# Class 12 RD Sharma Solutions – Chapter 31 Probability – Exercise 31.5 | Set 1

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

‘) = 1 – 80/100 = 20/100 = 1/5Now,

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

‘) = 3/4P(Tarun get rejected), P(T

‘) = 2/3P(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

‘) = 1/2A 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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