CBSE Class 12 Mathematics Probability MCQs Set D

Question. Two numbers are chosen from {1,2,3,4,5,6} one after the other without replacement. The probability that one of the smaller values is less than 4 is
(a) 4/5
(b) 1/15
(c) 1/5
(d) 14/15
Answer: a

Question. The probability of a student getting 1,2,3 division in an examination are 1/10, 3/5 and ¼ respectively. The probability that the student fails in the examination is
(a) 197/200
(b) 27/100
(c) 83/100
(d) None of the options
Answer: b

Question. Let A and B are two events. If P(A)=0.2 p(B)=0.4, P(AᴜB)=0.6, then P(A/B) is equal to …
(a) 0.8
(b) 0.5
(c) 0.3
(d) 0
Answer: d

Question. A speaks truth in 75% cases and B speaks truth in 80% cases. The probability that they contradict each other in a statement is
(a) 7/20
(b) 13/20
(c) 3/5
(d) 2/5
Answer: a

Question. Let A and B be two events such that P(A)=0.6, P(B)=0.2 and P (A/B) =0.5, then P(A’/B’) equals…
(a) 1/10
(b) 3/10
(c) 3/8
(d) 6/7
Answer: c

Question. The probability that a leap year will have 53 Fridays or 53 Saturdays
(a) 2/7
(b) 3/7
(c) 4/7
(d) 1/7
Answer: b

Question. If A and B are independent events the P(A∩B) = ……
Answer: P(A).P(B)

Question. If a fair die is rolling. The events are E={1,3,6}, F={4,6}. Then the probability P(E/F) is…
(a) 1/6
(b) 1/3
(c) 1/2
(d) 2/3
Answer: c

Question. A person writes 4 letters and addresses 4 envelopes. If the letters are placed in the envelopes at random, then the probability that all letters are not placed in the right envelopes, is
(a) 1/4
(b) 11/24
(c) 15/24
(d) 23/24
Answer: d

Very Short Answer Type

Question. The probabilities of A and B solving a problem independently are 1/3 and 1/4 respectively. If both of them try to solve the problem independently,what is the probability that the problem is solved?
Answer: 1- 2/3 x 3/4 = 1/2

Question. The probability that it will rain on any particular day is 50%. Find the probability that it rains only on first 4 days of the week.
Answer: (1/2)⁴(1/2)³ = (1/2)⁷

Question. Given two independent events A and B such that P(A¢) = 0.3 and P(B¢) = 0.6, find P(A¢Ç B¢).
Answer: P(A∩Ç B’) = P(A’) P(B’) 
= (0.7)(0.4) = 0.28

Question. Three distinct numbers are chosen randomly from the first 50 natural numbers. Find the probability that all the three numbers are divisible by both 2 and 3.
Answer :Since there are only 8 numbers (in first 50 natural numbers) which are divisible by 6.
∴ favourable number of outcomes are ⁸C₃.

 Question. If A and B are two independent events, prove that A’ and B are also independent.
Answer: P(A’ ∩ B) = P(B) – P(A ∩ B)
= P(B) – P(A) . P(B)
(∵ A and B are independent events)
= (1 – P(A)) P(B)
= P(A’) P(B) 1
Since, P(A’ ∩ B) = P(A’) P(B)
Therefore A’ and B are independent events.

 Question. If A and B are two independent events, then prove that the probability of occurrence of at least one of A and B is given by 1 – P(A’) · P(B’).
Answer: Required probability = P(A È B)
= P(A) + P(B) – P(A) · P(B)
= P(A) [1 – P(B)] + 1 – P(B’) 
= P(A) P(B’) – P(B’) + 1
= 1 – P(B’) [1 – P(A’)]
= 1 – P(A’) P(B’)

 Question. A couple has 2 children. Find the probability that both are boys, if it is known that (i) one of them is a boy (ii) the older child is a boy.
Answer: Sample space = {B1B2, B1G2, G1B2, G1G2},
B1 and G1 are the older boy and girl respectively.
Let E1 = both the children are boys;
E2 = one of the children is a boy;
E3 = the older child is a boy 

Question. The random variable X has a probability distribution P(X) of the following form, where ‘k’ is some number.
Answer: k + 2k + 3k = 1
⇒ k = 1/6

Question. A family has two children. What is the probability that both the children are boys given that at least one of them is a boy ?
Answer: Let b stand for boy and g for girl. The sample space of the experiment is
S = {(b, b), (g, b), (b, g), (g, g)}
Let E and F denote the following events :
E : ‘both the children are boys’
F : ‘at least one of the child is a boy’
Then E = {(b,b)} and F = {(b,b), (g,b), (b,g)}
Now E ∩ F = {(b,b)}
Thus P(F) = 3/ 4 and P (E ∩ F )= 1/ 4
Therefore P(E|F) = P(E∩F)/ P(F)=1/3

Question. A die is thrown twice and the sum of the numbers rising is noted to be 6. Calculate the is the conditional probability that the number 4 has arrived at least once?
Answer: If a dice is thrown twice, then the sample space obtained is:
S = {(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)
(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)
(3,1)(3,2)(3,3)(3,4)(3,5)(3,6)
(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)
(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)
(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)}
From the given data, it is needed to find the Probability that 4 has appeared at least once, given the sum of nos. is observed to be 6
Assume that, F: Addition of numbers is 6 and take E: 4 has appeared at least once
So, that, we need to find P(E|F)
Finding P (E):
The probability of getting 4 atleast once is:
E = {(1, 4), (2, 4), (3, 4), (4, 4), (5, 4), (6, 4), (4, 1), (4, 2), (4, 3), (4, 5), (4, 6)}
Thus , P(E) = 11/ 36
Finding P (F):
The probability to get the addition of numbers is 6 is:
F = {(1, 5), (5, 1), (2, 4), (4, 2), (3, 3)}
Thus, P(F) = 5/ 36
Also, E ∩ F = {(2,4), (4,2)}
P(E ∩ F) = 2/36
Thus, P(E|F) = (P(E ∩ F) ) / (P (F) )
Now, substitute the probability values obtained= (2/36)/ (5/36)
Hence, Required probability is 2/5.

Question. An fair die is thrown double times. Assume that the event A is “odd number on the first throw” and B the event “odd number on the second throw”. Compare the independence of the events A and B.
Answer: Let us consider two independent events A and B, then P(A ∩ B) = P(A). P(B)
when an unbiased die is thrown twice
S = {(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)
(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)
(3,1)(3,2)(3,3)(3,4)(3,5)(3,6)
(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)
(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)
(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)}
Let us describe two events as
A: odd number on the first throw
B: odd number on the second throw
To find P(A)
A = {(1, 1), (1, 2), (1, 3), …, (1, 6)
(3, 1), (3, 2), (3, 3), …, (3, 6)
(5, 1), (5, 2), (5, 3), …, (5, 6)}
Thus, P (A) = 18/36 = 1/2
To find P(B)
B = {(1, 1), (2, 1), (3, 1), …, (6, 1)
(1, 3), (2, 3), (3, 3), …, (6, 3)
(1, 5), (2, 5), (3, 5), …, (6, 5)}
Thus, P (B) = 18/36 = 1/2
A ∩ B = odd number on the first & second throw = { (1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5,
3), (5, 5)}
So, P(A ∩ B) = 9/36 = 1/ 4
Now, P(A). P(B) = (1/2) × (1/2) = 1/4
As P(A ∩ B) = P(A). P(B),
Hence, the two events A and B are independent events.

Question. 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
Answer: Given, P(A) = 1/2 ,
P (A ∪ B) = 3/5
and P(B) = p.
(1) For Mutually Exclusive
Given that, the sets A and B are mutually exclusive.
Thus, they do not have any common elements
Therefore, P(A ∩ B) = 0
We know that P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
Substitute the formulas in the above-given formula, we get
3/5 = (1/2) + p – 0
Simplify the expression, we get
(3/5) – (1/2) = p
(6 − 5)/10 = p
1/10 = p
Therefore, p = 1/10
Hence, the value of p is 1/10, if they are mutually exclusive.
(ii) For Independent events:
If the two events A & B are independent,
we can write it as P(A ∩ B) = P(A) P(B)
Substitute the values,
= (1/2) × p
= p/2
Now, P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
Now, substitute the values in the formula,
(3/5) = (1/2)+ p – (p/2)
(3/2)– (1/2)= p – (p/2)
(6 − 5)/10 = p/2
1/10 = p/2
p= 2/10
P = 1/5
Thus, the value of p is 1/5, if they are independent.

Question. The probability of solving the specific problem independently by the persons’ A and B are 1/2 and 1/3 respectively. In case, if both the persons try to solve the problem independently, then calculate the probability that the problem is solved.
Answer: Given that, the two events say A and B are independent if P(A ∩ B) = P(A). P(B)
From the given data, we can observe that P(A) = 1/2 & P(B) = 1/3
The probability that the problem is solved = Probability that person A solves the problem or the person B solves the Problem
This can be written as:
= P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
If A and B are independent, then P(A ∩ B) = P(A). P(B)
Now, substitute the values,
= (1/2) × (1/3)
P(A ∩ B) = 1/6
Now, the probability of problem solved is written as
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
Hence, the probability of the problem solved is 2/3.