These NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.2 Questions and Answers are prepared by our highly skilled subject experts.
NCERT Solutions for Class 12 Maths Chapter 13 Probability Exercise 13.2
Question 1.
If P(A) = \(\frac { 3 }{ 5 }\) and P(B) = \(\frac { 1 }{ 5 }\), find P(A∩B) if A and B are independent events.
Solution:
A and B are independent if
P (A ∩ B) = P(A) P(B) = \(\frac { 3 }{ 5 } \times \frac { 1 }{ 5 } =\frac { 3 }{ 25 } \)
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:
Let E : first card drawn is black
F : second card drawn is black
P(E ∩ F) = P(E) P(F|E)
P(E) = Probability of getting a black card \(\frac { 26 }{ 52 }\)
P(F|E) = Probability of getting a black card in the second draw given that the first card is black = \(\frac { 21 }{ 51 }\)
P(E ∩ F) = \(\frac { 26 }{ 52 }\) x \(\frac { 21 }{ 51 }\) = \(\frac { 25 }{ 102 }\)
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
P (a box is approved) = \(\frac { C(12,3) }{ C(15,3) } =\frac { 12\times 11\times 10 }{ 15\times 14\times 13 } =\frac { 44 }{ 91 }\)
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:
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:
Question 6.
Let E and F be the events with P(E) = \(\frac { 3 }{ 5 }\), P (F) = \(\frac { 3 }{ 10 }\) and P (E ∩ F) = \(\frac { 1 }{ 5 }\). Are E and F independent?
Solution:
P(E) = \(\frac { 3 }{ 5 }\), P(F) = \(\frac { 3 }{ 10 }\),
∴ P (E) x P (F) = \(\frac { 3 }{ 5 } \times \frac { 3 }{ 10 } =\frac { 9 }{ 50 }\)
P(E ∩ F) ≠ P(E) x P(F)
∴ The event A and B are not independent.
Question 7.
Given that the events A and B are such that P(A) = \(\frac { 1 }{ 2 }\), P(A∪B) = \(\frac { 3 }{ 5 }\) and P(B) = p. Find p if they are
(i) mutually exclusive
(ii) independent.
Solution:
Question 8.
Let A and B independent events 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:
P (A) = 0.3,
P (B) = 0.4
A and B are independent events
(i) ∴ P (A ∩ B) = P (A). P (B) = 0.3 x 0.4 = 0.12.
(ii) P(A∪B) = P(A) + P(B) – P(A).P(B)
= 0.3 + 0.4 – 0.3 x 0.4 = 0.7 – 0.12 = 0.58.
(iii) P (A | B) = \(\frac{P(A \cap B)}{P(B)}=\frac{0.12}{0.4}=0.3\)
(iv) P(B | A) = \(\frac{P(A \cap B)}{P(A)}=\frac{0.12}{0.3}=0.4\)
Question 9.
If A and B are two events, such that P (A) = \(\frac { 1 }{ 4 }\), P(B) = \(\frac { 1 }{ 2 }\), and P(A ∩B) = \(\frac { 1 }{ 8 }\). Find P (not A and not B)
Solution:
P (A) = \(\frac { 1 }{ 4 }\), P(B) = \(\frac { 1 }{ 2 }\), and P(A ∩B) = \(\frac { 1 }{ 8 }\)
P(A) P(B) = \(\frac { 1 }{ 4 }\) x \(\frac { 1 }{ 2 }\) = \(\frac { 1 }{ 8 }\)
P(A).P(B) = P(A ∩ B)
∴ A and B are independent events
Since A and B are independent events, A’ and B’ are also independent events.
∴ P (A’ ∩ B’) = P (A’). P (B’)
= [1 – P(A)] [1 – P(B)]
= \(\left(1-\frac{1}{4}\right)\left(1-\frac{1}{2}\right)=\frac{3}{4} \times \frac{1}{2}=\frac{3}{8}\)
Question 10
Events A and B are such that P(A) = \(\frac { 1 }{ 2 }\), P(B) = \(\frac { 7 }{ 12 }\) and P (not A or not B) = \(\frac { 1 }{ 4 }\). State whether A and Bare independent
Solution:
Hence A and B are not independent events.
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:
P(A) = 0.3, P(B) = 0.6
A and B are independent events.
i. P(A and B) = P(AB) = P(A).P(B)
= 0.3 x 0.6 = 0.18
ii. P(A and not B) = P(A B’ ) = P(A) P( B’)
= P(A) [1 – P(B)] = 0.3 (1 – 0.6)
= 0.3 x 0.4 = 0.12
iii P(A or B) = P(A ∪ B)
= P(A) + P(B) – P(AB)
= 0.3 + 0.6 – 0.18 from (i)
= 0.72
iv. P(neither A nor B) = P( A’ B’)
= P(A’)P(B’)
= [1 – P(A)] [1 – P(B)]
= [ 1 – 0.3] [1 – 0.6]
= 0.7 x 0.4 = 0.28
Another Method:
P(neither A nor B) = P(A’ B’)
= P(A’ ∩ B’) = P((A ∪ B)’)
= 1 – P(A ∪ B) = 1 – 0.72 from (iii) = 0.28
Question 12
A die is tossed thrice. Find the probability of getting an odd number at least once.
Solution:
P(getting an odd number when a die is 1 tossed) = \(\frac { 1 }{ 2 }\)
P(getting an even number when a die is 1 tossed) = \(\frac { 1 }{ 2 }\)
Let A : getting even number in all the 3 throws
P(A) = \(\frac { 1 }{ 2 }\) x \(\frac { 1 }{ 2 }\) x \(\frac { 1 }{ 2 }\) = \(\frac { 1 }{ 8 }\)
P(getting an odd number atleast once) = P(not getting even number in the 3 throws)
= 1 – P(A) = 1 – \(\frac { 1 }{ 8 }\) = \(\frac { 7 }{ 8 }\)
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:
Let R : red ball is drawn
B : black ball is drawn
P(R) = \(\frac { 8 }{ 18 }\), P(B) = \(\frac { 10 }{ 18 }\)
i. P(both are red) = P(1st ball is red and 2nd ball is red)
= P(RR) = P(R). P(R),
since the events are independent
= \(\frac { 8 }{ 18 }\) x \(\frac { 8 }{ 18 }\) = \(\frac { 6 }{ 81 }\)
ii. P(1st ball is black and 2nd ball is red)
= P(BR) = P(B) . P(R)
since the events are independent
\(\frac { 10 }{ 18 }\) x \(\frac { 8 }{ 18 }\) = \(\frac { 20 }{ 81 }\)
iii. P(one of them is black and the other is red) = P(1st ball is black & 2nd ball is red or 1st ball is red & 2nd ball is black) = P(BR or RB)
= P(BR) + P(RB)
= P(B) P(R) + P(R) P(B)
= 2 x \(\frac { 10 }{ 18 }\) x \(\frac { 8 }{ 18 }\) = \(\frac { 40 }{ 81 }\)
Question 14
Probability of solving specific problem independently by A and B are \(\frac { 1 }{ 2 }\) and \(\frac { 1 }{ 3 }\) 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:
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:
(i) E : card drawn is a spade
F : card drawn is an ace
E ∩ F : card drawn is an ace of spade
Hence E, F are independent events
(ii) E : card drawn is black
F : card drawn is a king
E ∩ F : card drawn is a black king
Hence E and F are independent events.
(iii) E : card drawn is king or queen
F : card drawn is queen or jack
E ∩ F : card drawn is queen
Hence E and F are not independent events.
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 newspapers.
(c) If she reads English newspaper, find the probability that she reads Hindi newspaper.
Solution:
Let H : student reads Hindi newspaper
E : student reads English newspaper
Question 17
The probability of obtaining an even prime number on each die when a pair of dice is rolled is
(a) 0
(b) \(\frac { 1 }{ 3 }\)
(c) \(\frac { 1 }{ 12 }\)
(d) \(\frac { 1 }{ 36 }\)
Solution:
(d) n(S) = 36
The sample space contains 36 simple events
Let E: getting even prime number on both dice
∴ E = {(2, 2)} ∴ P(E) = \(\frac { 1 }{ 36 }\)
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:
A and B are independent events if P(A ∩ B) = P(A) P(B)
P( A’B’) = P( A’ ∩ B’) = P((A ∪ B)’)
= 1 – P(A ∪ B)
= 1 – [P(A) + P(B) – P(A ∩ B)]
= 1 – P(A) – P(B) + P(A) P(B)
= [1 – P(A)] – P(B)[1 – P(A)]
= [1 – P(A)][1 – P(B)]
= P(A’)P(B’)
If A and B are mutually exclusive, then A and B are not independent events.
If P(A) = P(B), then A and B need not be independent events.
If P(A) + P(B) = 1, then A and B are complements of each other. Then also A and B are not independent events.