Class 12 Maths NCERT Solutions Chapter 13 Ex 13.3 – Bayes' Theorem | Boundless Maths

Ex 13.3 Class 12 Maths NCERT Solutions · Chapter 13

Class 12 Maths NCERT Solutions Chapter 13 Ex 13.3 – Bayes' Theorem

Free, step-by-step Class 12 Maths NCERT Solutions for Chapter 13 Ex 13.3 — all 14 questions solved, using the theorem of total probability and Bayes' theorem to work backwards from an observed outcome to its most likely cause.

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Question 1 uses the theorem of total probability, and Questions 2–12 apply Bayes' theorem across a wide variety of real-world settings — urns and bags of balls, a multiple-choice test where a student either knows the answer or guesses, a medical test with a false-positive rate, a two-headed coin mixed in with fair and biased ones, insured drivers split by vehicle type, two factory machines, two groups competing for a board seat, a girl who throws a die before deciding how to toss a coin, three machine operators with different defect rates, and a lost playing card. The method stays the same throughout: identify the partition of hypotheses (which bag, which machine, which operator), find each hypothesis's prior probability and the probability of the observed event given that hypothesis, then combine them with Bayes' theorem to get the posterior probability of the hypothesis being asked about. The exercise closes with two MCQs — a direct reverse-probability calculation about a coin toss, and a question on how P(A|B) compares with P(A) when A is a subset of B.

14Questions

Easy–HardDifficulty Mix

2026-27CBSE Syllabus

Class 12 Maths NCERT Solutions Chapter 13 Ex 13.3 — All 14 Questions

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?

Easy +

Solution

Let R_1: first ball drawn is red, B_1: first ball drawn is black. P(R_1)=\dfrac{5}{10}=\dfrac12, P(B_1)=\dfrac12.

If the first ball was red, the urn now has 7 red and 5 black (12 total), so P(\text{2nd red}|R_1)=\dfrac{7}{12}. If the first ball was black, the urn now has 5 red and 7 black (12 total), so P(\text{2nd red}|B_1)=\dfrac{5}{12}.

By the theorem of total probability, P(\text{2nd red})=\dfrac12\times\dfrac{7}{12}+\dfrac12\times\dfrac{5}{12}=\dfrac{7}{24}+\dfrac{5}{24}=\dfrac{12}{24}=\dfrac12.

Required probability = \dfrac12

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.

Easy +

Solution

Let E_1: bag I is chosen, E_2: bag II is chosen, both with P(E_1)=P(E_2)=\dfrac12. Let A: the ball drawn is red.

P(A|E_1)=\dfrac48=\dfrac12 and P(A|E_2)=\dfrac28=\dfrac14.

By Bayes' theorem, P(E_1|A)=\dfrac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1)+P(E_2)P(A|E_2)}=\dfrac{\frac12\times\frac12}{\frac12\times\frac12+\frac12\times\frac14}=\dfrac{1/4}{1/4+1/8}=\dfrac{1/4}{3/8}=\dfrac23.

Required probability = \dfrac23

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 hosteller?

Medium +

Solution

Let E_1: student is a hosteller, E_2: student is a day scholar. P(E_1)=0.6, P(E_2)=0.4.

Let A: student attains an A grade, with P(A|E_1)=0.3 and P(A|E_2)=0.2.

By Bayes' theorem, P(E_1|A)=\dfrac{0.6\times0.3}{0.6\times0.3+0.4\times0.2}=\dfrac{0.18}{0.18+0.08}=\dfrac{0.18}{0.26}=\dfrac{9}{13}.

Required probability = \dfrac{9}{13}

In answering a question on a multiple choice test, a student either knows the answer or guesses. Let \dfrac34 be the probability that he knows the answer and \dfrac14 be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability \dfrac14. What is the probability that the student knows the answer given that he answered it correctly?

Medium +

Solution

Let E_1: student knows the answer, E_2: student guesses. P(E_1)=\dfrac34, P(E_2)=\dfrac14.

Let A: the answer is correct, with P(A|E_1)=1 (a student who knows answers correctly) and P(A|E_2)=\dfrac14.

By Bayes' theorem, P(E_1|A)=\dfrac{\frac34\times1}{\frac34\times1+\frac14\times\frac14}=\dfrac{3/4}{3/4+1/16}=\dfrac{3/4}{13/16}=\dfrac34\times\dfrac{16}{13}=\dfrac{12}{13}.

Required probability = \dfrac{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 ?

Hard +

Solution

Let E: the person has the disease, so P(E)=0.001 and P(E')=0.999. Let A: the test is positive, with P(A|E)=0.99 and P(A|E')=0.005.

By Bayes' theorem, P(E|A)=\dfrac{0.001\times0.99}{0.001\times0.99+0.999\times0.005}=\dfrac{0.00099}{0.00099+0.004995}=\dfrac{0.00099}{0.005985}=\dfrac{990}{5985}=\dfrac{22}{133}.

Even with a highly accurate test, the disease is so rare that a positive result still makes actually having it only about \dfrac{22}{133}\approx0.165, i.e. 16.5% likely.

Required probability = \dfrac{22}{133}\approx0.165

There are three coins. One is a 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 ?

Medium +

Solution

Let E_1,E_2,E_3 be the events of choosing the two-headed, biased, and unbiased coin respectively, each with prior P(E_i)=\dfrac13.

Let A: the coin shows heads, with P(A|E_1)=1, P(A|E_2)=0.75, P(A|E_3)=0.5.

By Bayes' theorem, P(E_1|A)=\dfrac{\frac13\times1}{\frac13\times1+\frac13\times0.75+\frac13\times0.5}=\dfrac{1}{1+0.75+0.5}=\dfrac{1}{2.25}=\dfrac49.

Required probability = \dfrac{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?

Medium +

Solution

Out of 2000+4000+6000=12000 insured drivers, P(\text{scooter})=\dfrac16, P(\text{car})=\dfrac13, P(\text{truck})=\dfrac12.

The accident probabilities given each type are 0.01, 0.03 and 0.15 respectively.

By Bayes' theorem, P(\text{scooter}|\text{accident})=\dfrac{\frac16\times0.01}{\frac16\times0.01+\frac13\times0.03+\frac12\times0.15}.

Using a common denominator of 600: the numerator is \dfrac{1}{600}, and the denominator sums to \dfrac{1}{600}+\dfrac{6}{600}+\dfrac{45}{600}=\dfrac{52}{600}. So the probability is \dfrac{1}{52}.

Required probability = \dfrac{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?

Easy +

Solution

Let E_1: item made by A, E_2: item made by B. P(E_1)=0.6, P(E_2)=0.4. Let D: item is defective, with P(D|E_1)=0.02 and P(D|E_2)=0.01.

By Bayes' theorem, P(E_2|D)=\dfrac{0.4\times0.01}{0.6\times0.02+0.4\times0.01}=\dfrac{0.004}{0.012+0.004}=\dfrac{0.004}{0.016}=0.25.

Required probability = 0.25=\dfrac14

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.

Medium +

Solution

Let E_1: first group wins, E_2: second group wins. P(E_1)=0.6, P(E_2)=0.4.

Let A: a new product is introduced, with P(A|E_1)=0.7 and P(A|E_2)=0.3.

By Bayes' theorem, P(E_2|A)=\dfrac{0.4\times0.3}{0.6\times0.7+0.4\times0.3}=\dfrac{0.12}{0.42+0.12}=\dfrac{0.12}{0.54}=\dfrac29.

Required probability = \dfrac{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?

Hard +

Solution

Let E_1: die shows 5 or 6 (leading to three coin tosses), E_2: die shows 1, 2, 3 or 4 (leading to one coin toss). P(E_1)=\dfrac13, P(E_2)=\dfrac23.

Let A: exactly one head is obtained. In three tosses, P(A|E_1)=\binom31\left(\dfrac12\right)^3=\dfrac38. In one toss, "exactly one head" simply means the coin shows heads, so P(A|E_2)=\dfrac12.

By Bayes' theorem, P(E_2|A)=\dfrac{\frac23\times\frac12}{\frac13\times\frac38+\frac23\times\frac12}=\dfrac{1/3}{1/8+1/3}=\dfrac{1/3}{3/24+8/24}=\dfrac{1/3}{11/24}=\dfrac13\times\dfrac{24}{11}=\dfrac{8}{11}.

Required probability = \dfrac{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?

Medium +

Solution

Let E_1,E_2,E_3 be the events "item produced by A, B, C" with priors P(E_1)=0.5, P(E_2)=0.3, P(E_3)=0.2.

Let D: item is defective, with P(D|E_1)=0.01, P(D|E_2)=0.05, P(D|E_3)=0.07.

P(D)=0.5\times0.01+0.3\times0.05+0.2\times0.07=0.005+0.015+0.014=0.034.

By Bayes' theorem, P(E_1|D)=\dfrac{0.005}{0.034}=\dfrac{5}{34}.

Required probability = \dfrac{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.

Hard +

Solution

Let D: the lost card is a diamond, so P(D)=\dfrac{13}{52}=\dfrac14 and P(D')=\dfrac34. Let A: both cards drawn from the remaining 51 are diamonds.

If the lost card was a diamond, 12 diamonds remain among 51 cards, so P(A|D)=\dfrac{12}{51}\times\dfrac{11}{50}=\dfrac{132}{2550}. If not, 13 diamonds remain among 51, so P(A|D')=\dfrac{13}{51}\times\dfrac{12}{50}=\dfrac{156}{2550}.

By Bayes' theorem, P(D|A)=\dfrac{\frac14\times\frac{132}{2550}}{\frac14\times\frac{132}{2550}+\frac34\times\frac{156}{2550}}=\dfrac{132}{132+3\times156}=\dfrac{132}{132+468}=\dfrac{132}{600}=\dfrac{11}{50}.

Required probability = \dfrac{11}{50}

MCQ. Probability that A speaks truth is \dfrac45. A coin is tossed. A reports that a head appears. The probability that actually there was head is: (A) \dfrac45 (B) \dfrac12 (C) \dfrac15 (D) \dfrac25

Easy +

Solution

Let H: a head actually occurred, so P(H)=P(H')=\dfrac12. Let R: A reports a head.

If a head occurred, A truthfully reports it with probability \dfrac45; if a tail occurred, A would report a head only by lying, with probability \dfrac15.

By Bayes' theorem, P(H|R)=\dfrac{\frac12\times\frac45}{\frac12\times\frac45+\frac12\times\frac15}=\dfrac{4/10}{4/10+1/10}=\dfrac{4}{5}.

Answer: (A) \dfrac{4}{5}

MCQ. If A and B are two events such that A\subset B and P(B)\neq0, then which of the following is correct ? (A) P(A|B)=\dfrac{P(B)}{P(A)} (B) P(A|B) \lt P(A) (C) P(A|B)\ge P(A) (D) None of these

Medium +

Solution

Since A\subset B, every outcome of A is also in B, so A\cap B=A, giving P(A|B)=\dfrac{P(A\cap B)}{P(B)}=\dfrac{P(A)}{P(B)}.

Since 0 \lt P(B)\le1, dividing P(A) by P(B) can only keep it the same (if P(B)=1) or make it larger (if P(B) \lt 1). So P(A|B)\ge P(A) always.

Answer: (C) P(A|B)\ge P(A)

Common Questions

Class 12 Maths NCERT Solutions Chapter 13 Ex 13.3 — FAQs

How many questions are there in Exercise 13.3?

Exercise 13.3 has 14 questions (12 direct problems plus 2 MCQs), all built around the theorem of total probability and Bayes' theorem.

What concept does Exercise 13.3 test?

It tests Bayes' theorem — working backwards from an observed outcome to the probability of which "cause" or hypothesis produced it, using the priors P(E_i) and the likelihoods P(A|E_i) for a partition of hypotheses E_1, E_2, ..., E_n.

Where can I find the official NCERT textbook for this chapter?

Probability is Chapter 13 of the NCERT Class 12 Mathematics textbook (Part II), published by the National Council of Educational Research and Training (NCERT) and prescribed by CBSE. You can download the official textbook PDF directly from ncert.nic.in, NCERT's official website — the solutions on this page follow the exercise exactly as it appears there.