1}{2}$ $∴ P (F) = P (the card drawn is a king) = \frac{4}{52} = \frac{1}{13}$ In the pack of 52 cards, 2 cards are black as well as kings. $∴ P (E \cap F) = P$ (the card drawn is a black king)

$= \frac{2}{52} = \frac{1}{26}$

Also, $P (E) \times P (F) = \frac{1}{2} \cdot \frac{1}{13} = \frac{1}{26} = P (E \cap F)$ Therefore, the given events E and F are independent. (iii) In a deck of 52 cards, 4 cards are kings, 4 cards are queens, and 4 cards are jacks. $∴ P (E) = P$ (the card drawn is a king or a queen)

$= \frac{8}{52} = \frac{2}{13}$

$∴ P (F) = P ($ the card drawn is a queen or a jack $)$

$= \frac{8}{52} = \frac{2}{13}$

There are 4 cards which are either king or queen and either queen or jack. $∴ P (E \cap F) = P$ (the card drawn is either king or queen and either queen or jack) $= \frac{4}{52} = \frac{1}{13}$ Now, $P (E) \times P (F) = \frac{2}{13} \cdot \frac{2}{13} = \frac{4}{169} \neq = \frac{1}{13}$ $\Rightarrow P (E) \cdot P (F) \neq = P (E \cap F)$ Therefore, the given events E and F are not independent.

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. (i) Find the probability that she reads neither Hindi nor English newspapers. (ii) If she reads Hindi newspaper, find the probability that she reads English newspaper. (iii) If she reads English newspaper, find the probability that she reads Hindi newspaper. Sol. Let H denote the students who read Hindi newspaper and E denote the students who read English newspaper.

It is given that,

$P (H) = 60% = \frac{6}{10} = \frac{3}{5}; P (E) = 40% = \frac{40}{100} = \frac{2}{5}; P (H \cap E) = 20% = \frac{20}{100} = \frac{1}{5}$

(i) Probability that a student reads neither Hindi nor English newspaper is,

$P (H^{'} \cap E^{'}) = P (H \cup E)^{'} = 1 - P (H \cup E) = 1 - {P (H) + P (E) - P (H \cap E)} = 1 - (\frac{3}{5} + \frac{2}{5} - \frac{1}{5}) = 1 - \frac{4}{5} = \frac{1}{5}$

(ii) Probability that a randomly chosen student reads English newspaper, if she reads Hindi newspaper, is given by P (E|H).

$P (E ∣ H) = \frac{P ( E \cap H )}{P ( H )} = \frac{\frac{1}{5}}{\frac{3}{5}} = \frac{1}{3}$

(iii) Probability that a randomly chosen student reads Hindi newspaper, if she reads English newspaper, is given by P(H | E).

$P (H ∣ E) = \frac{P ( H \cap E )}{P ( E )} = \frac{\frac{1}{5}}{\frac{2}{5}} = \frac{1}{2}$

Choose the correct answer in the following Q. 17 and 18

The probability of obtaining an even prime number on each die, when a pair of dice is rolled is (A) 0 (B) $\frac{1}{3}$ (C) $\frac{1}{12}$ (D) $\frac{1}{36}$ Sol. (D) When two dice are rolled, the number of outcomes is 36. i.e. $\Rightarrow n (S) = 36$ The only even prime number is 2 . i.e. $\Rightarrow n (E) = 1$

Let E be the event of getting an even prime number on each die.

$∴ E = {(2, 2)} \Rightarrow P (E) = \frac{1}{36}$

Therefore, the correct Answer is 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) $P (A) + P (B) = 1$ Sol. (B) Two events A and B are said to be independent, if $P (A \cap B) = P (A) \times P (B)$ Consider the result given in alternative B. $P (A^{'} B^{'}) = [1 - P (A)] [1 - P (B)]$ $\Rightarrow P (A^{'} \cap B^{'}) = 1 - P (A) - P (B) + P (A) \cdot P (B)$ $\Rightarrow 1 - P (A \cup B) = 1 - P (A) - P (B) + P (A) \cdot P (B)$ $\Rightarrow P (A \cup B) = P (A) + P (B) - P (A) \cdot P (B)$ $\Rightarrow P (A) + P (B) - P (AB) = P (A) + P (B) - P (A) \cdot P (B)$ $\Rightarrow P (AB) = P (A) \cdot P (B)$ This implies that A and B are independent, if $P (A^{'} B^{'}) = [1 - P (A)] [1 - P (B)]$

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? Sol. The urn contains 5 red and 5 black balls. Let a red ball be drawn in the first attempt. $∴ P ($ drawing a red ball $) = \frac{5}{10} = \frac{1}{2}$ If two red balls are added to the urn, then the urn contains 7 red and 5 black balls. $P ($ drawing a red ball $) = \frac{7}{12}$ Let a black ball be drawn in the first attempt. ∴ P (drawing a black ball in the first attempt) $= \frac{5}{10} = \frac{1}{2}$ If two black balls are added to the urn, then the urn contains 5 red and 7 black balls. $P ($ drawing a red ball $) = \frac{5}{12}$ Therefore, probability of drawing second ball as red is

$= \frac{1}{2} \times \frac{7}{12} + \frac{1}{2} \times \frac{5}{12} = \frac{1}{2} (\frac{7}{12} + \frac{5}{12}) = \frac{1}{2} \times 1 = \frac{1}{2}$

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. Sol. Let $E_{1}$ and $E_{2}$ be the events of selecting first bag and second bag respectively. $∴ P (E_{1}) = P (E_{2}) = \frac{1}{2}$

Let A be the event of getting a red ball.

$\Rightarrow P (A ∣ E_{1}) = P ($ drawing a red ball from first bag $)$

$= \frac{4}{8} = \frac{1}{2}$

$\Rightarrow P (A ∣ E_{2}) = P$ (drawing a red ball from second bag)

$= \frac{2}{8} = \frac{1}{4}$

The probability of drawing a ball from the first bag, given that it is red, is given by $P (E_{1} / A)$ . By using Bayes' theorem, we obtain

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

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 hostlier? Sol. Let $E_{1}$ and $E_{2}$ be the events that the student is a hostlier and a day scholar respectively and A be the event that the chosen student gets grade A. Then, $E_{1}$ and $E_{2}$ are nuturally exclusive and exhaustive. $∴ P (E_{1}) = 60% = \frac{60}{100} = 0.6$ ;

$P (E_{2}) = 40% = \frac{40}{100} = 0.4$

$P (A ∣ E_{1}) = P ($ student getting an A grade is a hostlier) $= 30% = 0.3$ $P (A ∣ E_{2}) = P$ (student getting an A grade is a day scholar $) = 20% = 0.2$ The probability that a randomly chosen student is a hostlier, given that he has an A grade, is given by $P (E_{1} ∣ A)$ . By using Bayes' theorem, we obtain

$P (E_{1} ∣ A) = \frac{P ( E _{1} ) \cdot P ( A ∣ E _{1} )}{P ( E _{1} ) \cdot P ( A ∣ E _{1} ) + P ( E _{2} ) \cdot P ( A ∣ E _{2} )} = \frac{0.6 \times 0.3}{0.6 \times 0.3 + 0.4 \times 0.2} = \frac{0.18}{0.26} = \frac{18}{26} = \frac{9}{13}$

In answering a question on a multiple choice test, a student either knows the answer or guesses. Let $\frac{3}{4}$ be the probability that he knows the answer and $\frac{1}{4}$ be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability $\frac{1}{4}$ . What is the probability that the student knows the answer given that he answered it correctly? Sol. Let $E_{1}$ and $E_{2}$ be the events that the student knows the answer and he guesses the answer respectively. Let A be the event that the answer is correct. $∴ P (E_{1}) = \frac{3}{4}; P (E_{2}) = \frac{1}{4}$

The probability that the student answered correctly, given that he knows the answer, is 1.

$∴ P (A ∣ E_{1}) = 1$ Probability that the student answered correctly, given that he guessed, is $\frac{1}{4}$ . $∴ P (A ∣ E_{2}) = \frac{1}{4}$ The probability that the student knows the answer, given that he answered it correctly, is given by $P (E_{1} ∣ A)$ . By using Bayes' theorem, we obtain

$P (E_{1} ∣ A) = \frac{P ( E _{1} ) \cdot P ( A ∣ E _{1} )}{P ( E _{1} ) \cdot P ( A ∣ E _{1} ) + P ( E _{2} ) \cdot P ( A ∣ E _{2} )} = \frac{\frac{3}{4} \cdot 1}{\frac{3}{4} \cdot 1 + \frac{1}{4} \cdot \frac{1}{4}} = \frac{\frac{3}{4}}{\frac{3}{4} + \frac{1}{16}} = \frac{\frac{3}{4}}{\frac{13}{16}} = \frac{12}{13}$

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 (i.e., if a healthy person is tested, then, with probability 0.005, the test will imply he has the disease). 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? Sol. Let $E_{1}$ and $E_{2}$ be the events that a person has a disease and a person is healthy respectively. Since $E_{1}$ and $E_{2}$ are pairwise disjoint and exhaustive events. $∴ P (E_{1}) + P (E_{2}) = 1$

$\Rightarrow P (E_{2}) = 1 - P (E_{1}) = 1 - 0.001 = 0.999 [∵ P (E_{1}) = 0.1% = 0.001]$

Let A be the event that the blood test result is positive. $P (A ∣ E_{1}) = P$ (result is positive given the person has disease) $= 99% = 0.99$ $P (A ∣ E_{2}) = P$ (result is positive given that the person is healthy) $= 0.5% = 0.005$ Probability that a person has a disease, given that his test result is positive, is given by P ( $E_{1} ∣ A$ ).By using Bayes' theorem, we obtain

$P (E_{1} ∣ A) = \frac{P ( E _{1} ) \cdot P ( A ∣ E _{1} )}{P ( E _{1} ) \cdot P ( A ∣ E _{1} ) + P ( E _{2} ) \cdot P ( A ∣ E _{2} )} = \frac{0.001 \times 0.99}{0.001 \times 0.99 + 0.999 \times 0.005} = \frac{0.00099}{0.00099 + 0.004995} = \frac{0.00099}{0.005985} = \frac{990}{5985} = \frac{110}{665} = \frac{22}{133}$

There are three coins. One is two headed coin (having head on both faces), 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? Sol. Let $E_{1}, E_{2}$ and $E_{3}$ be the respective events of choosing a two headed coin, a biased coin and an unbiased coin. Then, $E_{1}, E_{2}, E_{3}$ are nuturally exclusive and exhaustive events. $∴ P (E_{1}) = P (E_{2}) = P (E_{3}) = \frac{1}{3}$ Let A be the event that the coin shows heads. A two-headed coin will always show heads. $∴ P (A ∣ E_{1}) = P$ (coin showing heads, given that it is a two-headed coin) = 1 Probability of heads coming up, given that it is a biased coin $= 75%$ $∴ P (A ∣ E_{2}) = P$ (coin showing heads, given that it is a biased coin) $= \frac{75}{100} = \frac{3}{4}$ Since the third coin is unbiased, the probability that it shows heads is always $\frac{1}{2}$ . $∴ P (A ∣ E_{3}) = P$ (coin showing heads, given that it is a biased coin) $= \frac{1}{2}$ The probability that the coin is two-headed, given that it shows heads, is given by $P (E_{1} ∣ A)$ .

By using Bayes' theorem, we obtain

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

An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of 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? Sol. Let $E_{1}, E_{2}$ , and $E_{3}$ be the events that the driver is a scooter driver, a car driver, and a truck driver respectively. Let A be the event that the person meets with an accident. There are 2000 scooter drivers, 4000 car drivers, and 6000 truck drivers. Total number of drivers $= 2000 + 4000 + 6000$

$= 12000$

$P (E_{1}) = P$ (driver is a scooter driver)

$= \frac{2000}{12000} = \frac{1}{6}$

$P (E_{2}) = P ($ driver is a car driver $) = \frac{4000}{12000} = \frac{1}{3}$

$P (E_{3}) = P (driver is a truck driver) = \frac{6000}{12000} = \frac{1}{2}$

$P (A ∣ E_{1}) = P (scooter driver met with an accident) = 0.01 = \frac{1}{100}$

$P (A ∣ E_{2}) = P (car driver met with an accident) = 0.03 = \frac{3}{100}$

$P (A ∣ E_{3}) = P (truck driver met with an accident) = 0.15 = \frac{15}{100}$

The probability that the driver is a scooter driver, given that he met with an accident, is given by $P (E_{1} ∣ A)$ . By using Bayes' theorem, we obtain

$P (E_{1} ∣ A) = \frac{P ( E _{1} ) \cdot P ( A ∣ E _{1} )}{P ( E _{1} ) \cdot P ( A ∣ E _{1} ) + P ( E _{2} ) \cdot P ( A ∣ E _{2} ) + P ( E _{3} ) \cdot P ( A ∣ E _{3} )} = \frac{\frac{1}{6} \cdot \frac{1}{100}}{\frac{1}{6} \cdot \frac{1}{100} + \frac{1}{3} \cdot \frac{3}{100} + \frac{1}{2} \cdot \frac{15}{100}} = \frac{\frac{1}{6} \cdot \frac{1}{100}}{\frac{1}{100} ( \frac{1}{6} + 1 + \frac{15}{2} )} = \frac{\frac{1}{6}}{\frac{104}{12}} = \frac{1}{6} \times \frac{12}{104} = \frac{1}{52}$

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 was produced by machine B? Sol. Let $E_{1}$ and $E_{2}$ be the events of items produced by machines A and B respectively. Let X be the event that the produced item was found to be defective. $∴$ Probability of items produced by machine A , P

$(E_{1}) = 60% = \frac{3}{5}$

Probability of items produced by machine B, P

$(E_{2}) = 40% = \frac{2}{5}$

Probability that machine A produced defective items, $P (X ∣ E_{1}) = 2% = \frac{2}{100}$ Probability that machine B produced defective items, $P (X ∣ E_{2}) = 1% = \frac{1}{100}$

The probability that the randomly selected item was from machine B, given that it is defective, is given by $P (E_{2} ∣ X)$ . By using Bayes' theorem, we obtain

$P (E_{2} ∣ X) = \frac{P ( E _{2} ) \cdot P ( X ∣ E _{2} )}{P ( E _{1} ) \cdot P ( X ∣ E _{1} ) + P ( E _{2} ) \cdot P ( X ∣ E _{2} )} = \frac{\frac{2}{5} \cdot \frac{1}{100}}{\frac{3}{5} \cdot \frac{2}{100} + \frac{2}{5} \cdot \frac{1}{100}} = \frac{\frac{2}{500}}{\frac{6}{500} + \frac{2}{500}} = \frac{2}{8} = \frac{1}{4}$

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. Sol. Let $E_{1}$ and $E_{2}$ be the events that the first group and the second group win the competition respectively. Let A be the event of introducing a new product. $P (E_{1}) =$ Probability that the first group wins the competition = 0.6 $P (E_{2}) =$ Probability that the second group wins the competition = 0.4 P $(A ∣ E_{1}) =$ Probability of introducing a new product if the first group wins = 0.7 P $(A ∣ E_{2}) =$ Probability of introducing a new product if the second group wins = 0.3 The probability that the new product is introduced by the second group is given by P (E $_{2} ∣$ A). By using Bayes' theorem, we obtain

$P (E_{2} ∣ A) = \frac{P ( E _{2} ) \cdot P ( A ∣ E _{2} )}{P ( E _{1} ) \cdot P ( A ∣ E _{1} ) + P ( E _{2} ) \cdot P ( A ∣ E _{2} )} = \frac{0.4 \times 0.3}{0.6 \times 0.7 + 0.4 \times 0.3} = \frac{0.12}{0.42 + 0.12} = \frac{0.12}{0.54} = \frac{12}{54} = \frac{2}{9}$

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? Sol. Let $E_{1}$ be the event that the outcome on the die is 5 or 6 and $E_{2}$ be the event that the outcome on the die is 1, 2, 3, or 4.

$∴ P (E_{1}) = \frac{2}{6} = \frac{1}{3} and (E_{2}) = \frac{4}{6} = \frac{2}{3}$

Let A be the event of getting exactly one head. $P (A ∣ E_{1}) =$ Probability of getting exactly one head by tossing the coin three times if she gets

$5 or 6 = \frac{3}{8}$

$P (A ∣ E_{2}) =$ Probability of getting exactly one head in a single throw of coin if she gets 1, 2, 3,

$or 4 = \frac{1}{2}$

The probability that the girl threw 1, 2, 3, or 4 with the die, if she obtained exactly one head, is given by $P (E_{2} ∣ A)$ . By using Bayes' theorem, we obtain

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

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 was produced by A? Sol. Let $E_{1}, E_{2}$ , and $E_{3}$ be the events of the time consumed by machine operator A, B, and C for the job respectively.

$P (E_{1}) = 50% = \frac{50}{100} = \frac{1}{2}; P (E_{2}) = 30% = \frac{30}{100} = \frac{3}{10}; P (E_{3}) = 20% = \frac{20}{100} = \frac{1}{5}$

Let X be the event of producing defective items.

$P (X ∣ E_{1}) = 1% = \frac{1}{100}; P (X ∣ E_{2}) = 5% = \frac{5}{100}; P (X ∣ E_{3}) = 7% = \frac{7}{100}$

The probability that the defective item was produced by A is given by $P (E_{1} ∣ A)$ . By using Bayes' theorem, we obtain

$P (E_{1} ∣ X) = \frac{P ( E _{1} ) \cdot P ( X ∣ E _{1} )}{P ( E _{1} ) \cdot P ( X ∣ E _{1} ) + P ( E _{2} ) \cdot P ( X ∣ E _{2} ) + P ( E _{3} ) \cdot P ( X ∣ E _{3} )} = \frac{\frac{1}{2} \cdot \frac{1}{100}}{\frac{1}{2} \cdot \frac{1}{100} + \frac{3}{10} \cdot \frac{5}{100} + \frac{1}{5} \cdot \frac{7}{100}} = \frac{\frac{1}{100} \cdot \frac{1}{2}}{\frac{1}{100} ( \frac{1}{2} + \frac{3}{2} + \frac{7}{5} )} = \frac{\frac{1}{2}}{\frac{17}{5}} = \frac{5}{34}$

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. Sol. Let $E_{1}$ and $E_{2}$ be the events that lost card is a diamond card and a card which is not diamond respectively. Let A be the event that two diamond cards are drawn. Out of 52 cards, 13 cards are diamond and 39 cards are not diamond.

$∴ P (E_{1}) = \frac{13}{52} = \frac{1}{4} P (E_{2}) = \frac{39}{52} = \frac{3}{4}$

When one diamond card is lost, there are 12 diamond cards out of 51 cards. Two diamond cards can be drawn out of 12 diamond cards in $^{12} C_{2}$ ways and 2 cards can be drawn out of 51 cards in $^{51} C_{2}$ ways. The probability of getting two diamond cards, when one diamond card is lost, is given by $P (A ∣ E_{1})$ .

$P (A ∣ E_{1}) = \frac{^{12} C _{2}}{^{51} C _{2}} = \frac{12 !}{2 ! \times 10 !} \times \frac{2 ! \times 49 !}{51 !} = \frac{11 \times 12}{50 \times 51} = \frac{22}{425}$

When the lost card is not a diamond, there are 13 diamond cards out of 51 cards. Two diamond cards can be drawn out of 13 diamond cards in $^{13} C_{2}$ ways whereas 2 cards can be drawn out of 51 cards in $^{51} C_{2}$ ways. The probability of getting two diamond cards, when one card is lost which is not diamond, is given by $P (A ∣ E_{2})$ .

$P (A ∣ E_{2}) = \frac{^{13} C _{2}}{^{51} C _{2}} = \frac{13 !}{2 ! \times 11 !} \times \frac{2 ! \times 49 !}{51 !} = \frac{12 \times 13}{50 \times 51} = \frac{26}{425}$

The probability that the lost card is diamond if drawn cards are found to be both diamond is given by $P (E_{1} ∣ A)$ . By using Bayes' theorem, we obtain

$P (E_{1} ∣ A) = \frac{P ( E _{1} ) \cdot P ( A ∣ E _{1} )}{P ( E _{1} ) \cdot P ( A ∣ E _{1} ) + P ( E _{2} ) \cdot P ( A ∣ E _{2} )} = \frac{\frac{1}{4} \cdot \frac{22}{425}}{\frac{1}{4} \cdot \frac{22}{425} + \frac{3}{4} \cdot \frac{26}{425}} = \frac{\frac{1}{425} ( \frac{22}{4} )}{\frac{1}{425} ( \frac{22}{4} + \frac{26 \times 3}{4} )} = \frac{\frac{11}{2}}{25} = \frac{11}{50}$

Probability that A speaks truth is $\frac{4}{5}$ . A coin is tossed. A reports that a head appears. The probability that actually there was head is (A) $\frac{4}{5}$ (B) $\frac{1}{2}$ (C) $\frac{1}{5}$ (D) $\frac{2}{5}$ Sol. (A) Let $E_{1}$ and $E_{2}$ be the events such that $E_{1}$ : A speaks truth, $E_{2}$ : A speaks lie Let X be the event that a head appears.

$P (E_{1}) = \frac{4}{5}$

$∴ P (E_{2}) = 1 - P (E_{1}) = 1 - \frac{4}{5} = \frac{1}{5}$

If a coin is tossed, then it may result in either head (H) or tail (T). The probability of getting a head is $\frac{1}{2}$ and probability of not getting head is also $\frac{1}{2}$

$∴ P (X ∣ E_{1}) = P (X ∣ E_{2}) = \frac{1}{2}$ The probability that there is actually a head is given by $P (E_{1} ∣ X)$ .

$P (E_{1} ∣ X) = \frac{P ( E _{1} ) \cdot P ( X ∣ E _{1} )}{P ( E _{1} ) \cdot P ( X ∣ E _{1} ) + P ( E _{2} ) \cdot P ( X ∣ E _{2} )} = \frac{\frac{4}{5} \cdot \frac{1}{2}}{\frac{4}{5} \cdot \frac{1}{2} + \frac{1}{5} \cdot \frac{1}{2}} = \frac{\frac{1}{2} \cdot \frac{4}{5}}{\frac{1}{2} ( \frac{4}{5} + \frac{1}{5} )} = \frac{\frac{4}{5}}{1} = \frac{4}{5}$

Therefore, the correct answer is A.

If A and B are two events such that A $\subset B$ and P $(B) \neq = 0$ , then which of the following is correct? (A) $P (A ∣$ $B) = \frac{P ( B )}{P ( A )}$ (B) $P (A ∣ B) < P (A)$ (C) $P (A ∣ B) \geq P (A)$ (D) None of these Sol. (C) If $A \subset B$ , then $A \cap B = A$ $\Rightarrow P (A \cap B) = P (A)$ Also, $P (A) < P (B)$ Consider

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

Again, consider

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

It is known that,

$P (B) \leq 1$

$\Rightarrow \frac{1}{P ( B )} \geq 1$ $\Rightarrow \frac{P ( A )}{P ( B )} \geq P (A)$ From (2), we obtain

$\Rightarrow P (A ∣ B) \geq P (A)$

$∴ P (A ∣ B)$ is not less than $P (A)$ . Thus, from (3), it can be concluded that the relation given in alternative C is correct.

3.0Class 12 Maths NCERT Solutions – Chapter-wise Links

Access NCERT Solutions for Class 12 Maths chapter-wise, covering NCERT exercises, key formulas, and easy-to-follow solutions for better understanding and regular question practice.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Relations and Functions
Chapter 2	Inverse Trigonometric Functions
Chapter 3	Matrices
Chapter 4	Determinants
Chapter 5	Continuity and Differentiability
Chapter 6	Application of Derivatives
Chapter 7	Integrals
Chapter 8	Application of Integrals
Chapter 9	Differential Equations
Chapter 10	Vector Algebra
Chapter 11	Three Dimensional Geometry
Chapter 12	Linear Programming

4.0Class 12 Maths Chapter 13 Probability Exercise-wise Solutions

Exercise	Number of Questions	Important Topics
Exercise 13.1	19 Questions and Solutions	Sample space, events and basic concepts of probability
Exercise 13.2	19 Questions and Solutions	Classical definition of probability and related problems
Exercise 13.3	14 Questions and Solutions	Axiomatic approach to probability and properties of probability

5.0Key Features and Benefits of Class 12 Maths Chapter 13 Probability

Regular practice of the solutions helps students improve accuracy and develop better problem-solving skills for probability questions.
Probability is an important topic for competitive examinations, including Olympiads. A clear understanding of the concepts can support exam preparation.
The solutions explain important probability concepts clearly and help students understand the steps involved in solving each question.
Step-by-step solutions make topics such as conditional probability and Bayes’ theorem easier to understand and apply in exams.
The exercises are based on the NCERT syllabus and provide practice for different types of questions asked in CBSE board examinations.

Frequently Asked Questions

Where can I find NCERT Solutions for Class 12 Maths Chapter 13 Probability?

How do NCERT Solutions for Class 12 Maths Chapter 13 explain conditional probability?

What is covered under the multiplication theorem in NCERT Class 12 Maths Chapter 13 Solutions?

When should I use Bayes’ theorem according to NCERT Solutions for Class 12 Maths Chapter 13?

Which exercises should I practise from NCERT Solutions for Class 12 Maths Chapter 13 Probability?

What do NCERT Class 12 Maths Chapter 13 Solutions explain about independent events?

How can NCERT Solutions for Class 12 Maths Chapter 13 help me prepare for the CBSE board exam?

Can NCERT Solutions for Class 12 Maths Chapter 13 Probability help with competitive exams like JEE?

Join ALLEN!

Frequently Asked Questions

Where can I find NCERT Solutions for Class 12 Maths Chapter 13 Probability?

How do NCERT Solutions for Class 12 Maths Chapter 13 explain conditional probability?

What is covered under the multiplication theorem in NCERT Class 12 Maths Chapter 13 Solutions?

When should I use Bayes’ theorem according to NCERT Solutions for Class 12 Maths Chapter 13?

Which exercises should I practise from NCERT Solutions for Class 12 Maths Chapter 13 Probability?

What do NCERT Class 12 Maths Chapter 13 Solutions explain about independent events?

How can NCERT Solutions for Class 12 Maths Chapter 13 help me prepare for the CBSE board exam?

Can NCERT Solutions for Class 12 Maths Chapter 13 Probability help with competitive exams like JEE?

Join ALLEN!

NCERT Solutions Class 12 Maths Chapter 13 Probability

1.0Key Concepts of Class 12 Maths Chapter 13 Probability

2.0NCERT Class 12 Maths Chapter 13 Probability : Detailed Solutions

EXERCISE - 13.1

EXERCISE 13.2

EXERCISE 13.3

3.0Class 12 Maths NCERT Solutions – Chapter-wise Links

4.0Class 12 Maths Chapter 13 Probability Exercise-wise Solutions

5.0Key Features and Benefits of Class 12 Maths Chapter 13 Probability

Table of Contents