NCERT Solutions for Class 12 Maths Ex 13.2 Chapter 13 Probability

NCERT Solutions for Class 12 Maths Chapter 13 Exercise 13.2 – Free PDF Download

Exercise 13.2 of NCERT Solutions for Class 12 Maths Chapter 13 – Probability is based on the following topics:

1. Multiplication Theorem on Probability
2. Independent Events: Two events such that the probability of occurrence of one of them is not affected by the occurrence of the other are called independent events.

The second exercise of this chapter contains problems based on the above-mentioned topics. Solving questions related to this help us understand the application of probability in real life.

NCERT Solutions for Class 12 Maths Chapter 13 – Probability Exercise 13.2

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Exercise 13.1 Solutions 17 Questions

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Exercise 13.4 Solutions 17 Questions

Exercise 13.5 Solutions 15 Questions

Miscellaneous Exercise On Chapter 13 Solutions 10 Questions

Access Answers of Maths NCERT Class 12 Chapter 13.2

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

Solution:

Given P (A) = 3/5 and P (B) = 1/5

As A and B are independent events.

⇒ P (A ∩ B) = P (A).P (B)

= 3/5 × 1/5 = 3/25

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:

Given a pack of 52 cards.

As we know, there are 26 cards in total, which are black. Let A and B denote respectively the events that the first and second drawn cards are black.

Now, P (A) = P (black card in first draw) = 26/52 = ½

Because the second card is drawn without replacement, now the total number of black cards will be 25, and the total cards will be 5, that is the conditional probability of B, given that A has already occurred.

Now, P (B/A) = P (black card in second draw) = 25/51

Thus the probability that both cards are black

⇒ P (A ∩ B) = ½ × 25/51 = 25/102

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

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:

Given a box of oranges.

Let A, B and C denotes the events respectively that the first, second and third drawn orange is good.

Now, P (A) = P (good orange in first draw) = 12/15

Because the second orange is drawn without replacement, now the total number of good oranges will be 11, and the total oranges will be 14; that is the conditional probability of B, given that A has already occurred.

Now, P (B/A) = P (good orange in second draw) = 11/14

Because the third orange is drawn without replacement, now the total number of good oranges will be 10, and the total oranges will be 13; that is the conditional probability of C, given that A and B have already occurred.

Now, P (C/AB) = P (good orange in third draw) = 10/13

Thus the probability that all the oranges are good.

⇒ P (A ∩ B ∩ C) = 12/15 × 11/14 × 10/13 = 44/91

Hence, the probability that a box will be approved for sale = 44/91

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:

Given a fair coin and an unbiased die are tossed.

We know that the sample space S:

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

Let A be the event head that appears on the coin:

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

⇒ P (A) = 6/12 = ½

Now, Let B be the event 3 on the die

⇒ B = {(H, 3), (T, 3)}

⇒ P (B) = 2/12 = 1/6

As, A ∩ B = {(H, 3)}

⇒ P (A ∩ B) = 1/12 …… (1)

And P (A). P (B) = ½ × 1/6 = 1/12 …… (2)

From (1) and (2) P (A ∩ B) = P (A). P (B)

Therefore, A and B are independent events.

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:

The sample space for the dice will be

S = {1, 2, 3, 4, 5, 6}

Let A be the event, the number is even:

⇒ A = {2, 4, 6}

⇒ P (A) = 3/6 = ½

Now, Let B be the event the number is red:

⇒ B = {1, 2, 3}

⇒ P (B) = 3/6 = 1/2

As, A ∩ B = {2}

⇒ P (A ∩ B) = 1/6 …….. (1)

And P (A). P (B) = ½ × ½ = ¼ ….. (2)

From (1) and (2) P (A ∩ B) ≠ P (A). P (B)

Therefore, A and B are not independent events.

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

Solution:

Given P (E) = 3/5, P (F) = 3/10 and P (E ∩ F) = 1/5

P (E). P (F) = 3/5 × 3/10 = 9/50 ≠ 1/5

⇒ P (E ∩ F) ≠ P (E). P (F)

Therefore, E and F are not independent events.

7. Given that the events A and B are such that P (A) = 1/2, P (A ∪ B) = 3/5 and P (B) = p. Find p if they are (i) mutually exclusive (ii) independent.

Solution:

Given P (A) = ½, P (A ∪ B) = 1/5 and P (B) = p

(i) Mutually exclusive

When A and B are mutually exclusive.

Then (A ∩ B) = ϕ

⇒ P (A ∩ B) = 0

As we know, P (A ∪ B) = P (A) + P (B) – P (A ∩ B)

⇒ 3/5 = ½ + p -0

⇒ P = 3/5 – ½ = 1/10

(ii) Independent

When A and B are independent.

⇒ P (A ∩ B) = P (A). P (B)

⇒ P (A ∩ B) = ½ p

As we know, P (A ∪ B) = P (A) + P (B) – P (A ∩ B)

⇒ 3/5 = ½ + 2 – p/2

⇒ p/2 = 3/5 – ½

⇒ p = 2 × 1/10 = 1/5

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:

Given P (A) = 0.3 and P(B) = 0.4

(i) P (A ∩ B)

When A and B are independent.

⇒ P (A ∩ B) = P (A). P (B)

⇒ P (A ∩ B) = 0.3 × 0.4

⇒ P (A ∩ B) = 0.12

(ii) P (A ∪ B)

As we know, P (A ∪ B) = P (A) + P (B) – P (A ∩ B)

⇒ P (A ∪ B) = 0.3 + 0.4 – 0.12

⇒ P (A ∪ B) = 0.58

(iv) P (B|A)

9. If A and B are two events such that P (A) = 1/4 , P (B) = 1/2 and P (A ∩ B) = 1/8, find P (not A and not B).

Solution:

Given P (A) = ¼, P (B) = ½ and P (A ∩ B) = 1/8

P (not A and not B) = P (A^’ ∩ B^’)

As, { A^’ ∩ B^’ = (A ∪ B)^’}

⇒ P (not A and not B) = P ((A ∪ B)^’)

= 1 – P (A ∪ B)

= 1- [P (A) + P (B) – P (A ∩ B)]

10. Events A and B are such that P (A) = 1/2, P (B) = 7/12 and P (not A or not B) = 1/4. State whether A and B are independent?

Solution:

Given P (A) = ½, P (B) =7/12 and P (not A or not B) = 1/4

⇒ P (A^’∪ B^’) = 1/4

⇒ P (A ∩ B)^’ = 1/4

⇒ 1 – P (A ∩ B) = 1/4

⇒ P (A ∩ B) = 1 – 1/4

⇒ P (A ∩ B) = 3/4…….. (1)

And P (A). P (B) = ½ × 7/12 = 7/24 …. (2)

From (1) and (2) P (A ∩ B) ≠ P (A). P (B)

Therefore, A and B are not independent events.

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:

Given P (A) = 0.3, P (B) = 0.6.

(i) P (A and B)

As A and B are independent events.

⇒ P (A and B) = P (A ∩ B) = P (A). P (B)

= 0.3 × 0.6

= 0.18

(ii) P (A and not B)

⇒ P (A and not B) = P (A ∩ B^’) = P (A) – P (A ∩ B)

= 0.3 – 0.18

= 0.12

(iii) P (A or B)

⇒ P (A or B) = P (A ∪ B)

As we know, P (A ∪ B) = P (A) + P (B) – P (A ∩ B)

⇒ P (A ∪ B) = 0.3 + 0.6 – 0.18

⇒ P (A ∪ B) = 0.72

(iv) P (neither A nor B)

P (neither A nor B) = P (A^’ ∩ B^’)

As, { A^’ ∩ B^’ = (A ∪ B)^’}

⇒ P (neither A nor B) = P ((A ∪ B)^’)

= 1 – P (A ∪ B)

= 1 – 0.72

= 0.28

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

Solution:

Given a die is tossed thrice.

Then the sample space S = {1, 2, 3, 4, 5, 6}

Let P (A) = probability of getting an odd number in first throw.

⇒ P (A) = 3/6 = ½.

Let P (B) = probability of getting an even number.

⇒ P (B) =3/6 = ½.

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

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

= 1 – probability of getting an odd number in no throw

= 1 – probability of getting an even number in three times

= 1 – 1/8

∴ Probability of getting an odd number at least once = 7/8

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:

Given A box containing 10 black and 8 red balls.

Total number of balls in box = 18

(i) Both balls are red.

Probability of getting a red ball in the first draw = 8/18 = 4/9

As the ball is replaced after the first throw,

Hence, the probability of getting a red ball in the second draw = 8/18 = 4/9

Now, the probability of getting both balls red = 4/9 × 4/9 = 16/81

(ii) the first ball is black, and the second is red.

Probability of getting a black ball in first draw = 10/18 = 5/9

As the ball is replaced after first throw,

Hence, the probability of getting a red ball in the second draw = 8/18 = 4/9

Now, the probability of getting the first ball black and the second red = 5/9 × 5/9 = 20/81

(iii) One of them is black, and the other is red.

Probability of getting a black ball in the first draw = 10/18 = 5/9

As the ball is replaced after the first throw,

Hence, the probability of getting a red ball in the second draw = 8/18 = 4/9

Now, the probability of getting the first ball black and the second red = 5/9 × 4/9 = 20/81

Probability of getting a red ball in the first draw = 8/18 = 4/9

As the ball is replaced after the first throw,

Hence, the probability of getting a black ball in the second draw = 10/18 = 5/9

Now, the probability of getting the first ball red and the second black = 5/9 × 4/9 = 20/81

Therefore, the probability of getting one of them black and the other red:

= Probability of getting the first ball black and second red + Probability of getting the first ball red and the second black

= 20/81 + 20 /81 = 40/81

14. Probability of solving specific problem independently by A and B are 1/2 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:

Given,

P (A) = Probability of solving the problem by A = 1/2

P (B) = Probability of solving the problem by B = 1/3

Because A and B are both independent.

⇒ P (A ∩ B) = P (A). P (B)

⇒ P (A ∩ B) = ½ × 1/3 = 1/6

P (A^’) = 1 – P (A) = 1 – 1/2 = 1/2

P (B^’) = 1 – P (B) = 1 – 1/3 = 2/3

(i) The problem is solved

The problem is solved, i.e. it is either solved by A, or it is solved by B.

= P (A ∪ B)

As we know, P (A ∪ B) = P (A) + P (B) – P (A ∩ B)

⇒ P (A ∪ B) = ½ + 1/3 – 1/6 = 4/6

⇒ P (A ∪ B) = 2/3

(ii) Exactly one of them solves the problem

That is either problem is solved by A but not by B or vice versa

That is P (A).P (B^’) + P (A^’).P (B)

= ½ (2/3) + ½ (1/3)

= 1/3 + 1/6 = 3/6

⇒ P (A).P (B^’) + P (A^’).P (B) = ½

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:

Given: A deck of 52 cards.

(i) In a deck of 52 cards, 13 cards are spade, and 4 cards are ace, and only one card is there which is spade and ace both.

Hence, P (E) = the card drawn is a spade = 13/52 = 1/4

P (F) = the card drawn is an ace = 4/52 = 1/13

P (E ∩ F) = the card drawn is a spade and ace both = 1/52….. (1)

And P (E). P (F)

= ¼ × 1/13 = 1/52…. (2)

From (1) and (2)

⇒ P (E ∩ F) = P (E). P (F)

Hence, E and F are independent events.

(ii) In a deck of 52 cards, 26 cards are black and 4 cards are king and only two card are there which are black and king both.

Hence, P (E) = the card drawn is of black = 26/52 = ½

P (F) = the card drawn is a king = 4/52 = 1/13

P (E ∩ F) = the card drawn is a black and king both = 2/52 = 1/26…. (1)

And P (E). P (F)

= ½ × 1/13 = 1/26…. (2)

From (1) and (2)

⇒ P (E ∩ F) = P (E). P (F)

Hence, E and F are independent events.

(iii) In a deck of 52 cards, 4 cards are queen, 4 cards are king and 4 cards are jack.

Hence, P (E) = the card drawn is either king or queen = 8/52 = 2/13

P (F) = the card drawn is either queen or jack = 8/52 = 2/13

There are 4 cards which are either king or queen and either queen or jack.

P (E ∩ F) = the card drawn is either king or queen and either queen or jack = 4/52 = 1/13 … (1)

And P (E). P (F)

= 2/13 × 2/13 = 4/169…. (2)

From (1) and (2)

⇒ P (E ∩ F) ≠ P (E). P (F)

Hence, E and F are not independent events.

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:

Given:

Let H and E denote the number of students who read Hindi and English newspapers respectively.

Hence, P (H) = Probability of students who read Hindi newspaper= 60/100 = 3/5

P (E) = Probability of students who read English newspaper = 40/100 = 2/5

P (H ∩ E) = Probability of students who read Hindi and English both newspapers = 20/100 = 1/5

(a) Find the probability that she reads neither Hindi nor English newspapers.

P (neither H nor E)

P (neither H nor E) = P (H^’ ∩ E^’)

As, { H^’ ∩ E^’ = (H ∪ E)^’}

⇒ P (neither A nor B) = P ((H ∪ E)^’)

= 1 – P (H ∪ E)

= 1- [P (H) + P (E) – P (H ∩ E)]

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. 1/36

Explanation:

Given A pair of dice is rolled.

Hence the number of outcomes = 36

Let P (E) be the probability of getting an even prime number on each die.

As we know, the only even prime number is 2.

So, E = {2, 2}

⇒ P € = 1/36

18. Two events A and B will 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) P (A′B′) = [1 – P (A)] [1 – P (B)]

Explanation:

P (A′B′) = [1 – P (A)] [1 – P (B)]

⇒ P (A′∩ B′) = 1 – P (A) – P (B) + P (A) P (B)

⇒ 1 – P (A ∪ B) =1 – P (A) – P (B) + P (A) P (B)

= – [P (A) + P (B) – P (A ∩ B)] = – P (A) – P (B) + P (A) P (B)

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

⇒ P (A ∩ B) = P (A). P (B)

Hence, it shows A and B are Independent events.

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