NCERT Solutions for Maths Class 12 Exercise 13.2

# NCERT Solutions for Maths Class 12 Exercise 13.2

Hello Students. In this post, you will find the complete NCERT Solutions for Maths Class 12 Exercise 13.2.

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NCERT Solutions for Maths Class 12 Exercise 13.1

NCERT Solutions for Maths Class 12 Exercise 13.3

## NCERT Solutions for Maths Class 12 Exercise 13.2

Maths Class 12 Ex 13.2 Question 1.

If P(A) = 3/5 and P(B) = 1/5, find P(A∩B) if A and B are independent events.

Solution:

We have given that P(A) = 3/5 and P(B) = 1/5

If A and B are independent event, then P(A∩B) = P(A).P(B)
= 3/5 × 1/5 = 3/25

Hence, the value of P(A∩B) is 3/25.

Maths Class 12 Ex 13.2 Question 2.

Two cards are drawn at random and without replacement from a pack of 52 playing cards. Find the probability that both the cards are black.

Solution:

Number of exhaustive cases = 52
Number of black cards = 26
One black card may be drawn in 26 ways.
Probability of getting a black card on first draw, P(A) = 26/52 = ½

After drawing one card, the number of cards left = 51

After drawing a black card, the number of black cards left = 25

Probability of getting a black card on second draw, P(B) = 25/51

Thus, the probability of getting both the cards black = P(A).P(B)

= ½ × 25/51 = 25/102

Hence, the probability that both the cards drawn are black is 25/102.

Maths Class 12 Ex 13.2 Question 3.

A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all the three oranges are good, the box is approved for sale, otherwise, it is rejected. Find the probability that a box containing 15 oranges out of which 12 are good and 3 are bad ones will be approved for sale.

Solution:

S = {12 good oranges, 3 bad oranges),
n(S) = 15

Probability that the first orange drawn is good, P(A) = 12/15

Probability that the second orange drawn is good, P(B) = 11/14

Probability that the third orange drawn is good, P(C) = 10/13

Thus, the probability of getting all the oranges good = P(A).P(B).P(C)

= 12/15 × 11/14 × 10/13 = 44/91

Hence, the probability that the box is approved for sale is 44/91.

Maths Class 12 Ex 13.2 Question 4.

A fair coin and an unbiased die are tossed. Let A be the event ‘head appears on the coin’ and B be the event ‘3 on the die’. Check whether A and B are independent events or not.

Solution:

When a coin is thrown, head or tail will occur.
Probability of getting head on the coin, P(A) = ½
When a die is tossed, (1, 2, 3, 4, 5, 6) one of them will appear
Probability of getting 3 on the die, P(B) = 1/6

P(A).P(B) = ½ × 1/6 = 1/12
When a die and coin is tossed, total number of cases are
H1, H2, H3, H4, H5, H6
T1, T2, T3, T4, T5, T6
Head and 3 will occur only in 1 way
Probability of getting head on coin and 3 on die = 1/12, i.e., P(A∩B) = 1/12

Therefore, P(A∩B) = P(A).P(B)

Hence, A and B are independent events.

Maths Class 12 Ex 13.2 Question 5.

A die marked 1, 2, 3 in red and 4, 5, 6 in green is tossed. Let A be the event, ‘the number is even’, and B be the event, ‘the number is red’. Are A and B independent?

Solution:

Even numbers on die are 2, 4, 6.
Probability of getting even number, P(A) = 3/6 = ½
There are two colours of the die: red and green.
Probability of getting red colour, P(B) = ½
Even number in red colour is 2.
Probability of getting red colour and even number = P(A∩B) = 1/6

P(A).P(B) = ½ × ½ = ¼

We have, P(A∩B) ≠ P(A).P(B)

Hence, A and B are not independent.

Maths Class 12 Ex 13.2 Question 6.

Let E and F be the events with P(E) = 3/5, P(F) = 3/10 and P(E∩F) = 1/5. Are E and F independent?

Solution:

We have, P(E) = 3/5, P(F) = 3/10 and P(E∩F) = 1/5.
P(E) × P(F) = 3/5 × 3/10 = 9/50
Here, P(E∩F) ≠ P(E) × P(F)
The events A and B are not independent.

Maths Class 12 Ex 13.2 Question 7.

Given that the events A and B are such that P(A) = ½, P(AB) = 3/5 and P(B) = p. Find p if they are
(i) mutually exclusive
(ii) independent.

Solution:

Maths Class 12 Ex 13.2 Question 8.

Let A and B be independent events with P(A) = 0.3 and P(B) = 0.4. Find
(i) P(A∩B)
(ii) P(A
B)
(iii) P(A|B)
(iv) P(B|A)

Solution:

We have, P(A) = 0.3 and P(B) = 0.4

(i) Since, A and B are independent events.
Therefore, P(A∩B) = P(A).P(B) = 0.3 × 0.4 = 0.12.

(ii) We know that, P(AB) = P(A) + P(B) – P(A).P(B)
= 0.3 + 0.4 – 0.3 × 0.4 = 0.7 – 0.12 = 0.58.

Maths Class 12 Ex 13.2 Question 9.

If A and B are two events such that P(A) = ¼, P(B) = ½ and P(A∩B) = 1/8, find P(not A and not B).

Solution:

Maths Class 12 Ex 13.2 Question 10.

Events A and B are such that P(A) = ½, P(B) = 7/12 and P(not A or not B) = ¼. State whether A and B are independent.

Solution:

Maths Class 12 Ex 13.2 Question 11.

Given two independent events A and B such that P(A) = 0.3, P(B) = 0.6. Find
(i) P(A and B)
(ii) P(A and not B)
(iii) P(A or B)
(iv) P(neither A nor B)

Solution:

(i) Since, A and B are independent events.
P(A and B) = P(A∩B) = P(A) × P(B)
= 0.3 × 0.6        [
P(A) = 0.3 and P(B) = 0.6]
P(A and B) = 0.18

Maths Class 12 Ex 13.2 Question 12.

A die is tossed thrice. Find the probability of getting an odd number at least once.

Solution:

We have, S = {1, 2, 3, 4, 5, 6}, n(S) = 6
The probability of getting an odd number in a single throw of die = 3/6 = ½

Similarly, the probability of getting an even number in a single throw of die = 3/6 = ½

Thus, the probability of getting an even number three times = ½ × ½ × ½ = 1/8

Therefore, the probability of getting an odd number at least once

= 1 – Probability of getting an odd number in none of the throws

= 1 – Probability of getting an even number thrice

= 1 – 1/8 = 7/8

Hence, the probability of getting an odd number at least once is 7/8.

Maths Class 12 Ex 13.2 Question 13.

Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that
(i) both balls are red.
(ii) first ball is black and second is red.
(iii) one of them is black and other is red.

Solution:

S = {10 black balls, 8 red balls}, n(S) = 18
Let drawing of a red ball be a success.
A = {8 red balls}, n(A) = 8

Maths Class 12 Ex 13.2 Question 14.

Probability of solving specific problem independently by A and B are ½ and 1/3 respectively. If both try to solve the problem independently, find the probability that
(i) the problem is solved
(ii) exactly one of them solves the problem.

Solution:

Probability that A solves the problem = ½
Probability that A does not solve the problem, P(A) =  1 – ½ = ½
Probability that B solves the problem = 1/3
Probability that B does not solve the problem, P(B) = 1 – 1/3 = 2/3

Maths Class 12 Ex 13.2 Question 15.

One card is drawn at random from a well shuffled deck of 52 cards. In which of the following cases are the events E and F independent?
(i) E: ‘the card drawn is a spade’
F: ‘the card drawn is an ace’
(ii) E: ‘the card drawn is black’
F: ‘the card drawn is a king’
(iii) E: ‘the card drawn is a king or queen’
F: ‘the card drawn is a queen or jack’

Solution:

Maths Class 12 Ex 13.2 Question 16.

In a hostel, 60% of the students read Hindi newspaper, 40% read English newspaper and 20% read both Hindi and English newspapers. A student is selected at random.
(a) Find the probability that she reads neither Hindi nor English newspapers.
(b) If she reads Hindi newspaper, find the probability that she reads English newspaper.
(c) If she reads English newspaper, find the probability that she reads Hindi newspaper.

Solution:

Choose the correct answer in Questions 17 and 18:

Maths Class 12 Ex 13.2 Question 17.

The probability of obtaining an even prime number on each die, when a pair of dice is rolled, is
(A) 0
(B) 1/3
(C) 1/12
(D) 1/36

Solution:

(D) When a pair of dice is rolled, then n(S) = 36.

The only even prime number is 2.
Let E be the event of getting an even prime number on each die.

Therefore, E = {2, 2}; n(E) = 1

P(E) = n(E)/n(S) = 1/36

Hence, the correct answer is option (D).

Maths Class 12 Ex 13.2 Question 18.

Two events A and B are said to be independent, if
(A) A and B are mutually exclusive
(B) P(A’B’) = [1 – P(A)] [1 – P(B)]
(C) P(A) = P(B)
(D) P(A) + P(B) = 1

Solution:

(B) Two events A and B are said to be independent, if P(A∩B) = P(A).P(B)

Thus, P(A’∩B’) = P(A’).P(B’)

We know that P(A’) = [1 – P(A)] and P(B’) = [1 – P(B)]

Therefore, P(A’∩B’) = [1 – P(A)].[1 – P(B)]

Or P(A’B’) = [1 – P(A)] [1 – P(B)]
Hence, the correct answer is option (B).