# RD Sharma Class 12 Ex 31.3 Solutions Chapter 31 Probability

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## RD Sharma Class 12 Ex 31.3 Solutions Chapter 31 Probability

### Question 1. If P(A) = 7 /13, P(B) = 9/13 and P(A ∩ B) = 4/13, find P(A/B).

Solution:

Given: P(A) = 7/13, P(B) = 9/13, and P(A∩ B) = 4/13

We know that, P(A/B) = P(A ∩ B)/P(B)

= (4/13) ÷ (9/13)

= 4/9

### Question 2. If A and B are events such that P(A) = 0.6, P(B) = 0.3, and P(A ∩ B) = 0.2, find P(A/B) and P(B/A).

Solution:

Given: P(A) = 0.6, P(B) = 0.3, P(A ∩ B) = 0.2

We know that P(A/B) = P(A ∩ B)/P(B)

= 0.2/0.3

= 2/3

and P(B/A) = P(A ∩ B)/P(A)

= 0.2/0.6

= 1/3

### Question 3. If A and B are two events such that P(A ∩ B) = 0.32 and P(B) = 0.5, find P(A/B).

Solution:

Given: P(A ∩ B) = 0.32 and P(B) = 0.5

We know that P(A/B) = P(A ∩ B)/P(B)

= 0.32/0.5

= 0.64

### Question 4. If P(A) = 0.4, P(B) = 0.8, P(B/A) = 0.6, find P(A/B) and P(A ∪ B).

Solution:

Given: P(A) = 0.4, P(B) = 0.8, P(B/A) = 0.6

We know that, P(B/A) = P(A ∩ B)/P(A)

P(A ∩ B) = P(B/A) × P(A)

P(A ∩ B) = 0.6 × 0.4

P(A ∩ B) = 0.24

Therefore, P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

= 0.4 + 0.8 – 0.24

= 0.96

And, P(A/B) = P(A ∩ B)/P(B)

= 0.24/0.8

= 0.3

### (i) P(A) = 1/3, P(B) = 1/4, and P(A ∪ B) = 5/12, find P(A/B) and P(B/A).

Solution:

We know that, P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

P(A ∩ B) = P(A) + P(B) – P(A ∪ B)

P(A ∩ B) = 1/3 + 1/4 – 5/12 = 2/12

Therefore, P(A/B) = P(A ∩ B)/P(B) = (2/12) ÷ (1/4) = 2/3

and, P(B/A) = P(A ∩ B)/P(A) = (2/12) ÷ (1/3) = 1/2

### (ii) P(A) = 6/11, P(B) = 5/11 and P(A ∪ B) = 7/11, find P(A ∩ B), P(A/B) and P(B/A).

Solution:

We know that, P(A∪ B) = P(A) + P(B) – P(A ∩ B)

P(A ∩ B)=P(A) + P(B) – P(A ∪ B)

P(A ∩ B) = 6/11 + 5/11 – 7/11

= 4/11

Therefore, P(A/B) = P(A ∩ B)/P(B)

= (4/11) ÷ (5/11) = 4/5

and, P(B/A) = P(A ∩ B)/P(A)

= (4/11) ÷ (6/11) = 2/3

### (iii) P(A) = 7/13, P(B) = 9/13 and P(A ∩ B) = 4/13, find P(A’/B).

Solution:

We know that, P(A’ ∩ B) = P(B) – P(A ∩ B)

= 9/13 – 4/13

= 5/13

Therefore, P(A’/B) = P(A’ ∩ B)/P(B)

= (5/13) ÷ (9/13)

= 5/9

### (iv) P(A) = 1/2, P(B) = 1/3 and  P(A ∩ B) = 1/4, find P(A/B), P(B/A), P(A’/B), and P(A’/B’).

Solution:

P(A/B) = P(A ∩ B)/P(B)

= (1/4) ÷ (1/3)

= 3/4

P(B/A) = P(A ∩ B)/P(A)

= (1/4) ÷ (1/2)

= 1/2

P(A’/B) = (P(B) – P(A ∩ B))/P(B)

= (1/3 – 1/4) ÷ (1/3)

= 1/4

P(A’/B’) = P(A’ ∩ B’)/P(B’)

= (1 – P(A’ ∪ B’))/(P(A) – P(A ∩ B))

= (1 – P(A) – P(B) + P(A ∩ B))/(P(A) – P(A ∩ B))

= (1 – 1/2 – 1/3 + 1/4)/(1/2 – 1/4)

= 5/4

### Question 6. If A and B are two events such that 2 × P(A) = P(B) = 5/13 and P(A/B) = 2/5, find P(A ∪ B).

Solution:

Given, 2 × P(A) = P(B) = 5/13

P(A) = 5/26

P(A/B) = P(A ∩ B)/P(B)

2/5 = P(A ∩ B) ÷ (5/13)

P(A ∩ B) = (2/5) ÷ (5/13)

P(A ∩ B) = 2/13

We know that,

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

= 5/26 + 5/13 – 2/13

= 11/26

### (i) P(A ∩ B)

Solution:

We know that, P(A∪ B) = P(A) + P(B) – P(A ∩ B)

P(A ∩ B) = P(A) + P(B) – P(A ∪ B)

= 6/11 + 5/11 – 7/11

= 4/11

### (ii) P(A/B)

Solution:

P(A/B) = P(A ∩ B)/P(B)

= (4/11) ÷ (5/11)

= 4/5

### (iii) P(B/A)

Solution:

P(B/A) = P(A ∩ B)/P(A)

= (4/11) ÷ (6/11)

= 2/3

### (iii) A = At most two tails, B = At least one tail.

Solution:

Sample space of three coins is given by

{HHH, HTH, THH, TTH, HHT, HTT, THT, TTT}

(i) A = Heads on third toss = {HHH, HTH, THH, TTH}

B = Heads on first two tosses = {HHH, HHT}

P(A ∩ B) = {HHH}

∴ P(A/B) = P(A ∩ B)/P(B) = 1/2

(ii) A = At least two heads = {HHH, HTH, THH, HHT}

B = At most two heads = {HHT, HTH, TTH, HHT, HTT, THT, TTT}

P(A ∩ B) = {HTH, THH, HHT}

∴ P(A/B) = P(A ∩ B)/P(B) = 3/7

(iii) A = At most two tails = {HHH, HTH, THH, TTH, HHT, HTT, THT}

B = At least one tail  = {HTH, THH, TTH, HHT, HTT, THT, TTT}

P(A ∩ B)  = {HTH, THH, TTH, HHT, HTT, THT}

∴  P(A/B) = P(A ∩ B)/P(B) = 6/7

### (ii) A = No tail appears, B = No head appears

Solution:

Sample space of two coins is given by

{HH, HT, TH, TT}

(i) A = Tail appears on one coin = {HT, TH}

B = One coin shows head = {HH, HT, TH}

P(A ∩ B)  = {HT, TH}

∴ P(A/B) = P(A ∩ B)/P(B)

= 2/2

= 1

(ii) A = No tail appears = {HH}

B = No head appears= {TT}

P(A ∩ B)  = { }

∴ P(A/B) = P(A ∩ B)/P(B) = 0

### A = 4 appears on the third toss, B = 6, and 5 appear respectively on the first two tosses.

Solution:

A = 4 appears on the third toss = { (1, 1, 4), (1, 2, 4), (1, 3, 4), (1, 4, 4), (1, 5, 4), (1, 6, 4),

(2, 1, 4), (2, 2, 4), (2, 3, 4), (2, 4, 4), (2, 5, 4), (2, 6, 4),

(3, 1, 4), (3, 2, 4), (3, 3, 4), (3, 4, 4), (3, 5, 4), (3, 6, 4),

(4, 1, 4), (4, 2, 4), (4, 3, 4), (4, 4, 4), (4, 5, 4), (4, 6, 4),

(5, 1, 4), (5, 2, 4), (5, 3, 4), (5, 4, 4), (5, 5, 4), (5, 6, 4)

(6, 1, 4), (6, 2, 4), (6, 3, 4), (6, 4, 4), (6, 5, 4), (6, 6, 4)}

B = 6 and 5 appear respectively on first two tosses

B = {(6, 5, 1), (6, 5, 2), (6, 5, 3), (6, 5, 4), (6, 5, 5), (6, 5, 6)}

(A ∩ B) = {(6, 5, 4)}

∴ P(A/B) = P(A ∩ B)/P(B) = 1/6

∴ P(B/A) = P(A ∩ B)/P(A) = 1/36

### Question 11. Mother, father, and son line up at random for a family picture. If A and B are two events given by A = Son on one end, B = Father in the middle, find P(A/B) and P(B/A).

Solution:

Let Mother = M, Father = F, and son = S

Sample Space = {FMS, FSM, MFS, MSF, SFM, SMF}

A = Son on one end = {FMS, MFS, SFM, SMF}

B = Father in the middle = {MFS, SFM}

P(A ∩ B) = {MFS, SFM}

∴ P(A/B) = P(A ∩ B)/P(B) = 2/2 = 1

∴ P(B/A) = P(A ∩ B)/P(A) = 2/4 = 1/2

### Question 12. A dice is thrown twice and the sum of the numbers appearing is observed to be 6. What is the conditional probability that the number 4 has appeared at least once?

Solution:

Let the sample space of the experiment is {(1, 1), (1, 2), (1, 3), . . . .,(6, 6)} consisting of 36 outcomes

P(A) =  P(Sum = 6) = 5/36

P(B) = P(4 has appeared at least once) = 11/36

P(B/A) = P(A ∩ B)/P(A) = (2/36) ÷ (5/36) = 2/5

### Question 13. Two dice are thrown. Find the probability that the numbers appeared has the sum 8, if it is known that the second die always exhibits 4.

Solution:

Two dice are thrown.

A = Sum on the dice is 8

A = {(2, 6), (6, 2), (3, 5), (5, 3), (4, 4)}

B = Second die always exhibits 4

= {(1, 4), (2, 4), (3, 4), (5, 4), (6, 4)}

P(A ∩ B) = {(4, 4)}

P(A/B) = P(A ∩ B)/P(B)

= 1/6

### Question 14. A pair of dice is thrown. Find the probability of getting 7 as the sum, if it is known that the second die always exhibits an odd number.

Solution:

A = Sum on two dice equals 7 = {(1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3)}

B = Second die always exhibits an odd number = {(1, 1), (2, 1), (3, 1), (4, 1), (5, 1), (6, 1)

(1, 3), (2, 3), (3, 3), (4, 3), (5, 3), (6, 3)

(1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5)}

(A ∩ B) = {(6, 1), (2, 5), (4, 3)}

P(A/B) = P(A ∩ B)/P(B)

= 3/18

= 1/6

### Question 15. A pair of dice is thrown. Find the probability of getting 7 as the sum if it is known that the second die always exhibits a prime number.

Solution:

A = Sum on two dice equals 7 = {(1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3)}

B = Second die always exhibits a prime number = {(1, 1), (2, 1), (3, 1), (4, 1), (5, 1), (6, 1),

(1, 3), (2, 3), (3, 3), (4, 3), (5, 3), (6, 3),

(1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5)}

(A ∩ B) = {(6, 1), (2, 5), (4, 3)}

P(A/B) = P(A ∩ B)/P(B)

= 3/18

= 1/6

### Question 16. A dice is rolled. If the outcome is an odd number, what is the probability that it is prime?

Solution:

A die is rolled

A = A prime number on die = {2, 3, 5}

B = An odd number on die = {1, 3, 5}

(A ∩ B) = {3, 5}

P(A/B) = P(A ∩ B)/P(B)

= 2/3

### Question 17. A pair of dice is thrown. Find the probability of getting the sum 8 or more, if 4 appears on the first die.

A pair of dice is thrown

A = Getting sum of 8 or more = {(2, 6), (6, 2), (3, 5), (5, 3), (4, 4), (3, 6), (6, 3)

(4, 5), (5, 4), (4, 6), (6, 4), (5, 5), (6, 5), (5, 6), (6, 6)}

B = 4 on first die = {(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)}

(A ∩ B) = {(4, 4), (4, 5), (4, 6)}

P(A/B) = P(A ∩ B)/P(B)

= 1/2

### Question 18. Find the probability that the sum of the numbers showing on the two dice is 8, given that at least one die does not show five.

Solution:

A pair of dice is thrown

A = Getting sum of 8 or more = {(2, 6), (6, 2), (3, 5), (5, 3), (4, 4)}

B = At least one die does not show 5 = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 6),

(2, 1), (2, 2), (2, 3), (2, 4), (2, 6),

(3, 1), (3, 2), (3, 3), (3, 4), (3, 6),

(4, 1), (4, 2), (4, 3), (4, 4), (4, 6),

(1, 1), (1, 2), (1, 3), (1, 4), (1, 6)}

(A ∩ B) = {(2, 6), (4, 6), (6, 2)}

P(A/B) = P(A ∩ B)/P(B)

= 3/25

### Question 19. Two numbers are selected at random from integers 1 through 9. If the sum is even, find the probability that both the numbers are odd.

Solution:

A = Both numbers are odd = {(3, 1), (5, 1), (7, 1), (9, 1), (3, 7), (5, 7), (9, 7)

(3, 9), (5, 9), (7, 9), (7, 3), (7, 5), (7, 9), (5, 3)

(9, 3), (1, 3), (1, 5), (1, 7)}

B = Sum of both numbers is even = {(1, 3), (1, 5), (2, 4), (1, 7), (2, 6), (3, 5), (1, 9),

(2, 8), (3, 7), (4, 6), (7, 5), (8, 4), (9, 3), (8, 6),

(9, 5), (9, 7)}

(A ∩ B) = {(1, 3), (1, 5), (1, 7), (3, 5), (1, 9), (3, 7), (7, 5), (9, 3), (9, 5), (9, 7)

P(A/B) = P(A ∩ B)/P(B)

= 10/16 = 5/8

### Question 20. A dice is thrown twice and the sum of the numbers appearing is observed to be 8. What is the conditional probability that the number 5 has appeared at least once?

A die is thrown twice

A = The number 5 has appeared at least once

A = {(1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 6)}

B =  Sum of numbers is 8 = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}

(A ∩ B) = {(3, 5), (5, 3)}

P(A/B) = P(A ∩ B)/P(B)

= 2/5

### Question 21. Two dice are thrown and it is known that the first die shows a 6. Find the probability that the sum of the numbers showing on two dice is 7.

Solution:

Two dice are thrown

A = Sum of the numbers is 7 = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}

B = First die shows a 6 = {(6, 1), ((6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

(A ∩ B) = {(6, 1)}

P(A/B) = P(A ∩ B)/P(B)

= 1/6

### Question 22. A pair of dice is thrown. Let A be the event that the sum is greater than or equal to 10 and F be the event “5 appears on the first die”. Find P(A/B). If B is the event “5 appears on at least one die”, fine P(A/B).

Solution:

A pair of dice is thrown

A = Sum is greater than or equal to 10 = {(4, 6), (6, 4), (5, 5), (5, 6), (6, 5), (6, 6)}

Case 1: B = 5 appears on first die = {(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}

(A ∩ B) = {(5, 5), (5, 6)}

P(A/B) = P(A ∩ B)/P(B)

= 2/6 = 1/3

Case 2: B = 5 appears on at least one die = {(1, 5), (2, 5), (3, 5), (4, 5), (6, 5),

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}

(A ∩ B) = {(5, 5), (5, 6), (6, 5)}

P(A/B) = P(A ∩ B)/P(B)

= 3/11

### Question 23. The probability that a student selected at random from a class will pass in Mathematics is 4/5, and the probability that he/she passes in Mathematics and Computer Science is 1/2. What is the probability that he/she will pass in Computer Science if it is known that he/she has passed in Mathematics?

Solution:

Probability to pass mathematics: P(M) = 4/5

Probability to pass Mathematics(M) and Computer Science (C) = P(M ∩ C) = 1/2

We know that, P(C/M) = P(M ∩ C) /P(M)

= (1/2) ÷ (5/4) = 5/8

### Question 24. The probability that a certain person will buy a shirt is 0.2, the probability that he will buy a trouser is 0.3, and the probability that he will buy a shirt given that he buys a trouser is 0.4. Find the probability that he will buy both a shirt and a trouser. Find also the probability that he will buy a trouser given that he buys a shirt.

Solution:

Probability that a person buys a shirt(S) = P(S) = 0.2

Probability that he buys a trouser(T) = P(T) = 0.3

P(S/T) = 0.4

We know that,

P(S/T) = P(S ∩ T)/P(T)

P(S ∩ T) = 0.4 × 0.3 = 0.12

P(T/S) = P(S ∩ T)/P(S)

= 0.12/0.2 = 0.6

### Question 25. In a school there are 1000 students, out of which 430 are girls. It is known that out of 430, 10% of the girls study in class XII. What is the probability that a student chosen randomly studies in class XII given that the chosen student is a girl.

Solution:

Total students = 1000

Number of girls = 430

Let A = Student chosen studies in class XII

B = Student chosen is a girl

Then P(B) = 430/1000

P(A ∩ B) = 43/1000

P(A/B) = P(A ∩ B)/P(B)

= 43/430 = 1/10

### Question 26. Ten cards numbered 1 through 10 are placed in a box, mixed up thoroughly and then one card is drawn randomly. If it is known that the number on the drawn card is more than 3, what is the probability that it is an even number?

Solution:

Total number of cards = 10

Let A = drawn number is more than 3

B = drawn number is even

P(B/A) = P(B ∩ A)/P(A)

P(A) = 7/10

P(A ∩ B) = 4/10

P(B/A) = 4/7

### (ii) at least one is girl.

Solution:

(i) Let ‘A’ be the event that both the children born are girls.

Let ‘B’ be the event that the youngest is a girl.

We have to find conditional probability P(A/B).

P(A/B) = P(A ∩ B)/P(B)

P(A ∩ B) = P(A) = P(GG)

= 1/2 × 1/2 = 1/4

P(B) = P(BG) + P(GG)

= 1/2 × 1/2 + 1/2 × 1/2 = 1/2

Hence, P(A/B) = (1/4) ÷ (1/2) = 1/2

(ii) Let ‘A’ be the event that both the children born are girls.

Let ‘B’ be the event that at least one is a girl.

We have to find the conditional probability P(A/B).

P(A/B) = P(A ∩ B)/P(B)

P(A ∩ B) = P(A) = P(GG)

= 1/2 × 1/2 = 1/4

P(B) = 1 – P(BB)

= 1 – 1/2 × 1/2 = 1 – 1/4 = 3/4

Hence, P(A/B) = (1/4) ÷ (3/4) = 1/3

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