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

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

Edited By Ramraj Saini | Updated on Dec 04, 2023 11:45 AM IST | #CBSE Class 12th
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CBSE Class 12th  Exam Date : 15 Feb' 2025 - 15 Feb' 2025

NCERT Solutions For Class 12 Maths Chapter 13 Exercise 13.2

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.

12th class Maths exercise 13.2 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise enumerated in NCERT Book together using the link provided below.

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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 eventsA 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 LetA 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}

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

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}

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.

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.

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

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 ]

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

(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.

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.

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Key Features Of NCERT Solutions for Exercise 13.2 Class 12 Maths Chapter 13

  • Comprehensive Coverage: The solutions encompass all the topics covered in ex 13.2 class 12, ensuring a thorough understanding of the concepts.
  • Step-by-Step Solutions: In this class 12 maths ex 13.2, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
  • Accuracy and Clarity: Solutions for class 12 ex 13.2 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
  • Conceptual Clarity: In this 12th class maths exercise 13.2 answers, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
  • Inclusive Approach: Solutions for ex 13.2 class 12 cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
  • Relevance to Curriculum: The solutions for class 12 maths ex 13.2 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.

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NCERT Solutions of Class 12 Subject Wise

Subject Wise NCERT Exampler Solutions

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Frequently Asked Questions (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.

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?
quartely question paper for mathematics cbse

Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

Read Complete Answer
Jefferson 25 September,2024
i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

Possible steps:

  1. Re-evaluate Your Study Strategies:

    • Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
    • Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
    • Practice Regularly: Consistent practice is key to mastering chemistry.

  2. Consider Professional Help:

    • Tutoring: A tutor can provide personalized guidance and support.
    • Counseling: If you're feeling overwhelmed or unsure about your path, counseling can help.

  3. Explore Alternative Options:

    • Retake the Exam: If you're confident in your ability to improve, consider retaking the chemistry compartment exam.
    • Change Course: If you're not interested in pursuing chemistry further, explore other academic options that align with your interests.

  4. Focus on NEET 2025 Preparation:

    • Stay Dedicated: Continue your NEET preparation with renewed determination.
    • Utilize Resources: Make use of study materials, online courses, and mock tests.

  5. Seek Support:

    • Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
    • Join Study Groups: Collaborating with peers can create a supportive learning environment.

Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.







Read Complete Answer
Kanishka kaushikii 17 September,2024
i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.

Scholarship Details:

Type A: For candidates scoring 60% or above in the exam.

Type B: For candidates scoring between 50% and 60%.

Type C: For candidates scoring between 40% and 50%.

Cash Scholarship:

Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

Read Complete Answer
Yuvan 30 August,2024
i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

Read Complete Answer
Ashvadha 12 July,2024
Kya mujhe class 12 ka important question milega jo exam mai per year puche jate hai

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

Read Complete Answer
kumardurgesh1802 10 July,2024
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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
Read More

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 3.8×107 J 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

 
Read More

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
Read More

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
Read More

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 .
Read More

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
Read More

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.
Read More

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.
Read More

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
Read More

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
Read More

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