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

• Difficulty Level : Medium
• Last Updated : 18 Mar, 2021

### 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) = 70/100

P(W) = 35/100, P(W) = 65/100

Now,

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) = 6/7

P(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) = 1/4

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

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

Now,

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) = 1/2

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

P(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) = 7/9

P(B passing the examination) = 5/9, P(B) = 5/9 and, P(B) = 4/9

Now,

(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

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