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Class 12 RD Sharma Solutions – Chapter 31 Probability – Exercise 31.5 | Set 1
  • Last Updated : 18 Mar, 2021

Question 1. A bag contains 6 black and 3 white balls. Another bag contains 5 black and 4 white balls. If one ball is drawn from each bag, find the probability that these two balls are of the same color.

Solution:

According to Question:

It is given that,

Bag 1 contains 6 black and 3 white balls.

Bag 2 contains 5 black and 4 white balls.



Now, 

One ball is drawn from each bag

Then, P(one black ball from bag 1) = 6/9 and, P(one white ball from bag 1) = 3/9

P(one black ball from bag 2) = 5/9 and, P(one white ball from bag 2) = 4/9

Now,

P(Two balls are of same color) = P(Both are black) + P(Both are white)

= 6/9 × 5/9 + 3/9 × 4/9

= 30/81 + 12/81 = 42/81

= 14/27

Hence, The required probability = 14/27

Question 2. A bag contains 3 red and 5 black balls and a second bag contains 6 red and 4 black balls. A ball is drawn from each bag. Find the probability that one is red and the other is black.

Solution:

According to Question:

It is given that,

Bag 1 contains 3 red and 5 black balls.

Bag 2 contains 6 red and 4 black balls.

Now,

One ball is drawn from each bag

Then, P(one red ball from bag 1), P(R1) = 3/8 and, P(one black ball from bag 1), P(B1) = 5/8



P(one red ball from bag 2), P(R2) = 6/10 and, P(one black ball from bag 2), P(B2) = 4/10

Now,

We have to find that,

P(one is red and other is black) 

= P(R1 ∩ B2) ∪ P(B1 ∩ R2)

= P(R1) × P(B2) + P(B1) × P(R2)

= 3/8 × 4/10 + 5/8 × 6/10 = 12/80 + 30/80 = 42/80

= 21/40

Hence, The required probability = 21/40

Question 3. Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that 

(i) both the balls are red (ii) the first ball is black and the second is red. (iii) one of them is black and the other is red.

Solution:

According to question:

It is given that,

A box contain 10 black and 8 red balls. And two balls are drawn at random with replacement.

Now,

(i) P(Both the balls are red)

= P(R1 ∩ R2)

= P(R1) × P(R2)

= 8/18 × 8/18 = 64/324

= 16/81

Required Probability = 16/81

(ii) P(The first ball is black and the second ball is red)

= P(B ∩ R)

= P(B) × P(R)

= 10/18 × 8/18 = 80/324

= 20/81

Required Probability = 20/81

(iii) P(One of them is black and the other is red)

= P((B ∩ R)∪ (R ∩ B))

= P(B ∩ R) + P(R ∩ B)

= P(B) × P(R) + P(R) × P(B)

 = 10/18 × 8/18 + 8/18 × 10/18 

= 20/81 + 20/81

= 40/81

Hence, The required probability = 40/81

Question 4. Two cards are drawn successively without replacement from a well-shuffled deck of cards. Find the probability of exactly one ace.

Solution:

According to question,

It is given that,

Two cards are drawn successively without replacement. In a well – shuffled deck of cards there are total 4 ace.

Now,

P(Exactly one ace) = P(first card is ace) + P(Second card is ace)

= 4/52 × 48/51 + 48/52 × 4/51

= 96/663

= 32/221

Hence, The required probability = 32/221

Question 5. A speaks truth in 75% and B in 80% of the cases. In what percentage of cases are they likely to contradict each other in narrating the same incident?

Solution:

According to question:

It is given that,

A speaks truth in 75% cases.

B speaks truth in 80% cases.

Now, P(A) = 75/100 = 3/4 and, P(A) = 1 – 75/100 = 25/100 = 1/4

P(B) = 80/100 = 4/5 and, P(B) = 1 – 80/100 = 20/100 = 1/5

Now,

P(A and B contradict each other)

= P(A ∩ B) + P(A∩ B)

= P(A) × P(B) + P(A) × P(B)

= 3/4 × 1/5 + 1/4 × 4/5

 = 3/20 + 4/20 = 7/20

= 0.35

 = 35 %

Hence, The required probability = 35 %.

Question 6. Kamal and Monica appeared for an interview for two vacancies. The probability of Kamal’s selection is 1/3 and that of Monika’s selection is 1/5. Find the probability that

(i) both of them will be selected (ii) none of them will be selected 

(iii) at least one of them will be selected (iv) only one of them will be selected.

Solution:

According to question:

It is given that,

P(K) = 1/3 and, P(M) = 1/5

Now,

(i) P(Both of them are selected)

= P(K ∩ M) = P(K) × P(M)

= 1/3 × 1/5 = 1/15

The required probability = 1/15

(ii) P(none of them will be selected)

= P(K∩ M) = P(K) × P(M)

= 1 – 1/3 × 1 – 1/5 = 2/3 × 4/5 

= 8/15

The required probability = 8/15

(iii) P(at least one of them will be selected)

= 1 – P(None of them is selected)

= 1 – 8/15                    [From eq(ii)]

= 7/15

The required probability = 7/15

(iv) P(only one of them will be selected) 

= P(K ∩ M) + P(K∩ M)

= P(K) × P(M) + P(K) × P(M)

= 1/3 × 4/5 + 2/3 × 1/5

= 4/15 + 2/15 = 6/15 = 2/5

Hence, The required probability = 2/5

Question 7. A bag contains 3 white, 4 red, and 5 black balls. Two balls are drawn one after the other, without replacement. What is the probability that one is white and the other is black?

Solution:

According to question:

It is given that,

A bag contains 3 white, 4 red, and 5 black balls. And Two balls are

drawn one after the other, without replacement.

Now,

P(One is white and other is black)

= P((W ∩ B) ∪ (B ∩ W))

= P(W ∩ B) + P(B ∩ W)

= P(W) × P(B/W) + P(B) × P(W/B)

= 3/12 × 5/11 + 5/12 × 3/11

= 15/132 + 15/132

= 30/132 = 5/22

Hence, The required probability = 5/22.

Question 8. A bag contains 8 red and 6 green balls. Three balls are drawn one after another without replacement. Find the probability that at least two balls drawn are green.

Solution:

According to question:

It is given that,

A bag contains 8 red and 6 green balls. And Three balls are 

drawn one after another without replacement.

Now,

P(at least two balls drawn are green)

= 1 – P(at most one ball is green)

= 1 – [P(first ball is green) + P(Second ball is green) + P(Third ball is green) + P(No green)]

= 1 – [6/14 × 8/13 × 7/12 + 8/14 × 6/13 × 7/12 + 8/14 × 7/13 × 6/12 + 8/14 × 7/13 × 6/12]

= 1 – [336/2184 + 336/2184 + 336/2184 + 336/2184]

= 1 – 1344/2184 = 840/2184

= 5/13

Hence, the required probability = 5/13

Question 9. Arun and Tarun appeared for an interview for two vacancies. The probability of Arun’s selection is 1/4 and that of Tarun’s rejection is 2/3. Find the probability that at least and of them will be selected.

Solution:

According to question:

It is given that,

P(Arun get selected), P(A) = 1/4 and, P(Arun get rejected), P(A) = 3/4

P(Tarun get rejected), P(T) = 2/3

P(Tarun get selected), P(T) = 1/3

Now,

P(at least one of them is selected)

= 1 – P(none of them is selected)

= 1 – P(A‘ ∩ T)

= 1 – P(A) × P(T)

= 1 – 3/4 × 2/3

= 1 – 6/12 = 1 – 1/2

= 1/2

Hence, The required probability is 1/2.

Question 10. A and B toss a coin alternatively till one of them gets a head and wins the game, if A starts the game, find the probability that B will win the game.

Solution:

According to question:

Let E be the event occurring head.

P(E) = 1/2 and P(E) = 1/2

A wins the game in first, third and fifth throw,

P(A wins in first throw) = P(E) = 1/2

P(A wins in third throw) = P(E) × P(E) × P(E) = 1/2 × 1/2 × 1/2 = (1/2)3

Similarly, P(A wins in fifth throw) = (1/2)5

Now,

P(Wining of A) 

= 1/2 + (1/2)3 + (1/2)5 + …

= 1/2 [1 + (1/2)2 + (1/2)4 + …]

= 1/2 [1/ (1 – (1/2)2]                            [Since, Sum of infinite term of g.p = a/1 – r]

= 1/2 [1/ 1 – 1/4]

= 1/2 × 4/3 = 2/3

Now, P(B wins) = 1 – P(A wins) 

= 1 – 2/3 = 1/3

Hence, The required probability = 1/3

Question 11. Two cards are drawn from a well-shuffled pack of 52 cards, one after another without replacement. Find the probability that one of these is a red card and the other a black card?

Solution:

According to question:

It is given that,

Two cards are drawn from a well-shuffled pack of 52 cards, 

one after another without replacement.

There are 26 red and 26 black cards.

Now, We have to find that,

P(one red and other black card)

 = P[(R ∩ B) ∪ (B ∩ R)] 

= P(R ∩ B) + P(B ∩ R)

= P(R) × P(B/R) + P(B) × P(R/B)

= 26/52 × 26/51 + 26/52 × 26/51 

= 26/51

Hence, the required probability 26/51.

Question 12. Tickets are numbered from 1 to 10. Two tickets are drawn one after the other at random. Find the probability that the number on one of the tickets is a multiple of 5 and on the other a multiple of 4.

Solution:

According to question:

It is given that,

Tickets are numbered from 1 to 10. And, Two tickets are drawn at random.

Now, Let us consider,

A = Ticket is a multiple of 5.

B = Ticket is a multiple of 4.

Since 5 and 10 are multiple of 5 . So, P(A) = 2/10 = 1/5.

and , 4, 8 are the multiple of 4. So, P(B) = 2/10 = 1/5.

Now, we have to find that,

P(one number is multiple of 5 and other is the multiple of 4)

= P[(A∩B) ∪ (B ∩ A)]

= P(A∩B) + P(B ∩ A)

= P(A)×P(B/A) + P(B) × P(A/B)

= 1/5 × 2/9 + 1/5 × 2/9

= 4/45

Hence, the required probability 4/45.

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