# NCERT Solutions for Exercise 13.2 Class 12 Maths Chapter 13 - Probability

Most of the questions in NCERT solutions for exercise 13.2 Class 12 Maths chapter 13 deal with multiplication theorem on probability, independent events, and its property. This NCERT book exercise is very important from the board's exam point of view as one question is generally asked from this exercise. There are 7 examples given before this exercise which you should try to solve before solving the exercises questions. You can take help from exercise 13.2 Class 12 Maths solutions if you are stuck while solving these NCERT problems. These Class 12 Maths chapter 13 exercise 13.2 solutions are created by the subject matter experts who know how best to answer in the board exams.

Also, see

• Probability Exercise 13.1
• Probability Exercise 13.3
• Probability Exercise 13.4
• Probability Exercise 13.5
• Probability Miscellaneous Exercise

## CBSE NCERT Solutions for Class 12 Maths Chapter 13 probability-Exercise: 13.2

Question:1 If $P(A)=\frac{3}{5}$ and $P(B)=\frac{1}{5},$ find $P(A\cap B)$ if $A$ and $B$ are independent events.

$P(A)=\frac{3}{5}$ and $P(B)=\frac{1}{5},$

Given : $A$ and $B$ are independent events.

So we have, $P(A\cap B)=P(A).P(B)$

$\Rightarrow \, \, \, \, P(A\cap B)=\frac{3}{5}\times \frac{1}{5}$

$\Rightarrow \, \, \, \, P(A\cap B)=\frac{3}{25}$

Question:2 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.

Two cards are drawn at random and without replacement from a pack of 52 playing cards.

There are 26 black cards in a pack of 52.

Let $P(A)$ be the probability that first cards is black.

Then, we have

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

Let $P(B)$ be the probability that second cards is black.

Then, we have

$P(B)= \frac{25}{51}$

The probability that both the cards are black $=P(A).P(B)$

$=\frac{1}{2}\times \frac{25}{51}$

$=\frac{25}{102}$

Question:3 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 $\inline 15$ oranges out of which $\inline 12$ are good and $\inline 3$ are bad ones will be approved for sale.

Total oranges = 15

Good oranges = 12

Let $P(A)$ be the probability that first orange is good.

The, we have

$P(A)= \frac{12}{15}=\frac{4}{5}$

Let $P(B)$ be the probability that second orange is good.

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

Let $P(C)$ be the probability that third orange is good.

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

The probability that a box will be approved for sale $=P(A).P(B).P(C)$

$=\frac{4}{5}.\frac{11}{14}.\frac{10}{13}$

$=\frac{44}{91}$

Question:4 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.

A fair coin and an unbiased die are tossed,then total outputs are:

$= \left \{ (H1),(H2),(H3),(H4),(H5),(H6),(T1),(T2),(T3),(T4),(T5),(T6) \right \}$

$=12$

A is the event ‘head appears on the coin’ .

Total outcomes of A are : $= \left \{ (H1),(H2),(H3),(H4),(H5),(H6) \right \}$

$P(A)=\frac{6}{12}=\frac{1}{2}$

B is the event ‘3 on the die’.

Total outcomes of B are : $= \left \{ (T3),(H3)\right \}$

$P(B)=\frac{2}{12}=\frac{1}{6}$

$\therefore A\cap B = (H3)$

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

Also, $P (A\cap B) = P(A).P(B)$

$P (A\cap B) = \frac{1}{2}\times \frac{1}{6}=\frac{1}{12}$

Hence, A and B are independent events.

Question:5 A die marked $\inline 1,2,3$ in red and $\inline 4,5,6$ in green is tossed. Let $\inline A$ be the event, ‘the number is even,’ and $\inline B$ be the event, ‘the number is red’. Are $\inline A$ and $\inline B$ independent?

Total outcomes $=\left \{ 1,2,3,4,5,6 \right \}=6$.

$\inline A$ is the event, ‘the number is even,’

Outcomes of A $=\left \{ 2,4,6 \right \}$

$n(A)=3.$

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

$\inline B$ is the event, ‘the number is red’.

Outcomes of B $=\left \{ 1,2,3 \right \}$

$n(B)=3.$

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

$\therefore (A\cap B)=\left \{ 2 \right \}$

$n(A\cap B)=1$

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

Also,

$P(A\cap B)=P(A).P(B)$

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

Thus, both the events A and B are not independent.

Question:6 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?

Given :

$P(E)=\frac{3}{5},P(F)=\frac{3}{10}$ and $P(E\cap F)=\frac{1}{5}.$

For events E and F to be independent , we need

$P(E\cap F)=P(E).P(F)$

$P(E\cap F)=\frac{3}{5}\times \frac{3}{10}=\frac{9}{50}\neq \frac{1}{5}$

Hence, E and F are not indepent events.

Question:7 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

Given,

$P(A)=\frac{1}{2},P(A\cup B)=\frac{3}{5}$

Also, A and B are mutually exclusive means $A\cap B=\phi$.

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

$\frac{3}{5}=\frac{1}{2}+P(B)-0$

$P(B)=\frac{3}{5}-\frac{1}{2}=\frac{1}{10}$

Question:7 Given that the events$A$ and $B$ are such that $P(A)=12,P(A\cup B)=\frac{3}{5}$ and $P(B)=p.$ Find p if they are

(ii) independent

Given,

$P(A)=\frac{1}{2},P(A\cup B)=\frac{3}{5}$

Also, A and B are independent events means

$P(A\cap B) = P(A).P(B)$. Also $P(B)=p.$

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

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

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

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

$p=\frac{2}{10}=\frac{1}{5}$

Question:8 Let A and B be independent events with $\inline P(A)=0.3$ and $\inline P(B)=0.4$ Find

(i) $\inline P(A\cap B)$

$\inline P(A)=0.3$ and $\inline P(B)=0.4$

Given : A and B be independent events

So, we have

$P(A\cap B)=P(A).P(B)$

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

Question:8 Let$\inline A$ and $\inline B$ be independent events with $\inline P(A)=0.3$ and $\inline P(B)=0.4$ Find

(ii) $\inline P(A\cup B)$

$\inline P(A)=0.3$ and $\inline P(B)=0.4$

Given : A and B be independent events

So, we have

$P(A\cap B)=P(A).P(B)$

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

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

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

Question:8 Let $\inline A$ and $\inline B$ be independent events with $\inline P(A)=0.3$ and $\inline P(B)=0.4$ Find

(iii) $\inline P(A\mid B)$

$\inline P(A)=0.3$ and $\inline P(B)=0.4$

Given : A and B be independent events

So, we have $P(A\cap B)=0.12$

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

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

Question:8 Let A and B be independent events with $P(A)=0.3$ and $P(B)=0.4$ Find

(iv) $P(B\mid A)$

$\inline P(A)=0.3$ and $\inline P(B)=0.4$

Given : A and B be independent events

So, we have $P(A\cap B)=0.12$

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

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

Question:9 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 $P(not\; A\; and\; not\; B).$

If $A$ and $B$ are two events such that $P(A)=\frac{1}{4},P(B)=\frac{1}{2}$ and $P(A\cap B)=\frac{1}{8},$

$P(not\; A\; and\; not\; B)= P(A'\cap B')$

$P(not\; A\; and\; not\; B)= P(A\cup B)'$ use, $(P(A'\cap B')= P(A\cup B)')$

$= 1-(P(A)+P(B)-P(A\cap B))$

$= 1-(\frac{1}{4}+\frac{1}{2}-\frac{1}{8})$

$= 1-(\frac{6}{8}-\frac{1}{8})$

$= 1-\frac{5}{8}$

$= \frac{3}{8}$

Question:10 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 ?

If $A$ and $B$ are two events such that $P(A)=\frac{1}{2},P(B)=\frac{7}{12}$ and $P(not \; A \; or\; not\; B)=\frac{1}{4}.$

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

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

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

$\Rightarrow \, \, \, P(A\cap B)=1-\frac{1}{4}=\frac{3}{4}$

$Also \, \, \, P(A\cap B)=P(A).P(B)$

$P(A\cap B)=\frac{1}{2}\times \frac{7}{12}=\frac{7}{24}$

As we can see $\frac{3}{4}\neq \frac{7}{24}$

Hence, A and B are not independent.

Question:11 Given two independent events $\inline A$ and $\inline B$ such that $\inline P(A)=0.3,P(B)=0.6,$ Find

(i) $\inline P(A \; and\; B)$

$\inline P(A)=0.3,P(B)=0.6,$

Given two independent events $\inline A$ and $\inline B$.

$P(A\cap B)=P(A).P(B)$

$P(A\cap B)=0.3\times 0.6=0.18$

Also , we know $P(A \, and \, B)=P(A\cap B)=0.18$

Question:11 Given two independent events A and B such that $\inline P(A)=0.3,P(B)=0.6,$ Find

(ii) $\inline P(A \; and \; not\; B)$

$\inline P(A)=0.3,P(B)=0.6,$

Given two independent events $\inline A$ and $\inline B$.

$\inline P(A \; and \; not\; B)$$=P(A)-P(A\cap B)$

$=0.3-0.18=0.12$

Question:11 Given two independent events A and B such that $P(A)=0.3,P(B)0.6,$ Find

(iii) $P(A\; or \; B)$

$\inline P(A)=0.3,P(B)=0.6,$

$P(A\cap B)=0.18$

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

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

$=0.3+0.6-0.18$

$=0.9-0.18$

$=0.72$

Question:11 Given two independent events $\inline A$ and $\inline B$ such that $\inline P(A)=0.3,P(B)=0.6,$ Find

(iv) $\inline P(neither\; A\; nor\; B)$

$\inline P(A)=0.3,P(B)=0.6,$

$P(A\cap B)=0.18$

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

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

$=0.3+0.6-0.18$

$=0.9-0.18$

$=0.72$

$\inline P(neither\; A\; nor\; B)$ $=P(A'\cap B')$

$= P((A\cup B)')$

$=1-P(A\cup B)$

$=1-0.72$

$=0.28$

Question:12 A die is tossed thrice. Find the probability of getting an odd number at least once.

A die is tossed thrice.

Outcomes $=\left \{ 1,2,3,4,5,6 \right \}$

Odd numbers $=\left \{ 1,3,5 \right \}$

The probability of getting an odd number at first throw

$=\frac{3}{6}=\frac{1}{2}$

The probability of getting an even number

$=\frac{3}{6}=\frac{1}{2}$

Probability of getting even number three times

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

The probability of getting an odd number at least once = 1 - the probability of getting an odd number in none of throw

= 1 - probability of getting even number three times

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

$=\frac{7}{8}$

Question:13 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.

Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls.

Total balls =18

Black balls = 10

Red balls = 8

The probability of getting a red ball in first draw

$=\frac{8}{18}=\frac{4}{9}$

The ball is repleced after drawing first ball.

The probability of getting a red ball in second draw

$=\frac{8}{18}=\frac{4}{9}$

the probability that both balls are red

$=\frac{4}{9}\times \frac{4}{9}=\frac{16}{81}$

Question:13 Two balls are drawn at random with replacement from a box containing $\inline 10$ black and $\inline 8$ red balls. Find the probability that

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

Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls.

Total balls =18

Black balls = 10

Red balls = 8

The probability of getting a black ball in the first draw

$=\frac{10}{18}=\frac{5}{9}$

The ball is replaced after drawing the first ball.

The probability of getting a red ball in the second draw

$=\frac{8}{18}=\frac{4}{9}$

the probability that the first ball is black and the second is red

$=\frac{5}{9}\times \frac{4}{9}=\frac{20}{81}$

Question:13 Two balls are drawn at random with replacement from a box containing $10$ black and $8$ red balls. Find the probability that

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

Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls.

Total balls =18

Black balls = 10

Red balls = 8

Let the first ball is black and the second ball is red.

The probability of getting a black ball in the first draw

$=\frac{10}{18}=\frac{5}{9}$

The ball is replaced after drawing the first ball.

The probability of getting a red ball in the second draw

$=\frac{8}{18}=\frac{4}{9}$

the probability that the first ball is black and the second is red

$=\frac{5}{9}\times \frac{4}{9}=\frac{20}{81}$ $...........................1$

Let the first ball is red and the second ball is black.

The probability of getting a red ball in the first draw

$=\frac{8}{18}=\frac{4}{9}$

The probability of getting a black ball in the second draw

$=\frac{10}{18}=\frac{5}{9}$

the probability that the first ball is red and the second is black

$=\frac{4}{9}\times \frac{5}{9}=\frac{20}{81}$ $...........................2$

Thus,

The probability that one of them is black and the other is red = the probability that the first ball is black and the second is red + the probability that the first ball is red and the second is black $=\frac{20}{81}+\frac{20}{81}=\frac{40}{81}$

Question:14 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

$P(A)=\frac{1}{2}$ and $P(B)=\frac{1}{3}$

Since, problem is solved independently by A and B,

$\therefore$ $P(A\cap B)=P(A).P(B)$

$P(A\cap B)=\frac{1}{2}\times \frac{1}{3}$

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

probability that the problem is solved $= P(A\cup B)$

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

$P(A\cup B)=\frac{1}{2}+\frac{1}{3}-\frac{1}{6}$

$P(A\cup B)=\frac{5}{6}-\frac{1}{6}$

$P(A\cup B)=\frac{4}{6}=\frac{2}{3}$

Question:14 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

(ii) exactly one of them solves the problem

$P(A)=\frac{1}{2}$ and $P(B)=\frac{1}{3}$

$P(A')=1-P(A)$, $P(B')=1-P(B)$

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

probability that exactly one of them solves the problem $=P(A\cap B') + P(A'\cap B)$

probability that exactly one of them solves the problem $=P(A).P(B')+P(A')P(B)$

$=\frac{1}{2}\times \frac{2}{3}+\frac{1}{2}\times \frac{1}{3}$

$= \frac{2}{6}+\frac{1}{6}$

$= \frac{3}{6}=\frac{1}{2}$

Question:15 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’

One card is drawn at random from a well shuffled deck of $52$ cards

Total ace = 4

E : ‘the card drawn is a spade

F : ‘the card drawn is an ace’

$P(E)=\frac{13}{52}=\frac{1}{4}$

$P(F)=\frac{4}{52}=\frac{1}{13}$

$E\cap F :$ a card which is spade and ace = 1

$P(E\cap F)=\frac{1}{52}$

$P(E).P(F)=\frac{1}{4}\times \frac{1}{13}=\frac{1}{52}$

$\Rightarrow P(E\cap F)=P(E).P(F)=\frac{1}{52}$

Hence, E and F are indepentdent events .

Question:15 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 ?

(ii) E : ‘the card drawn is

F : ‘the card drawn is a king’

One card is drawn at random from a well shuffled deck of $52$ cards

Total black card = 26

total king =4

E : ‘the card drawn is black’

F : ‘the card drawn is a king’

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

$P(F)=\frac{4}{52}=\frac{1}{13}$

$E\cap F :$ a card which is black and king = 2

$P(E\cap F)=\frac{2}{52}=\frac{1}{26}$

$P(E).P(F)=\frac{1}{2}\times \frac{1}{13}=\frac{1}{26}$

$\Rightarrow P(E\cap F)=P(E).P(F)=\frac{1}{26}$

Hence, E and F are indepentdent events .

Question:15 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 ?

(iii) E : ‘the card drawn is a king or queen’

F : ‘the card drawn is a queen or jack’.

One card is drawn at random from a well shuffled deck of $52$ cards

Total king or queen = 8

total queen or jack = 8

E : ‘the card drawn is a king or queen’

F : ‘the card drawn is a queen or jack’.

$P(E)=\frac{8}{52}=\frac{2}{13}$

$P(F)=\frac{8}{52}=\frac{2}{13}$

$E\cap F :$ a card which is queen = 4

$P(E\cap F)=\frac{4}{52}=\frac{1}{13}$

$P(E).P(F)=\frac{2}{13}\times \frac{2}{13}=\frac{4}{169}$

$\Rightarrow P(E\cap F)\neq P(E).P(F)$

Hence, E and F are not indepentdent events

Question:16 In a hostel, $\inline 60\; ^{o}/_{o}$ of the students read Hindi newspaper,$\inline 40\; ^{o}/_{o}$ read English newspaper and $\inline 20\; ^{o}/_{o}$ read both Hindi and English newspapers. A student is selected at random.

(a) Find the probability that she reads neither Hindi nor English newspapers

H : $\inline 60\; ^{o}/_{o}$ of the students read Hindi newspaper,

E : $\inline 40\; ^{o}/_{o}$ read English newspaper and

$H \cap E :$ $\inline 20\; ^{o}/_{o}$ read both Hindi and English newspapers.

$P(H)=\frac{60}{100}=\frac{6}{10}=\frac{3}{5}$

$P(E)=\frac{40}{100}=\frac{4}{10}=\frac{2}{5}$

$P(H\cap E)=\frac{20}{100}=\frac{2}{10}=\frac{1}{5}$

the probability that she reads neither Hindi nor English newspapers $=1-P(H\cup E)$

$=1-(P(H)+P(E)-P(H\cap E))$

$=1-(\frac{3}{5}+\frac{2}{5}-\frac{1}{5})$

$=1-\frac{4}{5}$

$=\frac{1}{5}$

Question:16 In a hostel, $60\; ^{o}/_{o}$ of the students read Hindi newspaper, $40\; ^{o}/_{o}$ read English newspaper and $20\; ^{o}/_{o}$ read both Hindi and English newspapers. A student is selected at random.

(b) If she reads Hindi newspaper, find the probability that she reads English newspaper.

H : $\inline 60\; ^{o}/_{o}$ of the students read Hindi newspaper,

E : $\inline 40\; ^{o}/_{o}$ read English newspaper and

$H \cap E :$ $\inline 20\; ^{o}/_{o}$ read both Hindi and English newspapers.

$P(H)=\frac{60}{100}=\frac{6}{10}=\frac{3}{5}$

$P(E)=\frac{40}{100}=\frac{4}{10}=\frac{2}{5}$

$P(H\cap E)=\frac{20}{100}=\frac{2}{10}=\frac{1}{5}$

The probability that she reads English newspape if she reads Hindi newspaper $=P(E|H)$

$P(E|H)=\frac{P(E\cap H)}{P(H)}$

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

$P(E|H)=\frac{1}{3}$

Question:16 In a hostel, $60^{o}/_{o}$ of the students read Hindi newspaper, $40^{o}/_{o}$ read English newspaper and $20^{o}/_{o}$ read both Hindi and English newspapers. A student is selected at random.

(c) If she reads English newspaper, find the probability that she reads Hindi newspaper.

H : $\inline 60\; ^{o}/_{o}$ of the students read Hindi newspaper,

E : $\inline 40\; ^{o}/_{o}$ read English newspaper and

$H \cap E :$ $\inline 20\; ^{o}/_{o}$ read both Hindi and English newspapers.

$P(H)=\frac{60}{100}=\frac{6}{10}=\frac{3}{5}$

$P(E)=\frac{40}{100}=\frac{4}{10}=\frac{2}{5}$

$P(H\cap E)=\frac{20}{100}=\frac{2}{10}=\frac{1}{5}$

the probability that she reads Hindi newspaper if she reads English newspaper $= P(H |E)$

$P(H |E)=\frac{P(H\cap E)}{P(E)}$

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

$P(H |E)=\frac{1}{2}$

Question:17 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) $\frac{1}{36}$

when a pair of dice is rolled, total outcomes $=6^2=36$

Even prime number $=\left \{ 2 \right \}$

$n(even \, \, prime\, \, number)=1$

The probability of obtaining an even prime number on each die $=P(E)$

$P(E)=\frac{1}{36}$

Option D is correct.

Question:18 Two events A and B will be independent, if

(A) $\inline A$ and $\inline B$ are mutually exclusive

(B) $\inline P(A'B')=\left [ 1-P(A) \right ]\left [ 1-P(B) \right ]$

(C) $P(A)=P(B)$

(D) $P(A)+P(B)=1$

Two events A and B will be independent, if

$P(A\cap B)=P(A).P(B)$

Or $P(A'\cap B')=P(A'B')=P(A').P(B')=(1-P(A)).(1-P(B))$

Option B is correct.

## More About NCERT Solutions for Class 12 Maths Chapter 13 Exercise 13.2:-

Class 12 Maths chapter 13 exercise 13.2 solutions are consist of questions related to multiplication theorem on probability. There are 7 examples given before the NCERT syllabus exercise questions. You can solve them before solving the NCERT problems. There are 18 questions questions given in this exercise which you can solve. All these questions are solved in NCERT Solutions for Class 12 Maths Chapter 13 Exercise 13.2 in a detailed manner. You can go through these solutions for reference.

Also Read| Probability Class 12th Notes

## Benefits of NCERT Solutions for Class 12 Maths Chapter 13 Exercise 13.2:-

• Class 12th Maths chapter 13 exercise 13.2 solutions are helpful for the students get conceptual clarity and hence performing well in the in the board exams.
• You must be thorough with the NCERT textbook problems.
• These Class 12 Maths chapter 13 exercise 13.2 solutions will give you conceptual clarity about probability multiplication rule and independent events.
• There are 18 questions in this exercise which are solved by the subject matter experts on which you can rely.
• You can quickly revise the important concepts before the final exam.

• ## NCERT Solutions for Class 12 Maths Chapter 13

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