Question 1. If P(A) = 3/5 and P(B) = 1/5, find P (A ∩ B) if A and B are independent events.
Solution:
If A and B are independent events.
Given, P(A) = 3/5 & P(B) = 1/5
Since events A & B are independent: P(A ∩ B) = P(A). P(B)
Putting values = 3/5 × 1/5 = 3/25
Question 2. Two cards are drawn at random & without replacement from a pack of 52 playing cards. Find the Probability that the cards are black.
Solution:
There are 26 black cards in a deck of 52 cards.
Probability both cards drawn are black = Probability first card drawn is black x Probability second black if first is black
= 26/52 × 25/51
= 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:
Given Total Number of oranges = 15
Number of Good oranges = 12
Number of Bad oranges = 3
Given that Box is approved for sale if all 3 oranges are good
Now, Probability that box is approved for sale = Probability that first orange is good × Probability that second orange is good, given first is good × Probability that third orange is good, given first two are good
= 12/15 × 11/14 × 10/13
= 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:
A fair coin and unbiased die are tossed
S = {(H, 1), (H, 2), ……….., (H, 6), (T, 1), (T, 2), ………….., (H, 6)}
Given:
A : head appears on the coin
B : 3 on the die
Event A = {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6)}
P(A) = 6/12 = 1/2
Event B = {(H, 3), (T, 3)}
P(B) = 2/12 = 1/6
Now, A ∩ B = {(H, 3)}
So, P(A ∩ B) = 1/12
Now, P(A) . P(B) = 1/2 × 1/6 = 1/12 = P(A ∩ B)
Since, P(A ∩ B) = P(A) . P(B),
Therefore, A & B are independent events
Question 5. A die marked 1, 2, 3 in red and 4, 5, 6 in green is tossed. Let A be the event, ‘number is even,’ and B be the event, ‘number is red’. Are A and B independent?
Solution:
Here, S = {1, 2, 3, 4, 5, 6}
Let two events be
A : the Number is even
B : the number is red
Event A = {"2, 4, 6" }
P(A) = 3/6=1/2
Event B = {"1, 2, 3" }
P(B) = 3/6=1/2
Also, A ∩ B = {2}
So, P(A ∩ B) = 1/6
Now, P(A) . P(B) = 1/2 ×1/2 = 1/4 ≠ P(A ∩ B)
Since P(A ∩ B) ≠ P(A) . P(B)
Therefore, A and B are not Independent
Question 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 & P(E ∩ F) = 1/5
Now, P(E) . P(F) = 3/5 × 3/10 = 9/50 ≠ P(E ∩ F)
Since P(E ∩ F) ≠ P(E) . P(F)
Therefore, E and F are not independent events
Question 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) = 1/2 , P (A ∪ B) = 3/5 and P(B) = p.
(i) Since sets A & B are mutually exclusive,
So, they have nothing in common ∴ P(A ∩ B) = 0
We know that P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
Putting values 3/5 = 1/2 + p – 0
3/5 – 1/2 = p
(6 − 5)/10 = p
1/10 = p
p = 1/10
(ii) Since events A & B are independent,
So, P(A ∩ B) = P(A) P(B)
1/2 × p = ?/2
Now, P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
Putting values 3/5 = 1/2 + p – ?/2
3/2 – 1/2 = p – ?/2
(6 − 5)/10 = ?/2
1/10 = ?/2
p = 2/10 = 1/5
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:
Given, P(A) = 0.3 , P(B) = 0.4
Since events A and B are independent
(i) P(A ∩ B) = P(A) . P(B) = 0.3 × 0.4 = 0.12
(ii) P(A ∪ B) = P(A) + P(B) – P(A ∩ B) = 0.3 + 0.4 – 0.12 = 0.70 – 0.12 = 0.58
(iii) P (A|B) = P (A ∩ B)/P(B) = P (A|B) = P (A) = 0.3
(iv) P (B|A) = P (A ∩ B)/P(A) = P (B|A) = P (B) = 0.4
Question 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) = 1/4 , P(B) = 1/2 and P(A ∩ B) = 1/8
P(not A & not B) = P(A’ ∩ B’) = P(A ∪ B)’
= 1 – P(A or B)
= 1 – P(A ∪ B)
= 1 – [P(A) + P(B) – P(A ∩ B)]
= 1 – (1/4 + 1/2 – 1/8)
=1 – 5/8
= 3/8
Question 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) = 1/2 , P(B) = 7/12 and P(not A or not B) = 1/4
Now, P(not A or not B) = P(A’ or B’)
1/4 = P(A’ ∪ B’)
1/4 = P(A ∩ B)’
1/4 = 1 – P(A ∩ B)
P(A ∩ B) = 1 – 1/4
P(A ∩ B) = 3/4
Also, P(A) . P(B) = 1/2 × 7/12 = 7/24 ≠ P(A ∩ B)
Since, P(A ∩ B) ≠ P(A) . P(B)
Therefore, the events A and B are not independent
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:
Given, P(A) = 0.3 & P(B) = 0.6 Since events A & B are independent
(i) P(A ∩ B) = P(A) . P(B) Now, 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 ∩ B’) = P(A) – P(A and B) = P(A) – P(A ∩ B) = 0.30 – 0.18 = 0.12
(iii) P(A or B) = P(A ∪ B) = P(A) + P(B) – P(A ∩ B) = 0.30 + 0.60 – 0.18 = 0.90 – 0.18 = 0.72
(iv) P(neither A nor B) = P(A’ and B’) = P(A’ ∩ B’) = P(A ∪ B)’ = 1 – P(A ∪ B) = 1 – 0.72 = 0.28
Question 12. A die is tossed thrice. Find the Probability of getting an odd number at least once.
Solution:
A die is tossed thrice. Probability of getting an odd number atleast once = Probability of getting an odd number once + Probability of getting an odd number twice + Probability of getting an odd number thrice
= 1 – Probability of getting an odd number 0 times
= 1 – Probability of getting an even number all 3 times
= 1 − 1/2 × 1/2 × 1/2
= 1 − 1/8 = 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.
Solution:
Two balls are drawn with replacement from a box P(both balls are red)
= P(first ball is red) × P(second ball is red given first is red)
= 8/18 × 8/18 = 4/9 × 4/9 = 16/18
(ii) first ball is black and second is red.
Solution:
P (first ball is black & second is red) = P(first ball is black) × P(second ball is red given first is black)
= 10/18 × 8/18 = 5/9 × 4/9
= 20/81
(iii) one of them is black and other is red.
Solution:
P(one ball is black & other is red)
= P(first ball is black & second is red) + P(first ball is red & second is black)
= (10/18×8/18)+(8/18×10/18)
= 2 × (10/18×8/18) = 40/81
Question 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.
Solution:
Given, P(A) = 1/2 & P(B) = 1/3 Probability that the problem is solved = Probability that A solves the problem or B solves the problem
= P(A ∪ B) = P(A) + P(B) – P(A ∩ B) Since A & B are independent,
P(A ∩ B) = P(A) . P(B) = 1/2 × 1/3 = 1/6
Now, P(Problem is solved) = P(A) + P(B) – P(A ∩ B) = 1/2 + 1/3 – 1/6 = 3/6 + 2/6 – 1/6 = 4/6 = 2/3
(ii) exactly one of them solves the problem.
Solution:
Probability that exactly one of them solves the problem = Probability that only A solves + Probability that only B solves
Therefore, P(exactly one of them solves) = P(A alone solves) + P(B alone solves)
= P(A ∩ B’) + P(B ∩ A’) = (P(A) – P(A ∩ B)) + (P(B) – P(B ∩ A)) = P(A) + P(B) – 2P(A ∩ B)
= 1/2 + 1/3 – 2 × 1/6 = 1/2 + 1/3 – 1/3 = 1/2
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’
Solution:
E : the card drawn is a spade
Number of spades = 13
Total number of cards = 52
P(E) = 13/52 = 1/4
F : the card drawn is an ace
Number of ace = 4
Total number of cards = 52
P(F) = 4/52 = 1/13
Now, E ∩ F = the card drawn is an ace of spades
Number of ace of spades = 1
So, P(E ∩ F) = 1/52
Now, P(E) . P(F) = 1/4 × 1/13 = 1/52 = P(E ∩ F)
Since, P(E ∩ F) = P(E) . P(F)
Therefore, E and F are independent events
(ii) E : ‘the card drawn is black’, F : ‘the card drawn is a king’
Solution:
E : the card drawn is a black
Number of black cards = 26
Total number of cards = 52
P(E) = 26/52 = 1/2
F : the card drawn is an king
Number of king cards = 4
Total number of cards = 52
P(F) = 4/52 = 1/13
Now, E ∩ F = the card drawn is a black king
Number of black king = 2
So, P(E ∩ F) = 2/52 = 1/26
Now, P(E) . P(F) = 1/2 × 1/13 = 1/26 = P(E ∩ F)
Since, P(E ∩ F) = P(E) . P(F)
Therefore, E and F are independent events
(iii) E : ‘the card drawn is a king or queen’, F : ‘the card drawn is a queen or jack’.
Solution:
E : the card drawn is a king or a queen
Number of king or a queen = 8
Total number of cards = 52
P(E) = 8/52 = 2/13
F : the card drawn is a queen or a jack
Number of queen or a jack = 8
Total number of cards = 52
P(F) = 8/52 = 2/13
Now, E ∩ F =the card drawn is (a king or a queen) and (a queen or a jack)
Number of queen = 4
So, P(E ∩ F) = 4/52 = 1/13
Now, P(E) . P(F) = 2/13 × 2/13 = 4/169 ≠ 1/13
Since, P(E ∩ F) ≠ P(E) . P(F)
Therefore, E and F are 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 newspaper.
(c) If she reads English newspaper, find the probability that she reads Hindi newspaper.
Solution:
Let H: Students reading Hindi newspaper & E: Students reading English newspaper
Given,
P(H) = 60% = 60/100 = 0.6
P(E) = 40% = 40/100 = 0.4
P(H ∩ E) = 20% = 20/100 = 0.2
(i) We need to find the Probability that she reads neither Hindi nor English newspapers. i.e. P(H’ ∩ E’)
Now, P(H’ ∩ E’) = 1 – Probability that she reads both the newspapers.
= 1 – P(H ∪ E)
= 1 − [P(H) + P(E) – P(H ∩ E)]
= 1 − [0.6 + 0.4 – 0.2]
= 1 – 0.8 = 0.2 = 1/5
(ii) We need to find the Probability that she reads English newspaper, given she reads Hindi newspaper i.e. we need to find P(E|H)
Now, P(E|H) = P(E ∩ H) / P(H) = (0. 2)/(0. 6) = 2/6 = 1/3
(iii) We need to find the Probability that she reads Hindi newspaper, given she reads English i.e. P(H|E)
Now, P(H|E) = P(H ∩ E) / P(E) = (0. 2)/(0. 4) = 2/4 = 1/2
Question 17. The Probability of obtaining an even prime number on each die, when a pair of die is rolled is
(A) 0 (B) 1/3 (C) 1/12 (D) 1/36
Solution:
A pair of die is rolled, S = Even Prime number is 2
So, getting an even prime number on each die means Getting (2, 2)
Thus, Probability of obtaining even prime number on each die = Probability of getting (2, 2) = 1/36
So, D is the correct answer
Question 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:
Two events A & B are independent if P(A ∩ B) = P(A) . P(B)
Replacing A with A’ and B with B’
P(A’ ∩ B’) = P(A’) . P(B’) = [1 – P(A)] [1 – P(B)]
Hence, B is the correct answer
Summary
Exercise 13.2 focuses on conditional probability and its applications. It covers the concept of dependent and independent events, the multiplication rule of probability for dependent events, and how to calculate conditional probabilities. Students learn to solve problems involving real-life scenarios where the occurrence of one event affects the probability of another event.
Similar Reads
Class 12 NCERT Solutions- Mathematics Part ii – Chapter 13 – Probability Exercise 13.3 NCERT Solutions for Class 12 Maths, Chapter 13, Exercise 13.3, have been created by GeeksforGeeks to assist students in their academic journey. There are a total of 14 questions in Exercise 13.3 of Chapter 13 in the Class 12 Maths NCERT textbook. In this article, we will demonstrate how to solve all
15+ min read
Class 12 NCERT Solutions- Mathematics Part ii – Chapter 13 – Probability Exercise 13.1 NCERT Solutions for Class 12 Maths Chapter 13 Exercise 13.1 provides solutions for all the questions in the NCERT textbook with simple and easy to understand solution. Our comprehensive solutions will guide you step-by-step through each question in Exercise 13.1, making complex concepts easy to gras
13 min read
Class 12 NCERT Solutions- Mathematics Part ii – Chapter 13 – Probability Miscellaneous Exercise on Chapter 13 Question 1. A and B are two events such that P (A) ≠0. Find P (B|A), if:(i) A is a subset of B ​(ii) A ∩ B = φSolution: It is given that, A and B are two events such that P (A) ≠0 We have, A ∩ B = A P( A ∩ B) = P(B ∩ A) = P(A) Hence , P(B | A)= P(B ∩ A) /P(A) P(A) /P(A) = 1 (ii) We have, P (A ∩ B)
9 min read
NCERT Solutions Class 11 - Chapter 14 Probability - Exercise 14.2 Question 1. Which of the following cannot be a valid assignment of probabilities for outcomes of sample Space S = {ω1, ω2, ω3, ω4, ω5, ω6, ω7}Assignment: ω1ω2ω3ω4ω5ω6ω7(a)0.10.010.050.030.010.20.6(b)1/71/71/71/71/71/71/7(c)0.10.20.30.40.50.60.7(d)-0.10.20.30.4-0.20.10.3(e)1/142/143/144/145/146/1415/
13 min read
Class 10 NCERT Solutions - Chapter 15 Probability - Exercise 15.2 Question 1. Two customers Shyam and Ekta are visiting a particular shop in the same week (Tuesday to Saturday). Each is equally likely to visit the shop on any day as on another day. What is the probability that both will visit the shop on (i) the same day? (ii) consecutive days? (iii) different day
7 min read