NCERT Solutions class 12 Maths Exercise Miscellaneous (Ex Misc.) Chapter 13 Probability

NCERT Solutions for Class 12 Maths Exercise Miscellaneous Chapter 13 Probability – FREE PDF Download

Free PDF download of NCERT Solutions for Class 12 Maths Chapter 13 Exercise Miscellaneous (Ex Misc) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 12 Maths Chapter 13 Probability Exercise Miscellaneous Questions with Solutions to help you to revise complete Syllabus and Score More marks.

NCERT Solutions for Class 12 Maths Chapter 13 Probability (Ex Misc) Exercise Miscellaneous

1.A and B are two events such that P (A)  0. Find P (B/A) if:

(i)A is a subset of B

(ii)A  B =  


Ans. A and B are two events such that P (A)  0 

To find: P

(i) A is a subset of B

 A  B


(ii)  B = 


2.A couple has two children.


(i)Find the probability that both children are males, it is known that at least one of the children is male.

(ii)Find the probability that both children are females if it is known that the elder child is a female.


Ans. (i) Let B1 and G1 stand for male and female respectively. 

Now the sample space is (S) = {B1B2, B1G2, B2G1, G1G2}

Let us consider the following events,

A = both are males

B = at least one is a male

 A = {B1B2} and B = {B1B2, B1G2, B2G1}

P (B) = , A  B = {B1B2}  and 

Required probability = P

(ii) Let A = both are females, A = {G1G2} and C = the older is a girl

C = {B1B2, B1G2}P (C) = 

Required probability = P

3.Suppose that 5% of men and 0.25% of women have grey hair. A grey haired person is selected at random. What is the probability of this person being male? Assume that there are equal number of males and females.



Ans. Men are represented by E1 and women are represented by E2

P (E1) =  and P (E2) = 

A represents grey hair persons.

P and P


 =  =  =  5100×10000525=20215100×10000525=2021

4.Suppose that 90% of people are right-handed. What is the probability that at most of 6 of a random sample of 10 people are right-handed?



Ans.  and  

P (at most 6 successes) = 1 – [P (X = 7) + P (X = 8) + P (X = 9) + P (X = 10)]

= 1 – 

5.An urn contains 25 balls of which 10 balls bear a mark ‘X’ and the remaining 15 bear a mark ‘Y’. A ball is drawn at random from the urn, its mark noted down and it is replaced. If 6 balls are drawn in this way, find the probability that:


(i)all will bear ‘X’ mark.

(ii)not more than 2 will bear ‘Y’ mark.

(iii)at least one ball will bear ‘Y’ mark.

(iv)the number of balls with ‘X’ mark and ‘Y’ mark will be equal.


Ans. S = {10 balls with mark X, 15 balls with mark Y}  = 25 

Let a ball drawn with mark X be denoted by A.

A = {10 balls with mark X} = 10

 and  q=1p=125=35q=1−p=1−25=35

(i) All will bear X mark, i.e.,  = 6

P (X = ) = 

P (X = 6) = 

(ii) Not more than 2 will bear mark Y.

 For Y mark,  = 0, 1, 2 and For X mark,  = 6, 5, 4

P (not more than 2 will bear mark Y) = P (X = 4) + P (X = 5) + P (X =6)


(iii) P (at least one ball will bear Y mark) = P (not more than 5 balls will bear mark X)

= 1 – P (6)

= 1 – 

(iv) P (equal number of balls will bear mark X and Y) = P (3)

6.In a hurdle race, a player has to cross 10 hurdles. The probability that he will clear each hurdle is  What is the probability that he will knock down fewer than 2 hurdles?



Ans.  = Probability of knocking down a hurdle =  

 = Probability of clearing a hurdle = 

P (He will knock down fewer than 2 hurdles) =  = P (X = 0) + P (X = 1)

7.A die is thrown again and again until three sixes are obtained. Find the probability of obtaining the third six in the sixth throw of the die.



Ans. S = {1, 2, 3, 4, 5, 6}  = 6 

A = {6} = 1


Since the 3rd six is obtained in the 6th throw of die, there are two sixes i.e., two successes in the first 5 throws.

  Required probability = P (2 success in the first 5 throws) x P (success in the 6th throw)

8.If a leap year is selected at random, what is the change that it will contain 52 Tuesday?



Ans. A leap year has 366 days which means 52 complete weeks and 2 days. If any one of these two days in a Tuesday then the year will have 53 Tuesdays. 

Number of total days in a week = 7

Number of favourable days = 2

Therefore, P (the year will have 53 Tuesday) = 

9.An experiment succeeds twice as often as it fails. Find the probability that in the next six trails, there will be at least 4 successes.







P (at least 4 successes) = P (X = 4) + P (X = 5) + P (X = 6)

10.How many times must a man toss a fair coin so that the probability of having at least one head is more than 90%?



Ans. For one fair coin, probability of head  = , therefore, probability of tail =  



Therefore, it is clear that 

Since the minimum value of nn is 4, the man must loss the coin at least 4 times.

11.In a game a man wins a rupee for a six and looses a rupee for any other number when a fair die is thrown. The man decided to throw a die thrice but to quit as and when he gets a six. Find the expected value of the amounts he wins/looses.



Ans. When a die is thrown, probability of getting a six  =  


(i) If he gets a six in first throw, then,
Probability of getting a six = 

(ii) If he does not get a six, in first throw, but he gets a six in the second throw, then
Its probability = 
Probability that he does not get a six in first two throws and he gets a six in third throw

Probability that he does not get a six in any of the three throws = 
In first throw he gets a six, will receive Re.1
If he gets a six in second throw, he will receive Rs. (1 – 1) = 0
If he gets a six in third throw, he will receive Rs. (– 1 – 1 + 1) = Rs. – 1
= he will loss Rs. 1
If he does not get a six in all three throws, he will receive Rs. (–1 –1 –1) = Rs. –3
Expected value = 16×1+(56×16)×0+(56×56×16)×(1)+(56×56×56)×316×1+(56×16)×0+(56×56×16)×(−1)+(56×56×56)×−3
162521637521616−25216−375216364216=9154−364216=−9154 = Rs. 91549154Loss

12.Suppose we have four boxes A, B, C and D containing coloured marbles as given below:


BoxMarble colour
















One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from box A?, box B? box C?


Ans. Let R represents the drawing of red ball and the four boxes are represented by A, B, C and D. 

So , ,

Since there are 4 bags.

Therefore, P (A) =  P (B) =  P (C) =  P (D) = 

P(A|R)=P(A|R)= P(A).P(R|A)P(A).P(R|A)+P(B).P(R|B)+P(C).P(R|C)+P(D).P(R|D)P(A).P(R|A)P(A).P(R|A)+P(B).P(R|B)+P(C).P(R|C)+P(D).P(R|D)



13.Assume that the chances of a patient having a heart attack is 40%. It is also assumed that a meditation and yoga course reduces the risk of heart attack by 30% and prescription of certain drug reduces its chances by 25%. At a time a patient can choose any one of the two options with equal probabilities. It is given that after going through one of the two options the patient selected at random suffers a heart attack. Find the probability that the patient followed a course of meditation and yoga.



Ans. A patient has options to have the treatment of yoga and meditation and that of prescription of drugs. 

Let these events be denoted by E1 and E2 i.e.,

E1 = Treatment of yoga and meditation

E= Treatment of prescription of certain drugs

P (E1) =  and P (E2) = 

Let A denotes that a person has heart attack, then P (A) = 40% = 0.40

Yoga and meditation reduces heart attack by 30.

 Inspite of getting yoga and meditation treatment heart risk is 70% of 0.40

  = 0.40 x 0.70 = 0.28

Also, Drug prescription reduces the heart attack rick by 25%

Even after adopting the drug prescription hear rick is 75% of 0.40

  = 0.40 x 0.75 = 0.30

14.If each element of a second order determinant is either zero or one, what is the probability that the value of the determinant is positive? (Assume that the individual entries of the determinant are chosen independently, each value being assumed with probability ).



Ans. There are four entries in a determinant of 2 x 2 order. Each entry may be filled up in two ways with 0 or 1. 

 Number of determinants that can be formed =  = 16

The value of determinants is positive in the following cases:

  = 3

Therefore, the probability that the determinant is positive = 

15. An electronic assembly consists of two sub-systems say A and B. From previous testing procedures, the following probabilities are assumed to be known:


P (A fails) = 0.2

P (B fails alone) = 0.15

P (A and B fail) = 0.15

Evaluate the following probabilities.


Ans. Event A fails and B fails denoted by  and  respectively. 

 and P (A and B fails) = 0.15

  = 0.15

P( above) = 

0.15 =  0.15

 = 0.30

(i) P = 0.150.30=12=

(ii) P (A fails alone) = P ( alone) =  = 0.20 – 0.15 = 0.05

16.Bag I contains 3 Red and 4 Black balls and B II contains 4 Red and % Black balls. One ball is transferred from Bag I to bag II and then a ball is drawn from bag II. The ball so drawn is found to be Red in colour. Find the probability that the transferred ball is Black.



Ans. Let E1 = Ball transferred from Bag I to Bag II is red 

E2 = Ball transferred from Bag I to Bag Ii is black

A = Ball drawn from Bag II is red in colour

P (E1) =  and P (E2) = 

  =  and = 



Choose the correct answer in each of the following:


17.If A and B are two events such that P (A)  0 and P = 1, then:

(A) A  B

(B) B  A

(C) B =  

(D) A =  


Ans. A and B are two events such that P (A)  0 and P = 1 



Therefore, option (A) is correct.

18. If P > P (A), then which of the following is correct:


(A) P < P (B)

(B) P (A  B) < P (A) . P (B)

(C) P  > P (B)

(D) P  = P (B)



Therefore, option (C) is correct.

19.If A and B are any two events such that P (A) + P (B) – P (A and B) = P (A), then


(A) P  = 1

(B) P  = 1

(C) P  = 0

(D) P  = 0


Ans. P (A) + P (B) – P (A and B) = P (A) 

P (B) – P (A and B)

 = 1

Therefore, option (B) is correct.