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Class 12 RD Sharma Solutions – Chapter 31 Probability – Exercise 31.5 | Set 3
  • Difficulty Level : Medium
  • Last Updated : 18 Mar, 2021

Question 24. X is taking up subjects – Mathematics, Physics, and Chemistry in the examination. His probabilities of getting grade A in these subjects are 0.2, 0.3, and 0.5 respectively. Find the probability that he gets.

(i) Grade A in all subjects (ii) Grade A in no subject (iii) Grade A in two subjects.

Solution:

According to question,

It is given that,

P(Getting grade A in mathematics), P(A) = 0.2 and, P(A) = 0.8



P(Getting grade A in physics), P(B) = 0.3 and, P(B) = 0.7

P(Getting grade A in chemistry), P(C) = 0.5 and, P(C) = 0.5

Now,

(i) P(Grade A in all subjects)

= P(A) × P(B) × P(C)

= 0.2 × 0.3 × 0.5

= 0.03

Hence, Required probability = 0.03

(ii) P(Grade A in no subject)

= P(A) × P(B) × P(C)

= 0.8 × 0.7 × 0.5

= 0.28

Required probability = 0.28

(iii) P(Grade A in two subjects)

= P(Not grade A in mathematics) + P(Not grade A in physics) + P(Not grade A in chemistry)

 = P(A) × P(B) × P(C) + P(A) × P(B) × P(C) + P(A) × P(B) × P(C)

= 0.8 × 0.3 × 0.5 + 0.2 × 0.7 × 0.5 + 0.2 × 0.3 × 0.5

 = 0.12 + 0. 07 + 0.03



= 0.22

Hence, The required probability = 0.22.

Question 25. A and B take turns in throwing two dice, the first to throw 9 being awarded the prize. Show that their chance of winning are in the ratio 9 : 8.

Solution:

According to question,

It is given that,

A and B take turns in throwing two dice.

Now, The sum of 9 can be obtained by

E = {(3, 6), (4, 5), (5, 4), (6, 3)}

P(E) = 4/36 = 1/9 and P(E) = 8/9

Now, P(A) = 1/9 and, P(A) = 8/9

P(B) = 1/9 and P(B) = 8/9

Now, let A starts the game 

P(A wins the game)

= P(getting 9 in first throw) + P(getting 9 in third throw) + P(getting 9 in fifth throw) + ….

= 1/9 + 8/9 × 8/9 × 1/9 + 8/9 × 8/9 × 8/9 × 8/9 × 1/9 + …..

= 1/9 ×[1+ (8/9)2 + (8/9)4 + …]

= 1/9 ×[1/ (1 – (8/9)2)]                 [since, sum of infinite term of G.P = a/1-r]

= 9/17

P(B wins the game) = 1 – P(A wins the game) = 1 – 9/17 = 8/17

Chances of winning A:B is

= 9/17 : 8/17

= 9 : 8

 Hence, chances of winning of A : B is 9 : 8.

Question 26. A, B and C in order to toss a coin. The one to throw a head wins. What are their respective chances of winning assuming that the game may continue indefinitely?

Solution:

According to question,

It is given that,

P(Getting head) = 1/2

P(Not getting head) = 1/2

P(A wins the game)

= P(getting head in first toss) + P(getting head in fourth toss) + P(getting head in 7th toss) + ….

= 1/2 + 1/2 × 1/2 × 1/2 × 1/2 + 1/2 × 1/2 × 1/2 × 1/2 ×1/2 ×1/2 × 1/2 + …..

= 1/2 ×[1 + (1/2)3 + (1/2)6 + …]

= 1/2 ×[1/ (1 – (1/2)3)]                 [since, sum of infinite term of G.P = a/1-r]

= 4/7

Now,

P(B wins the game)

= P(getting head in second toss) + P(getting head in fifth toss) + P(getting head in 8th toss) + ….

= 1/2×1/2 + 1/2 × 1/2 × 1/2 ×1/2 × 1/2 × 1/2 + 1/2 × 1/2 × 1/2 × 1/2 ×1/2 ×1/2 × 1/2 × 1/2 + …..

= 1/4 ×[1+ (1/2)3 + (1/2)6 + …]

= 1/4 × [1/ (1 – (1/2)3)]                 [since, sum of infinite term of G.P = a/1-r]

= 2/7

Now,

P(C wins the game) = 1 – P(A wins) – P(B wins)

= 1 – 4/7 – 2/7

= 1/7

Hence, the required probability of wining of A, B and C is 4/7, 2/7 and 1/7.

Question 27. Three persons A, B, C throw a die in succession till one gets a ‘six’ and wins the game. Find their respective probabilities of winning.

Solution:

According to question,

It is given that,

P(Getting six) = 1/6

P(Not getting six) = 5/6

P(A wins the game)

= P(getting 6 in first throw) + P(getting 6 in fourth throw) + P(getting 6 in 7th throw) + ….

= 1/6 + 5/6 × 5/6 × 5/6 × 1/6 + 5/6 × 5/6 × 5/6 × 5/6 ×5/6 ×5/6 × 1/6 + …..

= 1/6 ×[1 + (5/6)3 + (5/6)6 + …]

= 1/6 × [1/ (1 – (5/6)3)]                 [since, sum of infinite term of G.P = a/1-r]

= 36/91

Now,

P(B wins the game) 

= P(getting 6 in second throw) + P(getting 6 in fifth throw) + P(getting 6 in 8th throw) + ….

= 5/6 × 1/6 + 5/6 × 5/6 × 5/6 ×5/6 × 5/6 × 1/6 + 5/6 × 5/6 × 5/6 × 5/6 ×5/6 ×5/6 × 5/6 × 1/6 + …..

= 5/36 × [1 + (5/6)3 + (5/6)6 + …]

= 5/36 × [1/ (1 – (5/6)3)]                 [Since, sum of infinite term of G.P = a/1-r]

= 30/91

Now,

P(C wins the game) = 1 – P(A wins) – P(B wins)

= 1 – 36/91 – 30/91

= 25/91

Hence, the required probability of wining of A, B and C is 36/91, 30/91 and 25/91.

Question 28. A and B take turns in throwing two dice, the first to throw 10 being awarded the prize, show that if A has the first throw, their chance of winning are in the ratio 12:11.

Solution:

According to question,

It is given that,

A and B take turns in throwing two dice.

Now, The sum of 10 can be obtained by

E = {(4, 6), (5, 5), (6, 4)}

P(E) = 3/36 = 1/12 and P(E) = 11/12

Now, P(A) = 1/12 and, P(A) = 11/12

P(B) = 1/12 and P(B) = 11/12

Now, let A starts the game

P(A wins the game)

= P(getting 10 in first throw) + P(getting 10 in third throw) + P(getting 10 in fifth throw) + ….

= 1/12 + 11/12 × 11/12 × 1/12 + 11/12 × 11/12 × 11/12 × 11/12 × 1/12 + …..

= 1/12 × [1 + (11/12)2 + (11/12)4 + …]

= 1/12 × [1/ (1 – (11/12)2)]                 [since, sum of infinite term of G.P = a/1-r]

= 12/23

P(B wins the game) = 1 – P(A wins the game) = 1 – 12/23 = 11/23

Chances of winning A:B is

= 12/23 : 11/23

= 12 : 11

Hence, chances of winning of A : B is 12 : 11.

Question 29. There are 3 red and 5 black balls in the bag ‘A’ and 2 red and 3 black balls in bag ‘B’. One ball is drawn from bag ‘A’ and two from bag ‘B’. Find the probability that out of the 3 balls drawn one is red and 2 are black.

Solution:

According to question,

It is given that,

There are 3 red and 5 black balls in the bag ‘A’ and 2 red and 3 black 

balls in bag ‘B’. And, One ball is drawn from bag ‘A’ and two from bag ‘B’.

Now,

P(one red ball from bag A and 2 black ball from bag B) + P(one black ball from bag A and 

                                                                                                    one red ball from bag A and 

                                                                                                    one black ball from bag B) 

= P(R1 ∩ (2B2)) + P(B1 ∩ R2 ∩ B2) 

= 3/8 × 3/5 × 2/4 + 5/8 × 2/5 × 3/4 × 2

= 18/160 + 30/160 = 48/160

Required probability = 3/10.

Question 30. Fatima and John appear in an interview for two vacancies in the same post. The probability of Fatima’s selection is 1/7 and that of John’s selection is 1/5. What is the probability that

(i) both of them will be selected?

(ii) only one of them will be selected?

(iii) none of them will be selected?

Solution:

According to Question,

It is given that,

P(F) = 1/7 and, P(F) = 6/7

P(J) = 1/5 and, P(J) = 4/5

(i) P(Both of them will be selected)

= P(F ∩ J)

= P(F) × P(J)

= 1/7 × 1/5 = 1/35

Hence, The required probability = 1/35.

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

= P[(F ∩ J) ∪ (F ∩ J)]

= P(F ∩ J) + P(F ∩ J)

= P(F) × P(J) + P(F) × P(J)

= 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(F∩J)

= P(F) × P(J)

= 6/7 × 4/5  = 24/35

Hence, The required probability = 24/35

Question 31. A bag contains 8 marbles of which 3 are blue and 5 are red. One marble is drawn at random, its colour is noted and the marble is replaced in the bag. A marble is again drawn from the bag and its colour is noted. Find the probability that the marble will be

(i) blue followed by red.

(ii) blue and red in any order.

(iii) of the same colour.

Solution:

According to question,

It is given that,

A bag contains 8 marbles of which 3 are blue and 5 are red. And, One marble is drawn at random, its colour is noted and the marble is replaced in the bag. 

Now,

(i) P(Getting blue followed by red)

= P(B) × P(R)

= 3/8 × 5/8 = 15/64

Required probability = 15/64

(ii) P(Getting blue and red in any order)

= P(B) × P(R) + P(R) × P(B)

= 3/8 × 5/8 + 5/8 × 3/8

= 30/64 = 15/32

Required probability = 15/32.

(iii) P(of same color)

= P(R1) × P(R2) + P(B1) × P(B2)

= 5/8 × 5/8 + 3/8 × 3/8

= 25/64 + 9/64 = 34/64 = 17/32

Required probability = 17/32

Question 32. An urn contains 7 red and 4 blue balls. Two balls are drawn at random with replacement. Find the probability of getting

(i) 2 red balls

(ii) 2 blue balls

(iii) One red and one blue ball.

Solution:

According to question,

It is given that,

An urn contains 7 red and 4 blue balls. And, Two balls are drawn at random with replacement.

Now,

(i) P(Getting 2 red balls)

= P(R1) × P(R2)

= 7/11 × 7/11 = 49/121

Required probability = 49/121

(ii) P(Getting 2 blue balls)

= P(B1) × P(B2)

= 4/11 × 4/11 = 16/121

Required probability = 16/121

(iii) P(Getting one red and one blue balls)

= P(R) × P(B) + P(B) × P(R)

= 7/11 × 4/11 + 4/11 × 7/11

= 28/121 + 28/121 = 56/121

Required probability = 56/121

Question 33.  A card is drawn from a well-shuffled deck of 52 cards. The outcome is noted, the card is replaced and the deck reshuffled. Another card is then drawn from the deck.

(i) What is the probability that both the cards are of the same suit?

(ii) What is the probability that the first card is an ace and the second card is a red queen?

Solution:

According to question,

It is given that,

A card is drawn from a well-shuffled deck of 52 cards. The outcome is noted, the card is replaced and the deck reshuffled.

Now,

(i) We know that, There are four suit are club, spade, diamond and heart.

P(both the cards are of the same suit)

= P(Both cards are diamonds) + P(Both cards are spades) + 

   P(Both cards are clubs) + P(Both cards are hearts)

= 13/52 × 13/52 + 13/52 × 13/52 + 13/52 × 13/52 + 13/52 × 13/52 

= 1/16 + 1/16 + 1/16 + 1/16 = 4/16 = 1/4

Required probability = 1/4

(ii) We know that, There are four ace cards and 2 red queens.

= P(Getting an ace card) × P(Getting a red queen)

= 4/52 × 2/52 = 1/338

Required probability = 1/338.

Question 34. Out of 100 students, two sections of 40 and 60 are formed. If you and your friend are among 100 students, what is the probability that: (i) you both enter the same section? (ii) you both enter the different sections?

Solution:

According to question,

It is given that,

Out of 100 students, two sections of 40 and 60 are formed.

Now,

(i) P(Both enter the same section) 

= P(Both enter same section A) + P(Both enter the same section B)

= 40/100 × 40/100 + 60/100 × 60/100

= 4/25 + 9/25 = 13/25.

Hence, The required probability = 13/25.

(ii) P(Both enter different section)

= 1 – P(Both enter the same section)

 = 1 – 13/25

 = 12/25

Hence, The required probability = 12/25

Question 35. In a hockey match, both teams A and B scored the same number of goals up to the end of the game, so to decide the winner, the referee asked both the captains to throw a die alternately and decide that the team, whose captain gets a first six, will be declared the winner. If the captain of team A was asked to start, find their respective probabilities of winning the match and state whether the decision of the referee was fair or not.

Solution:

According to question,

It is given that,

P(Getting six) = 1/6

P(Not getting six) = 5/6

P(A wins the game)

= P(getting 6 in first throw) + P(getting 6 in third throw) + P(getting 6 in 5th throw) + ….

= 1/6 + 5/6 × 5/6 × 1/6 + 5/6 × 5/6 × 5/6 × 5/6 × 1/6 + …..

= 1/6 × [1 + (5/6)2 + (5/6)4 + …]

= 1/6 × [1/ (1 – (5/6)2)]                 [Since, sum of infinite term of G.P = a/1-r]

= 6/11

Now,

P(B wins the game)

= P(getting 6 in second throw) + P(getting 6 in fourth throw) + P(getting 6 in 6th throw) + ….

= 5/6 × 1/6 + 5/6 × 5/6 × 5/6 × 1/6 + 5/6 × 5/6 × 5/6 × 5/6 × 5/6 × 1/6 + …..

= 5/36 × [1 + (5/6)2 + (5/6)4 + …]

= 5/36 × [1/ (1 – (5/6)2)]                 [since, sum of infinite term of G.P = a/1-r]

= 5/11

Here we can see that Probabilities are not equal. So, the decision of the referee was not a fair one.

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