NCERT Solutions for Exercise 13.2 Class 12 Maths Chapter 13 Probability are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. 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.

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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.

Answer:

$P(A)=\frac{3}{5}$ 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.

Answer:

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

Answer:

Total oranges = 15

Good oranges = 12

Bad oranges = 3

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.

Answer:

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 $1,2,3$ 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?

Answer:

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

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

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

$n(A)=3.$

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

$B$ 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?

Answer:

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

Answer:

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

Answer:

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 $P(A)=0.3$ and $P(B)=0.4$ Find

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

Answer:

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

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 $A$ and $B$ be independent events with $P(A)=0.3$ and $P(B)=0.4$ Find

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

Answer:

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

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 $A$ and $B$ be independent events with $P(A)=0.3$ and $P(B)=0.4$ Find

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

Answer:

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

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) $P(B\mid A)$

Answer:

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

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 $P(not\; A\; and\; not\; B).$

Answer:

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 ?

Answer:

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 $A$ and $B$ such that $P(A)=0.3,P(B)=0.6,$ Find

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

Answer:

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

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

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

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

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

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

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

Answer:

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

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

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

$=0.3-0.18=0.12$

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

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

Answer:

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

$P(A\cap B)=0.18$

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

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

$=0.3+0.6-0.18$

$=0.9-0.18$

$=0.72$

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

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

Answer:

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

$P(A\cap B)=0.18$

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

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

$=0.3+0.6-0.18$

$=0.9-0.18$

$=0.72$

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

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

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

$=1-0.72$

$=0.28$

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

Answer:

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.

Answer:

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 $10$ black and $8$ red balls. Find the probability that

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

Answer:

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.

Answer:

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

Answer:

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

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

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

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

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

(ii) exactly one of them solves the problem

Answer:

$P(A)=\frac{1}{2}$ 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’

Answer:

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

Total ace = 4

total spades =13

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’

Answer:

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’.

Answer:

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, $60\; ^{o}/_{o}$ 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.

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

Answer:

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

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

$H \cap E :$ $20\; ^{o}/_{o}$ 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 $=1-P(H\cup E)$

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

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

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

$=\frac{1}{5}$

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

Answer:

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

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

$H \cap E :$ $20\; ^{o}/_{o}$ 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.

Answer:

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

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

$H \cap E :$ $20\; ^{o}/_{o}$ 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) $\frac{1}{3}$

(D) $\frac{1}{36}$

Answer:

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) $A$ and $B$ are mutually exclusive

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

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

Answer:

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.

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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.

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NCERT Solutions of Class 12 Subject Wise

Subject Wise NCERT Exampler Solutions

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Frequently Asked Question (FAQs)

1. What is the number of questions in Probability exercise 13.2 ?

There are 18 questions in NCERT Class 12 Maths chapter 13 exercise 13.2.

2. What does it means by Independent Events ?

If the occurrence of one event doesn’t affect the probability of the other event such events are called independent events.

3. What is the probability of getting a tail when we toss a fair coin ?

The probability of getting a tail is 0.5 when we toss a fair coin.

4. Can I get free NCERT Exemplar solutions for Class 12 Maths probability?

You will get NCERT Exemplar Solutions for Class 12 Maths Probability by clicking on the given link.

5. What is an Impossible Event ?

An event is called an impossible event if the number of favorable outcomes is zero.

6. What is the probability of impossible event ?

The probability of an impossible event is zero.

7. Can I get free NCERT solutions for Class 12 Maths?

Click here to get NCERT Solutions for Class 12 Maths for other subjects.

8. If the probability of getting tail on tossing of a biased coin is 0.47 then what is the probability of getting head ?

The probability of getting head of this biased coin is 0.53.

Also Read

NCERT Class 12 Physics Chapter 9 Notes Ray Optics and Optical Instruments - Download PDF

NCERT Class 12 Physics Chapter 8 Notes Electromagnetic Waves - Download PDF

NCERT Class 12 Physics Chapter 7 Notes Alternating Current - Download PDF

NCERT Class 12 Physics Chapter 6 Notes Electromagnetic Induction - Download PDF

NCERT Class 12 Physics Chapter 5 Notes Magnetism and Matter - Download PDF

NCERT Class 12 Physics Chapter 4 Notes Moving Charges and Magnetism - Download PDF

NCERT Books for Class 12 Hindi 2024 - Download chapter wise PDFs here

NCERT Books for Class 12 Biology 2024 - Download Free PDF

NCERT Books for Class 12 Chemistry 2024 - Download Free Pdf

NCERT Book for Class 12 Maths 2024 - Download Free PDF

NCERT Book for Class 12 Physics 2024 - Download Free PDF

NCERT Books for Class 12 - Download All Subjects PDFs Here

NCERT Syllabus for Class 12 English 2024 - Download Syllabus PDF

NCERT Syllabus for Class 12 2024 (All Subjects) - Download PDF Here

NCERT Syllabus for Class 6 to 12 (2023-24) - All Subjects PDF Download

NCERT Solutions for Miscellaneous Exercise Chapter 7 Class 12 - Integrals

NCERT Solutions for Exercise 4.1 Class 11 Maths Chapter 4 - Principle of Mathematical Induction

NCERT Solutions for Class 12 Biology - Check CBSE 12 Biology NCERT Chapterwise Solution

NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry

NCERT Solutions for Class 12

NCERT Solutions for Class 12 Maths (Updated 2023-24)

NCERT Exemplar Class 12 Physics Solutions Chapter 8 Electromagnetic Waves

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

I am interested in graphic designing and coding. how do I make a career .which stream should I go.what engineering subject to take

Hello aspirant,

The purpose of graphic design extends beyond the brand's look. Nevertheless, by conveying what the brand stands for, it significantly aids in the development of a sense of understanding between a company and its audience. The future in the field of graphic designing is very promising.

There are various courses available for graphic designing. To know more information about these courses and much more details, you can visit our website by clicking on the link given below.

https://www.careers360.com/courses/graphic-designing-course

Thank you

Hope this information helps you.

Read Complete Answer

Sajal Trivedi 29 January,2024

i couldn't clear 1st compartment (which held in August 2022) cbse class 12th so can I give improvement in all subjects which to be held on March 2023. tell me a solution i need more than 75% in class 12th

hello,

Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.

I hope this was helpful!

Good Luck

Read Complete Answer

Upasana Barman 24 August,2022

i was not able to clear 1st compartment and now I have to give 2nd compartment so can I give improvement exam next year? plz tell me very much tensed(cbse class 12th)

Hello dear,

If you was not able to clear 1st compartment and now you giving second compartment so YES, you can go for your improvement exam next year but if a student receives an improvement, they are given the opportunity to retake the boards as a private candidate the following year, but there are some requirements. First, the student must pass all of their subjects; if they received a compartment in any subject, they must then pass the compartment exam before being eligible for the improvement.

As you can registered yourself as private candidate for giving your improvement exam of 12 standard CBSE(Central Board of Secondary Education).For that you have to wait for a whole year which is bit difficult for you.

Positive side of waiting for whole year is you have a whole year to preparing yourself for your examination. You have no distraction or something which may causes your failure in the exams. In whole year you have to stay focused on your 12 standard examination for doing well in it. By this you get a highest marks as a comparison of others.

Believe in Yourself! You can make anything happen

All the very best.

Read Complete Answer

Mayankjha.student2007 23 August,2022

if i am giving improvement in one subject this year, will i be able to give improvement in more subjects next year?

Hello Student,

I appreciate your Interest in education. See the improvement is not restricted to one subject or multiple subjects and we cannot say if improvement in one subject in one year leads to improvement in more subjects in coming year.

You just need to have a revision of all subjects what you have completed in the school. have a revision and practice of subjects and concepts helps you better.

All the best.

Read Complete Answer

Nuthalapati Vyshnavi 21 August,2022

I got an RT for 2 subjects, so do I have to give exams for all the subjects in the coming year or just the 2. pls help me

Hi,

You just need to give the exams for the concerned two subjects in which you have got RT. There is no need to give exam for all of your subjects, you can just fill the form for the two subjects only.

Read Complete Answer

Muskan Dhalwal 17 August,2022

View All

Popular CBSE Class 12th Questions

A block of mass 0.50 kg is moving with a speed of 2.00 ms^-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up ? Fat supplies 3.8×10⁷J of energy per kg which is converted to mechanical energy with a 20% efficiency rate. Take g = 9.8 ms⁻² :

Option 1)

2.45×10⁻³ kg

Option 2)

6.45×10⁻³ kg

Option 3)

9.89×10⁻³ kg

Option 4)

12.89×10⁻³ kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

A particle is projected at 60⁰ to the horizontal with a kinetic energy $K$ . The kinetic energy at the highest point

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

In the reaction,

$2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, $Mg_{3}(PO_{4})_{2}$ will contain 0.25 mole of oxygen atoms?

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol^-1) is

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t²) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m² , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

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3 Jobs Available

Journalist

Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.

3 Jobs Available

Editor

Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.

3 Jobs Available

Reporter

Individuals who opt for a career as a reporter may often be at work on national holidays and festivities. He or she pitches various story ideas and covers news stories in risky situations. Students can pursue a BMC (Bachelor of Mass Communication), B.M.M. (Bachelor of Mass Media), or MAJMC (MA in Journalism and Mass Communication) to become a reporter. While we sit at home reporters travel to locations to collect information that carries a news value.

2 Jobs Available

Corporate Executive

Are you searching for a Corporate Executive job description? A Corporate Executive role comes with administrative duties. He or she provides support to the leadership of the organisation. A Corporate Executive fulfils the business purpose and ensures its financial stability. In this article, we are going to discuss how to become corporate executive.

2 Jobs Available

Multimedia Specialist

A multimedia specialist is a media professional who creates, audio, videos, graphic image files, computer animations for multimedia applications. He or she is responsible for planning, producing, and maintaining websites and applications.

2 Jobs Available

Welding Engineer

5 Jobs Available

QA Manager

4 Jobs Available

Quality Controller

A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product.

A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

3 Jobs Available

Production Manager

3 Jobs Available

Product Manager

3 Jobs Available

QA Lead

A QA Lead is in charge of the QA Team. The role of QA Lead comes with the responsibility of assessing services and products in order to determine that he or she meets the quality standards. He or she develops, implements and manages test plans.

2 Jobs Available

Structural Engineer

2 Jobs Available

Process Development Engineer

The Process Development Engineers design, implement, manufacture, mine, and other production systems using technical knowledge and expertise in the industry. They use computer modeling software to test technologies and machinery. An individual who is opting career as Process Development Engineer is responsible for developing cost-effective and efficient processes. They also monitor the production process and ensure it functions smoothly and efficiently.

2 Jobs Available

QA Manager

4 Jobs Available

AWS Solution Architect

An AWS Solution Architect is someone who specializes in developing and implementing cloud computing systems. He or she has a good understanding of the various aspects of cloud computing and can confidently deploy and manage their systems. He or she troubleshoots the issues and evaluates the risk from the third party.

4 Jobs Available

Azure Administrator

An Azure Administrator is a professional responsible for implementing, monitoring, and maintaining Azure Solutions. He or she manages cloud infrastructure service instances and various cloud servers as well as sets up public and private cloud systems.

4 Jobs Available

Computer Programmer

Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

3 Jobs Available

Product Manager

3 Jobs Available

Information Security Manager

Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack

3 Jobs Available

ITSM Manager

3 Jobs Available

Automation Test Engineer

An Automation Test Engineer job involves executing automated test scripts. He or she identifies the project’s problems and troubleshoots them. The role involves documenting the defect using management tools. He or she works with the application team in order to resolve any issues arising during the testing process.

2 Jobs Available