NCERT Solutions Miscellaneous Exercise: Probability | Mathematics (Maths) Class 12 - JEE PDF Download

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NCERT Solutions for Class 12 Maths Chapter 13 – 
Probability  
Miscellaneous Exercise                                                  Page No: 582 
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 n B = f 
Solution: 
 
 
2. A couple has two children, 
(i) Find the probability that both children are males, if 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. 
Solution: 
(i) According to the question, if the couple has two children, then the sample space is 
S = {(b, b), (b, g), (g, b), (g, g)} 
Assume that A denote the event of both children having male and B denote the event of having at least one of the male 
children. 
Thus, we have 
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NCERT Solutions for Class 12 Maths Chapter 13 – 
Probability  
Miscellaneous Exercise                                                  Page No: 582 
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 n B = f 
Solution: 
 
 
2. A couple has two children, 
(i) Find the probability that both children are males, if 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. 
Solution: 
(i) According to the question, if the couple has two children, then the sample space is 
S = {(b, b), (b, g), (g, b), (g, g)} 
Assume that A denote the event of both children having male and B denote the event of having at least one of the male 
children. 
Thus, we have 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 13 – 
Probability  
 
(ii) Assume that C denotes the event of having both children as females and D denotes the event of having an elder 
child as female. 
? C = {(g, g)} 
P (C) = ¼ 
And, D = {(g, b), (g, g)} 
P (D) = (2/4) 
 
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 is an equal number of males and females. 
Solution: 
Given that 5% of men and 0.25% of women have grey hair. 
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NCERT Solutions for Class 12 Maths Chapter 13 – 
Probability  
Miscellaneous Exercise                                                  Page No: 582 
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 n B = f 
Solution: 
 
 
2. A couple has two children, 
(i) Find the probability that both children are males, if 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. 
Solution: 
(i) According to the question, if the couple has two children, then the sample space is 
S = {(b, b), (b, g), (g, b), (g, g)} 
Assume that A denote the event of both children having male and B denote the event of having at least one of the male 
children. 
Thus, we have 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 13 – 
Probability  
 
(ii) Assume that C denotes the event of having both children as females and D denotes the event of having an elder 
child as female. 
? C = {(g, g)} 
P (C) = ¼ 
And, D = {(g, b), (g, g)} 
P (D) = (2/4) 
 
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 is an equal number of males and females. 
Solution: 
Given that 5% of men and 0.25% of women have grey hair. 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 13 – 
Probability  
? Total % of people having grey hair = 5 + 0.25 
= 5.25 % 
Hence, the probability of having a selected male person having grey hair, P = 5/25 = 20/21 
4. Suppose that 90% of people are right-handed. What is the probability that at most 6 of a random sample of 10 
people are right-handed? 
Solution: 
Given that 90% of the people are right-handed. 
Let p denotes the probability of people that are right-handed, and q denotes the probability of people that are left-
handed. 
p = 9/10 and q = 1 – 9/10 = 1/10 
Now, by using the binomial distribution probability of having more than 6 right-handed people can be given as: 
 
5. An urn contains 25 balls of which 10 balls bear the mark ‘X’ and the remaining 15 bear the mark ‘Y’. A ball 
is drawn at random from the urn, its mark is 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. 
Solution: 
(i) It is given in the question that 
Total number of balls in the urn = 25 
Number of balls bearing mark ‘X’ = 10 
Number of balls bearing mark ‘Y’ = 15 
Let p denotes the probability of balls bearing the mark ‘X’, and q denotes the probability of balls bearing the mark ‘Y’. 
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NCERT Solutions for Class 12 Maths Chapter 13 – 
Probability  
Miscellaneous Exercise                                                  Page No: 582 
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 n B = f 
Solution: 
 
 
2. A couple has two children, 
(i) Find the probability that both children are males, if 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. 
Solution: 
(i) According to the question, if the couple has two children, then the sample space is 
S = {(b, b), (b, g), (g, b), (g, g)} 
Assume that A denote the event of both children having male and B denote the event of having at least one of the male 
children. 
Thus, we have 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 13 – 
Probability  
 
(ii) Assume that C denotes the event of having both children as females and D denotes the event of having an elder 
child as female. 
? C = {(g, g)} 
P (C) = ¼ 
And, D = {(g, b), (g, g)} 
P (D) = (2/4) 
 
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 is an equal number of males and females. 
Solution: 
Given that 5% of men and 0.25% of women have grey hair. 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 13 – 
Probability  
? Total % of people having grey hair = 5 + 0.25 
= 5.25 % 
Hence, the probability of having a selected male person having grey hair, P = 5/25 = 20/21 
4. Suppose that 90% of people are right-handed. What is the probability that at most 6 of a random sample of 10 
people are right-handed? 
Solution: 
Given that 90% of the people are right-handed. 
Let p denotes the probability of people that are right-handed, and q denotes the probability of people that are left-
handed. 
p = 9/10 and q = 1 – 9/10 = 1/10 
Now, by using the binomial distribution probability of having more than 6 right-handed people can be given as: 
 
5. An urn contains 25 balls of which 10 balls bear the mark ‘X’ and the remaining 15 bear the mark ‘Y’. A ball 
is drawn at random from the urn, its mark is 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. 
Solution: 
(i) It is given in the question that 
Total number of balls in the urn = 25 
Number of balls bearing mark ‘X’ = 10 
Number of balls bearing mark ‘Y’ = 15 
Let p denotes the probability of balls bearing the mark ‘X’, and q denotes the probability of balls bearing the mark ‘Y’. 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 13 – 
Probability  
p = 10/25 = 2/5 and q = 15/25 = 3/5 
Now, 6 balls are drawn with replacement. Hence, the number of trials are Bernoulli triangle. 
Assume Z be the random variable that represents the number of balls bearing the ‘Y’ mark in the trials. 
? Z has a binomial distribution where n = 6 and p = 2/5 
 
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NCERT Solutions for Class 12 Maths Chapter 13 – 
Probability  
Miscellaneous Exercise                                                  Page No: 582 
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 n B = f 
Solution: 
 
 
2. A couple has two children, 
(i) Find the probability that both children are males, if 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. 
Solution: 
(i) According to the question, if the couple has two children, then the sample space is 
S = {(b, b), (b, g), (g, b), (g, g)} 
Assume that A denote the event of both children having male and B denote the event of having at least one of the male 
children. 
Thus, we have 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 13 – 
Probability  
 
(ii) Assume that C denotes the event of having both children as females and D denotes the event of having an elder 
child as female. 
? C = {(g, g)} 
P (C) = ¼ 
And, D = {(g, b), (g, g)} 
P (D) = (2/4) 
 
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 is an equal number of males and females. 
Solution: 
Given that 5% of men and 0.25% of women have grey hair. 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 13 – 
Probability  
? Total % of people having grey hair = 5 + 0.25 
= 5.25 % 
Hence, the probability of having a selected male person having grey hair, P = 5/25 = 20/21 
4. Suppose that 90% of people are right-handed. What is the probability that at most 6 of a random sample of 10 
people are right-handed? 
Solution: 
Given that 90% of the people are right-handed. 
Let p denotes the probability of people that are right-handed, and q denotes the probability of people that are left-
handed. 
p = 9/10 and q = 1 – 9/10 = 1/10 
Now, by using the binomial distribution probability of having more than 6 right-handed people can be given as: 
 
5. An urn contains 25 balls of which 10 balls bear the mark ‘X’ and the remaining 15 bear the mark ‘Y’. A ball 
is drawn at random from the urn, its mark is 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. 
Solution: 
(i) It is given in the question that 
Total number of balls in the urn = 25 
Number of balls bearing mark ‘X’ = 10 
Number of balls bearing mark ‘Y’ = 15 
Let p denotes the probability of balls bearing the mark ‘X’, and q denotes the probability of balls bearing the mark ‘Y’. 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 13 – 
Probability  
p = 10/25 = 2/5 and q = 15/25 = 3/5 
Now, 6 balls are drawn with replacement. Hence, the number of trials are Bernoulli triangle. 
Assume Z be the random variable that represents the number of balls bearing the ‘Y’ mark in the trials. 
? Z has a binomial distribution where n = 6 and p = 2/5 
 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 13 – 
Probability  
 
6. In a hurdle race, a player has to cross 10 hurdles. The probability that he will clear each hurdle is 5/6. What is 
the probability that he will knock down fewer than 2 hurdles? 
Solution: 
Assume that p be the probability that the player will clear the hurdle, while q be the probability that the player will 
knock down the hurdle. 
? p = 5/6 and q = 1 – 5/6 = 1/6 
Let us also assume X be the random variable that represents the number of times the player will knock down the hurdle.