NCERT Solutions - Exercise 13.2: Probability

``` Page 1

NCERT Solutions for Class 12 Maths Chapter 13 –
Probability
Exercise 13.2                                                                        Page No: 546
1. If P (A) = 3/5 and P (B) = 1/5, find P (A n 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 n 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
number of cards will be 51, which 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 n 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 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 number of oranges will be 14, which is the conditional probability of B, given that A has already occurred.
Page 2

NCERT Solutions for Class 12 Maths Chapter 13 –
Probability
Exercise 13.2                                                                        Page No: 546
1. If P (A) = 3/5 and P (B) = 1/5, find P (A n 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 n 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
number of cards will be 51, which 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 n 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 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 number of oranges will be 14, which is the conditional probability of B, given that A has already occurred.

NCERT Solutions for Class 12 Maths Chapter 13 –
Probability
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, which 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 n B n 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 n B = {(H, 3)}
? P (A n B) = 1/12 …… (1)
And P (A). P (B) = ½ × 1/6 = 1/12 …… (2)
From (1) and (2) P (A n 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
Page 3

NCERT Solutions for Class 12 Maths Chapter 13 –
Probability
Exercise 13.2                                                                        Page No: 546
1. If P (A) = 3/5 and P (B) = 1/5, find P (A n 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 n 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
number of cards will be 51, which 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 n 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 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 number of oranges will be 14, which is the conditional probability of B, given that A has already occurred.

NCERT Solutions for Class 12 Maths Chapter 13 –
Probability
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, which 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 n B n 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 n B = {(H, 3)}
? P (A n B) = 1/12 …… (1)
And P (A). P (B) = ½ × 1/6 = 1/12 …… (2)
From (1) and (2) P (A n 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

NCERT Solutions for Class 12 Maths Chapter 13 –
Probability
S = {1, 2, 3, 4, 5, 6}
Let A be the event, and the number is even.
? A = {2, 4, 6}
? P (A) = 3/6 = ½
Now, let B be the event, and the number is red.
? B = {1, 2, 3}
? P (B) = 3/6 = 1/2
As, A n B = {2}
? P (A n B) = 1/6 …….. (1)
And P (A). P (B) = ½ × ½ = ¼ ….. (2)
From (1) and (2) P (A n 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 n F) = 1/5. Are E and F independent?
Solution:
Given P (E) = 3/5, P (F) = 3/10 and P (E n F) = 1/5
P (E). P (F) = 3/5 × 3/10 = 9/50 ? 1/5
? P (E n 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 n B) = ?
? P (A n B) = 0
As we know, P (A ? B) = P (A) + P (B) – P (A n B)
Page 4

NCERT Solutions for Class 12 Maths Chapter 13 –
Probability
Exercise 13.2                                                                        Page No: 546
1. If P (A) = 3/5 and P (B) = 1/5, find P (A n 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 n 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
number of cards will be 51, which 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 n 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 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 number of oranges will be 14, which is the conditional probability of B, given that A has already occurred.

NCERT Solutions for Class 12 Maths Chapter 13 –
Probability
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, which 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 n B n 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 n B = {(H, 3)}
? P (A n B) = 1/12 …… (1)
And P (A). P (B) = ½ × 1/6 = 1/12 …… (2)
From (1) and (2) P (A n 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

NCERT Solutions for Class 12 Maths Chapter 13 –
Probability
S = {1, 2, 3, 4, 5, 6}
Let A be the event, and the number is even.
? A = {2, 4, 6}
? P (A) = 3/6 = ½
Now, let B be the event, and the number is red.
? B = {1, 2, 3}
? P (B) = 3/6 = 1/2
As, A n B = {2}
? P (A n B) = 1/6 …….. (1)
And P (A). P (B) = ½ × ½ = ¼ ….. (2)
From (1) and (2) P (A n 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 n F) = 1/5. Are E and F independent?
Solution:
Given P (E) = 3/5, P (F) = 3/10 and P (E n F) = 1/5
P (E). P (F) = 3/5 × 3/10 = 9/50 ? 1/5
? P (E n 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 n B) = ?
? P (A n B) = 0
As we know, P (A ? B) = P (A) + P (B) – P (A n B)

NCERT Solutions for Class 12 Maths Chapter 13 –
Probability
? 3/5 = ½ + p -0
? P = 3/5 – ½ = 1/10
(ii) Independent
When A and B are independent.
? P (A n B) = P (A). P (B)
? P (A n B) = ½ p
As we know, P (A ? B) = P (A) + P (B) – P (A n 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 n 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 n B)
When A and B are independent.
? P (A n B) = P (A). P (B)
? P (A n B) = 0.3 × 0.4
? P (A n B) = 0.12
(ii) P (A ? B)
As we know, P (A ? B) = P (A) + P (B) – P (A n B)
? P (A ? B) = 0.3 + 0.4 – 0.12
? P (A ? B) = 0.58
Page 5

NCERT Solutions for Class 12 Maths Chapter 13 –
Probability
Exercise 13.2                                                                        Page No: 546
1. If P (A) = 3/5 and P (B) = 1/5, find P (A n 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 n 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
number of cards will be 51, which 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 n 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 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 number of oranges will be 14, which is the conditional probability of B, given that A has already occurred.

NCERT Solutions for Class 12 Maths Chapter 13 –
Probability
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, which 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 n B n 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 n B = {(H, 3)}
? P (A n B) = 1/12 …… (1)
And P (A). P (B) = ½ × 1/6 = 1/12 …… (2)
From (1) and (2) P (A n 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

NCERT Solutions for Class 12 Maths Chapter 13 –
Probability
S = {1, 2, 3, 4, 5, 6}
Let A be the event, and the number is even.
? A = {2, 4, 6}
? P (A) = 3/6 = ½
Now, let B be the event, and the number is red.
? B = {1, 2, 3}
? P (B) = 3/6 = 1/2
As, A n B = {2}
? P (A n B) = 1/6 …….. (1)
And P (A). P (B) = ½ × ½ = ¼ ….. (2)
From (1) and (2) P (A n 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 n F) = 1/5. Are E and F independent?
Solution:
Given P (E) = 3/5, P (F) = 3/10 and P (E n F) = 1/5
P (E). P (F) = 3/5 × 3/10 = 9/50 ? 1/5
? P (E n 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 n B) = ?
? P (A n B) = 0
As we know, P (A ? B) = P (A) + P (B) – P (A n B)

NCERT Solutions for Class 12 Maths Chapter 13 –
Probability
? 3/5 = ½ + p -0
? P = 3/5 – ½ = 1/10
(ii) Independent
When A and B are independent.
? P (A n B) = P (A). P (B)
? P (A n B) = ½ p
As we know, P (A ? B) = P (A) + P (B) – P (A n 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 n 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 n B)
When A and B are independent.
? P (A n B) = P (A). P (B)
? P (A n B) = 0.3 × 0.4
? P (A n B) = 0.12
(ii) P (A ? B)
As we know, P (A ? B) = P (A) + P (B) – P (A n B)
? P (A ? B) = 0.3 + 0.4 – 0.12
? P (A ? B) = 0.58

NCERT Solutions for Class 12 Maths Chapter 13 –
Probability

(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 n B) = 1/8, find P (not A and not B).
Solution:
Given P (A) = ¼, P (B) = ½ and P (A n B) = 1/8
P (not A and not B) = P (A
’
n B
’
)
As, { A
’
n B
’
= (A ? B)
’
}
? P (not A and not B) = P ((A ? B)
’
)
= 1 – P (A ? B)
= 1- [P (A) + P (B) – P (A n 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
```

## Mathematics (Maths) for JEE Main & Advanced

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## Mathematics (Maths) for JEE Main & Advanced

209 videos|443 docs|143 tests

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