Here we provide RD Sharma Class 12 Ex 31.5 Solutions Chapter 31 Probability for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 12 Ex 31.5 Solutions Chapter 31 Probability book pdf download. Now you will get step-by-step solutions to each question.
| Textbook | NCERT |
| Class | Class 12th |
| Subject | Maths |
| Chapter | 31 |
| Exercise | 31.5 |
| Category | RD Sharma Solutions |
RD Sharma Class 12 Ex 31.5 Solutions Chapter 31 Probability
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.
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
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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