Class 12 Maths NCERT Solutions Chapter 13 Ex 13.1 – Conditional Probability | Boundless Maths
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Ex 13.1 Class 12 Maths NCERT Solutions · Chapter 13

Class 12 Maths NCERT Solutions Chapter 13 Ex 13.1 – Conditional Probability

Free, step-by-step Class 12 Maths NCERT Solutions for Chapter 13 Ex 13.1 — all 17 questions solved, applying the conditional probability formula P(E|F) = P(E ∩ F) / P(F) both algebraically and directly from a sample space.

Questions 1–5 work directly from the conditional probability formula using given numeric probabilities — substitute the known values and rearrange to find an unknown like P(A∪B) or P(B|A), with no sample space needed. Questions 6–13 shift to computing P(E|F) from an actual equally-likely sample space — coin tosses, dice throws, a family line-up, and a question bank — where you list the outcomes of E and F directly and count them. Questions 14 and 15 test two genuinely different scenarios: a reduced sample space (only outcomes where the two dice show different numbers) and a two-stage experiment (a die, followed by either a second die or a coin) where spotting which outcomes can actually satisfy both events prevents a wrong answer. The exercise closes with two MCQs on the formal properties of conditional probability itself — what P(A|B) equals when P(B) = 0, and what P(A|B) = P(B|A) tells you about A and B.

17Questions
Easy–HardDifficulty Mix
2026-27CBSE Syllabus

Class 12 Maths NCERT Solutions Chapter 13 Ex 13.1 — All 17 Questions

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).

Easy +
Solution

By the definition P(E|F)=\dfrac{P(E\cap F)}{P(F)}, so P(E|F)=\dfrac{0.2}{0.3}=\dfrac{2}{3}.

Similarly, P(F|E)=\dfrac{P(E\cap F)}{P(E)}=\dfrac{0.2}{0.6}=\dfrac{1}{3}.

P(E|F) = \dfrac{2}{3},   P(F|E) = \dfrac{1}{3}
2

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

Easy +
Solution

P(A|B)=\dfrac{P(A\cap B)}{P(B)}=\dfrac{0.32}{0.5}=0.64.

P(A|B) = 0.64
3

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).

Medium +
Solution

(i) From P(B|A)=\dfrac{P(A\cap B)}{P(A)}, P(A\cap B)=P(A)\cdot P(B|A)=0.8\times0.4=0.32.

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

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

P(A∩B) = 0.32,   P(A|B) = 0.64,   P(A∪B) = 0.98
4

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

Medium +
Solution

From 2P(A)=\dfrac{5}{13}, P(A)=\dfrac{5}{26}, and P(B)=\dfrac{5}{13}.

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

P(A\cup B)=P(A)+P(B)-P(A\cap B)=\dfrac{5}{26}+\dfrac{10}{26}-\dfrac{4}{26}=\dfrac{11}{26}.

P(A∪B) = \dfrac{11}{26}
5

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

Easy +
Solution

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

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

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

P(A∩B) = \dfrac{4}{11},   P(A|B) = \dfrac{4}{5},   P(B|A) = \dfrac{2}{3}
6

A coin is tossed three times. Determine P(E|F) 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.

Medium +
Solution

The sample space has 8 equally likely outcomes: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT, each with probability \dfrac18.

(i) E=\{HHH,HTH,THH,TTH\}, so P(E)=\dfrac12. F=\{HHH,HHT\}, so P(F)=\dfrac14.

E\cap F=\{HHH\}, so P(E\cap F)=\dfrac18. Thus P(E|F)=\dfrac{1/8}{1/4}=\dfrac12.

(ii) E=\{HHH,HHT,HTH,THH\}, P(E)=\dfrac12. F excludes only TTT, so P(F)=\dfrac78.

E\cap F is exactly-two-heads outcomes \{HHT,HTH,THH\}, so P(E\cap F)=\dfrac38. Thus P(E|F)=\dfrac{3/8}{7/8}=\dfrac37.

(iii) E excludes only TTT, so P(E)=\dfrac78. F excludes only HHH, so P(F)=\dfrac78.

E\cap F excludes both HHH and TTT, so P(E\cap F)=\dfrac68. Thus P(E|F)=\dfrac{6/8}{7/8}=\dfrac67.

(i) \dfrac12   (ii) \dfrac37   (iii) \dfrac67
7

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

Easy +
Solution

The sample space is S=\{HH,HT,TH,TT\}, each with probability \dfrac14.

(i) E=\{HT,TH\} and F=\{HT,TH\}, so E\cap F=\{HT,TH\}. Then P(E)=P(F)=P(E\cap F)=\dfrac12, giving P(E|F)=1.

(ii) E=\{HH\} and F=\{TT\}, so E\cap F=\varnothing and P(E\cap F)=0. Since P(F)=\dfrac14\neq0, P(E|F)=0.

(i) 1   (ii) 0
8

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

Medium +
Solution

The sample space has 6^3=216 equally likely outcomes. E fixes only the third toss as 4, so n(E)=6\times6\times1=36 and P(E)=\dfrac{36}{216}=\dfrac16.

F fixes the first two tosses as 6 and 5, with the third toss free, so n(F)=6 and P(F)=\dfrac{6}{216}=\dfrac{1}{36}.

E\cap F=\{(6,5,4)\}, a single outcome, so P(E\cap F)=\dfrac{1}{216}. Thus P(E|F)=\dfrac{1/216}{6/216}=\dfrac16.

P(E|F) = \dfrac{1}{6}
9

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

Medium +
Solution

There are 3!=6 equally likely line-ups: MFS, MSF, FMS, FSM, SMF, SFM (M = mother, F = father, S = son).

E (son at either end) is \{SMF,SFM,MFS,FMS\}, so P(E)=\dfrac46=\dfrac23.

F (father in the middle) is \{MFS,SFM\}, so P(F)=\dfrac26=\dfrac13.

E\cap F=\{MFS,SFM\} — both father-in-middle arrangements already have the son on an end — so P(E\cap F)=\dfrac26=\dfrac13. Thus P(E|F)=\dfrac{1/3}{1/3}=1.

P(E|F) = 1
10

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

Medium +
Solution

(a) Given the black die shows 5, the red die can be any of 1–6 (6 equally likely outcomes). Sum > 9 needs red > 4, i.e. red = 5 or 6 — 2 favourable outcomes. So the conditional probability is \dfrac26=\dfrac13.

(b) Given the red die shows a number less than 4 (1, 2 or 3), there are 3\times6=18 equally likely outcomes for (black, red). Sum = 8 with red < 4 needs (black, red) = (7,1) — impossible — (6,2) or (5,3), giving 2 favourable outcomes. So the conditional probability is \dfrac{2}{18}=\dfrac19.

(a) \dfrac13   (b) \dfrac19
11

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).

Hard +
Solution

P(E)=\dfrac36=\dfrac12, P(F)=\dfrac26=\dfrac13, P(G)=\dfrac46=\dfrac23.

(i) E\cap F=\{3\}, so P(E\cap F)=\dfrac16. Then P(E|F)=\dfrac{1/6}{1/3}=\dfrac12 and P(F|E)=\dfrac{1/6}{1/2}=\dfrac13.

(ii) E\cap G=\{3,5\}, so P(E\cap G)=\dfrac26=\dfrac13. Then P(E|G)=\dfrac{1/3}{2/3}=\dfrac12 and P(G|E)=\dfrac{1/3}{1/2}=\dfrac23.

(iii) E\cup F=\{1,2,3,5\}, so (E\cup F)\cap G=\{2,3,5\} and P((E\cup F)\cap G)=\dfrac36=\dfrac12, giving P((E\cup F)|G)=\dfrac{1/2}{2/3}=\dfrac34.

Also (E\cap F)\cap G=\{3\}\cap G=\{3\}, so P((E\cap F)\cap G)=\dfrac16, giving P((E\cap F)|G)=\dfrac{1/6}{2/3}=\dfrac14.

(i) \dfrac12,\ \dfrac13   (ii) \dfrac12,\ \dfrac23   (iii) \dfrac34,\ \dfrac14
12

Assume 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
(i) the youngest is a girl,
(ii) at least one is a girl?

Medium +
Solution

Writing each outcome as (elder, younger), the sample space is S=\{bb,bg,gb,gg\}, each with probability \dfrac14. Let E=\{gg\} be "both are girls."

(i) "Youngest is a girl" is F=\{bg,gg\}, so P(F)=\dfrac12.

E\cap F=\{gg\}, so P(E\cap F)=\dfrac14. Thus P(E|F)=\dfrac{1/4}{1/2}=\dfrac12.

(ii) "At least one is a girl" is F=\{bg,gb,gg\}, so P(F)=\dfrac34.

E\cap F=\{gg\}, so P(E\cap F)=\dfrac14. Thus P(E|F)=\dfrac{1/4}{3/4}=\dfrac13.

(i) \dfrac12   (ii) \dfrac13
13

An instructor's question bank has 300 easy True/False, 200 difficult True/False, 500 easy multiple-choice and 400 difficult multiple-choice questions. A question is selected at random. What is the probability it is an easy question, given that it is a multiple-choice question?

Easy +
Solution

Let E: "the question is easy" and F: "the question is multiple-choice." There are 500+400=900 multiple-choice questions in total, of which 500 are easy.

Since every outcome here is equally likely, P(E|F) is simply the proportion of multiple-choice questions that are easy: P(E|F)=\dfrac{500}{900}=\dfrac59.

P(E|F) = \dfrac{5}{9}
14

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

Medium +
Solution

Let F: "the two numbers appearing are different." Out of the 36 outcomes of two dice, the 6 doubles are excluded, leaving n(F)=30 equally likely outcomes, so P(F)=\dfrac{30}{36}=\dfrac56.

Let E: "the sum is 4." The pairs summing to 4 are (1,3), (2,2) and (3,1); since (2,2) has equal numbers, it doesn't lie in F, leaving only (1,3) and (3,1) in E\cap F. So P(E\cap F)=\dfrac{2}{36}=\dfrac{1}{18}.

Thus P(E|F)=\dfrac{P(E\cap F)}{P(F)}=\dfrac{1/18}{5/6}=\dfrac{1}{18}\times\dfrac65=\dfrac{1}{15}.

Required probability = \dfrac{1}{15}
15

Consider the experiment of throwing a die: if a multiple of 3 comes up, throw the die again; if any other number comes up, toss a coin. Find the conditional probability of the event "the coin shows a tail," given that "at least one die shows a 3."

Hard +
Solution

The multiples of 3 on a die are 3 and 6, so a second die throw happens only when the first throw is 3 or 6; for any other first throw (1, 2, 4 or 5), a coin is tossed instead — and no second die is involved in that branch.

Let F: "at least one die shows a 3." This can only happen inside the die-die branch — either the first throw is 3 (any second throw), or the first throw is 6 and the second throw is 3. So P(F)=\dfrac16+\dfrac{1}{36}=\dfrac{7}{36}.

Let E: "the coin shows a tail." This event can only occur in the branch where the first throw is 1, 2, 4 or 5 — a branch that, by definition, never shows a 3 on any die. So E and F share no outcomes at all: E\cap F=\varnothing, giving P(E\cap F)=0.

This is a genuine "gotcha": since a coin is tossed only when the die does not show a multiple of 3 on the first throw, it's impossible for the coin to show a tail in the same run where a 3 has appeared on a die. The two events are mutually exclusive by construction.
P(E|F) = \dfrac{0}{7/36}=0

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16

MCQ. If P(A)=\dfrac12, P(B)=0, then P(A|B) is:   (A) 0   (B) \dfrac12   (C) not defined   (D) 1

Easy +
Solution

The conditional probability formula P(A|B)=\dfrac{P(A\cap B)}{P(B)} requires P(B)\neq0. Since P(B)=0 here, division by zero makes the expression not defined — no numerical value can be assigned.

Answer: (C) not defined
17

MCQ. 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=\varnothing   (D) P(A)=P(B)

Medium +
Solution

Writing out both sides, P(A|B)=P(B|A) means \dfrac{P(A\cap B)}{P(B)}=\dfrac{P(A\cap B)}{P(A)}. Provided P(A\cap B)\neq0, this can be cancelled to give P(A)=P(B) — the events need not be equal as sets, only equally likely.

Answer: (D) P(A)=P(B)
Common Questions

Class 12 Maths NCERT Solutions Chapter 13 Ex 13.1 — FAQs

How many questions are there in Exercise 13.1?
Exercise 13.1 has 17 questions (15 direct problems plus 2 MCQs), all built around the conditional probability of one event given that another has occurred.

What concept does Exercise 13.1 test?
It tests the conditional probability formula P(E|F) = P(E ∩ F) / P(F), provided P(F) ≠ 0 — applied both algebraically, from given probability values, and directly from an equally-likely sample space by listing and counting outcomes.

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.

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