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

Determine P(E|F). 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

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=\{HHT,HTH,THH,HTT,THT,TTH,TTT\} (all outcomes except HHH), so P(F)=\dfrac78.

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

(iii) E=\{HHH,HHT,HTH,THH,HTT,THT,TTH\} (all outcomes except TTT), so P(E)=\dfrac78. F=\{HHT,HTH,THH,HTT,THT,TTH,TTT\} (all outcomes except HHH), so P(F)=\dfrac78.

E\cap F=\{HHT,HTH,THH,HTT,THT,TTH\}, so P(E\cap F)=\dfrac68. Thus P(E|F)=\dfrac{6/8}{7/8}=\dfrac67.

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

Determine P(E|F). 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

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

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

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

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

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 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.

Medium +
Solution

Writing each outcome as (black, red), the sample space has 6\times6=36 equally likely outcomes.

(a) Let E : sum greater than 9 and F : black die shows 5. Then F=\{(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)\}, so P(F)=\dfrac{6}{36}.

E\cap F=\{(5,5),(5,6)\}, so P(E\cap F)=\dfrac{2}{36}. Thus P(E|F)=\dfrac{P(E\cap F)}{P(F)}=\dfrac{2/36}{6/36}=\dfrac13.

(b) Let E : sum is 8 and F : red die shows a number less than 4. Then F contains all outcomes with red = 1, 2 or 3, so n(F)=3\times6=18 and P(F)=\dfrac{18}{36}.

E\cap F=\{(6,2),(5,3)\}, so P(E\cap F)=\dfrac{2}{36}. Thus P(E|F)=\dfrac{2/36}{18/36}=\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 that
(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 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?

Easy +
Solution

Total number of questions = 300+200+500+400=1400. Let E : the question is easy and F : the question is a multiple choice question.

P(F)=\dfrac{500+400}{1400}=\dfrac{900}{1400} and P(E\cap F)=\dfrac{500}{1400}.

Thus P(E|F)=\dfrac{P(E\cap F)}{P(F)}=\dfrac{500/1400}{900/1400}=\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 'the sum of 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 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'.

Hard +
Solution

If the first throw is 3 or 6 (a multiple of 3), the die is thrown again; if the first throw is 1, 2, 4 or 5, a coin is tossed. The sample space is

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

Each of the 12 die–die outcomes has probability \dfrac16\times\dfrac16=\dfrac{1}{36} and each of the 8 die–coin outcomes has probability \dfrac16\times\dfrac12=\dfrac{1}{12}.

Let E : the coin shows a tail and F : at least one die shows a 3. Then E=\{(1,T),(2,T),(4,T),(5,T)\} and F=\{(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(6,3)\}, so P(F)=\dfrac{7}{36}.

Clearly E\cap F=\varnothing, so P(E\cap F)=0. Thus P(E|F)=\dfrac{P(E\cap F)}{P(F)}=\dfrac{0}{7/36}=0.

P(E|F) = 0
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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