# Class 12 RD Sharma Solutions – Chapter 31 Probability – Exercise 31.5 | Set 2

### Question 13. In a family, Husband (H) tells a lie in 30% cases and Wife(W) tells a lie in 35% cases. Find the probability that both contradict on the same fact.

**Solution:**

According to question:

It is given that,

In a family, Husband (H) tells a lie in 30% cases and Wife(W) tells a lie in 35% cases.

P(H) = 30/100, P(H

‘) = 70/100P(W) = 35/100, P(W

‘) = 65/100Now,

P(Both contradict on the same fact)

= [P(H ∩ W

‘) ∪ P(H‘∩ W)]= P(H ∩ W

‘) + P(H‘∩ W)= P(H) × P(W

‘) + P(H‘) × P(W)= 30/100 × 65/100 + 70/100 × 35/100

= 4400/10000 = 0.44

Hence, The required probability is 0.44 i.e., 44%

### Question 14. A husband and wife appear in an interview for two vacancies in the same post. The probability of the husband’s selection is 1/7 and that of the wife’s selection is 1/5. What is the probability that

**(a) both of them will be selected?**

**(b) only one of them will be selected?**

**(c) none of them will be selected?**

**Solution:**

According to Question,

It is given that,

P(H) = 1/7 and, P(H

‘) = 6/7P(W) = 1/5 and, P(W

‘) = 4/5

(i)P(Both of them will be selected)= P(H∩W)

= P(H) × P(W)

= 1/7 × 1/5 = 1/35

Hence, The required probability = 1/35.

(ii)P(only one of them will be selected)= P[(H ∩ W

‘) ∪ (H‘∩ W)]= P(H ∩ W

‘) + P(H‘∩ W)= P(H) × P(W

‘) + P(H‘) × P(W)= 1/7 × 4/5 + 6/7 × 1/5

= 4/35 + 6/35 = 10/35 = 2/7

Hence, The required probability = 2/7

(iii)P(none of them will be selected)= P(H

‘∩ W‘)= P(H

‘) × P(W‘)= 6/7 × 4/5 = 24/35

Hence, The required probability = 24/35

### Question 15. A bag contains 7 white, 5 black, and 4 red balls. Four balls are drawn without replacement. Find the probability that at least three balls are black.

**Solution:**

According to question,

It is given that,

A bag contains 7 white, 5 black, and 4 red balls.

And, four balls are drawn without replacement

Now,

P(at least three balls are black)

= P(exactly 3 black balls) + P(all 4 black balls)

= (11/16 × 5/15 × 4/14 × 3/13 × 4) + (5/16 × 4/15 × 3/14 × 2/13)

= 11/ (13 × 14) + 1/ (2 × 13× 14)

= 23/364

Hence, the required probability = 23/364

### Question 16. A, B, and C are independent witness of an event which is known to have occurred. A speaks the truth three times out of four, B four times out of five, and C five times out of six. What is the probability that the occurrence will be reported truthfully by the majority of three witnesses?

**Solution:**

According to question,

It is given that,

P(A speaks truth) = 3/4 and, P(A

‘) = 1/4P(B speaks truth) = 4/5 and, P(B

‘) = 1/5P(C speaks truth) = 5/6, and P(C

‘) = 1/6Now,

P(Reported truthfully by the majority of three witnesses)

= P(A) × P(B) × P(C

‘) + P(A) × P(B‘) × P(C) + P(A‘) × P(B) × P(C)= 3/4 × 4/5 × 1/6 + 3/4 × 1/5 × 5/6 + 1/4 × 4/5 × 5/6

= 107/120

Hence, the required probability = 107/120

### Question 17. A bag contains 4 white balls and 2 black balls. Another contains 3 white balls and 5 black balls. If one ball is drawn from each bag, find the probability that

**(i) Both are white**

**(ii) Both are black**

**(iii) One is white and one is black**

**Solution:**

According to question,

It is given that,

Bag 1 contains 4 white balls and 2 black balls.

Bag 2 contains 3 white balls and 5 black balls.

(i)P(Both are white)= 4/6 × 3/8 = 12/48

= 1/4

Required probability = 1/4

(ii)P(Both are black)= 2/6 × 5/8 = 10/48

= 5/24

Required probability = 5/24

(iii)P(One is white and one is black)= 4/6 × 5/8 + 3/8 × 2/6

= 20/48 + 6/ 48

= 13/24

Required probability = 13/24

### Question 18. A bag contains 4 white, 7 black, and 5 red balls. 4 balls are drawn with replacement. What is the probability that at least two are white?

**Solution:**

According to question,

It is given that,

A bag contains 4 white, 7 black, and 5 red balls.

Now,

P(at least two are white)

= 1 – P(Maximum 1 white ball)

= 1 – [P(no white) + P(exactly 1 white ball)]

= 1 – [12/16 × 12/16 × 12/16 × 12/16 + 4/16 × 12/16 × 12/16 × 12/16 × 4]

= 1 – [81/256 + 108/256]

= 1 – 189/256 = 67/256

Hence, the required probability = 67/256

### Question 19. Three cards are drawn with replacement from a well-shuffled pack of cards. Find the probability that the cards are a king, a queen, and a jack.

**Solution:**

According to question,

It is given that,

Three cards are drawn with replacement from a well-shuffled pack of cards.

Now,

P(King) = P(A) = 4/52

P(Queen) = P(B) = 4/52

P(Jack) = P(C) = 4/52

Now,

P(King, Queen and a Jack)

= 3! × P(A) × P(B) × P(C)

= 3 × 2 × 4/52 × 4/52 × 4/52

= 6/2197

Hence, the required probability = 6/2197.

### Question 20. A bag contains 4 red and 5 black balls, a second bag contains 3 red and 7 black balls. One ball is drawn at random from each bag; find the probability that the (i)balls are of different colors (ii) balls are of the same color.

**Solution:**

According to question,

It is given that,

A bag contains 4 red and 5 black balls and,

a second bag contains 3 red and 7 black balls.

(i)P(balls are of different colors)= P[(R1 ∩ B2) ∪ (B1 ∩ R2)]

= P(R1 ∩ B2) + P(B1 ∩ R2)

= P(R1) × P(B2) + P(B1) × P(R2)

= 4/9 × 7/10 + 5/9 × 3/10

= 28/90 + 15/90 = 43/90

Required probability = 43/90

(ii)P(balls are of the same color)= P[(B1 ∩ B2) ∪ (R1 ∩ R2)]

= P(B1 ∩ B2) + P(R1 ∩ R2)

= P(B1) × P(B2) + P(R1) × P(R2)

= 5/9 × 7/10 + 4/9 × 3/10

= 47/90

Required probability = 47/90

### Question 21. A can hit a target 3 times in 6 shots, B: 2 times in 6 shots, and C: 4 times in 4 shots. They fix a volley. What is the probability that at least 2 shots hit?

**Solution:**

According to question:

It is given that,

P(A hits a target) = 3/6 = 1/2, and, P(A

‘) = 1/2P(B hits a target) = 2/6 = 1/3 and, P(B

‘) = 2/3P(C hits a target) = 4/4 = 1

Now, we have to find that,

P(At least 2 shots hit) = P(Exactly two shot hit) + P(all three shot hit)

= 1/2 × 2/3 × 1 + 1/2 × 1/3 × 1 + 1/2 × 1/3 × (1 – 1) + 1/2 × 1/3 × 1

= 2/6 + 1/6 + 1/6 = 4/6 = 2/3

Hence, The required probability = 2/3

### Question 22. The probability of student A passing an examination is 2/9 and of the student, B passing is 5/9. Assuming the two events: ‘A passes’, ‘B passes as independent, find the probability of (i) only A passing the examination (ii) only one of them passing the examination.

**Solution:**

According to question,

It is given that,

P(A passing the examination) = 2/9, P(A) = 2/9 and, P(A

‘) = 7/9P(B passing the examination) = 5/9, P(B) = 5/9 and, P(B

‘) = 4/9Now,

(i)P(Only A passing the examination)= P(A ∩ B

‘)= P(A) × P(B

‘)= 2/9 × 4/9 = 8/81

Required probability = 8/81

(ii)P(Only one of them passing the examination)= P[(A ∩ B

‘) ∪ (A‘ ∩B)]= P(A ∩ B’) + P(A

‘∩ B)= P(A)× P(B

‘) + P(A‘) × P(B)= 2/9 × 4/9 + 7/9 × 5/9

= 8/81 + 35/81 = 43/81

Required probability = 43/81

### Question 23. There are three urns A, B, and C. Urn A contains 4 red balls and 3 black balls. Urn B contains 5 red balls and 4 black balls. Urn C contains 4 red and 4 black balls. One ball is drawn from each of these urns. What is the probability that 3 balls drawn consist of 2 red balls and a black ball?

**Solution:**

According to question,

It is given that,

Urn A contains 4 red balls and 3 black balls.

Urn B contains 5 red balls and 4 black balls.

Urn C contains 4 red and 4 black balls.

Now,

P(3 balls drawn consist of 2 red balls and a black ball)

= P(Black from urn A) + P(Black from urn B) + P(Black from Urn C)

= 3/7 × 5/9 × 4/8 + 4/7 × 4/9 × 4/8 + 4/7 × 5/9 × 4/8

= 204/504 = 17/42

Required probability = 17/42