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Textbook | NCERT |

Class | Class 12th |

Subject | Maths |

Chapter | 31 |

Exercise | 31.1 |

Category | RD Sharma Solutions |

Table of Contents

**RD Sharma Class 12 Ex 31.1 Solutions Chapter 31 Probability**

**Question 1: 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:**

Here,

Sample space (S) = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

Let, A = the number on the drawn card is even

So, A = {2, 4, 6, 8, 10}

n(A) = 5

and B = the number on the drawn card is more than 3

So, B = {4, 5, 6, 7, 8, 9, 10}

n(B) = 7

Now, A ∩ B = {4, 6, 8, 10}

n(A ∩ B) = 4

Thus, the required probability is –

P(A/B) = n(A ∩ B)/ n(B) = 4/7

**Question 2: Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that – **

**(i) the youngest is a girl.**

**(ii) at least one is a girl.**

**Solution:**

Let b and g represent the boy and the girl child respectively. If a family has two children, the sample space will be –

S = {(b, b), (b, g), (g, b), (g, g)}

n(S) = 4

Let A be the event that both children are girls.

So, A = {(g, g)}

n(A) = 1

(i)Let B be the event that the youngest child is a girl.

So, B = {(b, g), (g, g)}

n(B) = 2

Now, A ∩ B = {(g, g)}

n(A ∩ B) = 1

So, P(B) = n(B)/ n(S) = 2/4 = 1/2

and P(A ∩ B) = n(A ∩ B)/ n(S) = 1/4Now, the conditional probability that both are girls, given that the youngest child is a girl, is –

P(A/B) = P(A ∩ B)/ P(B) = (1/4)/ (1/2) = 1/2

Thus, the required probability is 1/2.

(ii)Let C the event that at least one child is a girl.

So, C = {(b, g), (g, b), (g, g)}

n(C) = 3

Now, A ∩ C = {(g, g)}

n(A ∩ C) = 1

So, P(C) = n(C)/ n(S) = 3/4

and P(A ∩ C) = n(A ∩ C)/ n(S) = 1/4Now, the conditional probability that both are girls, given that the youngest child is a girl, is –

P(A/C) = P(A ∩ C)/ P(C) = (1/4)/ (3/4) = 1/3

Thus, the required probability is 1/3.

**Question 3: Given that the two numbers appearing on throwing two dice are different. Find the probability of the event ‘the sum of numbers on the dice is 4’.**

**Solution:**

Let A be the event of having two different numbers on the dice.

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

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

n(A) = 30

and B be the event getting a sum of 4 on the dice.

So, B = {(1, 3), (2, 2), (3, 1)}

n(B) = 3

Now, A ∩ B = {(1, 3), (3, 1)}

n(A ∩ B) = 2

Thus, the required conditional probability is –

P(B/A) = n(A ∩ B)/ n(A) = 2/30 = 1/15

**Question 4: A coin is tossed three times, if head occurs on first two tosses, find the probability of getting head on third toss.**

**Solution:**

Let A be the event of a head appearing on the first two tosses.

So, A = {HHT, HHH}

n(A) = 2

and B be the event of getting a head on the third toss.

So, B = {HHH, HTH, THH, TTH}

n(B) = 4

Now, A ∩ B = {HHH}

n(A ∩ B) = 1

Thus, the required conditional probability is –

P(B/A) = n(A ∩ B)/ n(A) = 1/2

**Question 5: A die is thrown three times, find the probability that 4 appears on the third toss if it is given that 6 and 5 appear respectively on first two tosses.**

**Solution:**

Let A be the event of 4 appearing on the third toss, if a die is thrown three times.

So, A = {(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)}

n(A) = 36

and B be the event of 6 and 5 appearing respectively on first two tosses, if the die is tossed three times.

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

n(B) = 6

Now, A ∩ B = {(6, 5, 4)}

n(A ∩ B) = 1

Thus, the required probability is –

P(A/B) = n(A ∩ B)/ n(B) = 1/6

**Question 6: Compute P(A/B), if P(B) = 0.5 and P(A ∩ B) = 0.32**

**Solution:**

Given, P(B) = 0.5 and P(A ∩ B) = 0.32

We know that, P(A/B) = P(A ∩ B)/ P(B) = 0.32/ 0.5 = 16/25

Thus, P(A/B) = 16/25

**Question 7: If P(A) = 0.4, P(B) = 0.3 and P(B/A) = 0.5, Find P(A ∩ B) and P(A/B).**

**Solution:**

Given, P(A) = 0.4, P(B) = 0.3 and P(B/A) = 0.5

We know that,

P(B/A) = P(A ∩ B)/ P(A)

0.5 = P(A ∩ B)/ 0.4

P(A ∩ B) = 0.5 × 0.4

Thus, P(A ∩ B) = 0.2Now, P(A/B) = P(A ∩ B)/ P(B)= 0.2/ 0.3

Thus, P(A/B) = 2/3

**Question 8: If A and B are two events such that P(A) = 1/3, P(B) = 1/5 and P(A ∪ B) = 11/30, find P(A/B) and P(B/A).**

**Solution: **

Given, P(A) = 1/3, P(B) = 1/5 and P(A ∪ B) = 11/30

We know that,

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

11/30 = 1/3 + 1/5 – P(A ∩ B)

P(A ∩ B) = 1/3 + 1/5 – 11/30

P(A ∩ B) = (10 + 6 – 11)/ 30

P(A ∩ B) = 5/30 = 1/6

Now,

P(A/B) = P(A ∩ B)/ P(B)

P(A/B) = (1/6)/ (1/5)

P(A/B)= 5/6and P(B/A) = P(A ∩ B)/ P(A)

P(B/A) = (1/6)/ (1/3)

P(B/A) = 3/6 = 1/2Thus, P(A/B) = 5/6 and P(B/A) = 1/2.

**Question 9: A couple has two children. Find the probability that both the children are – **

**(i) males, if it is known that at least one of the children is male.**

**(ii) females, if it is known that elder child is a female.**

**Solution: **

Let m and f represent the male and the female child respectively.

(i)Let A be the event that both are males.

So, A = {(m, m)}

n(A) = 1

and B be the event that at least one is male.

So, B = {(m, m), (m, f), (f, m)}

n(B) = 3

Now, A ∩ B = {(m, m)}

n(A ∩ B) = 1

Thus, the required probability is –

P(A/B) = n(A ∩ B) / n(B) = 1/3

(ii)Let C be the event that both are females.

So, C = {(f, f)}

n(C) = 1

and D be the event that elder child is female.

So, D = {(f, m), (f, f)}

n(D) = 2

Now, C ∩ D = {(f, f)}

n(C ∩ D) = 1

Thus, the required probability is –

P(C/D) = n(C ∩ D) / n(D) = 1/2

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