Probability - Solved Questions and Answers

Last Updated : 02 Aug, 2025

Probability measures how likely an event is to occur, expressed as a value between 0 and 1.

Formula:

P(E)=\frac{Number of favorable outcomes}{Total possible outcome}

Probability questions and answers are provided below for you to learn and practice.

Question 1: Three unbiased coins are tossed. What is the probability that at most one had occurred?

Solution:

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
Favorable outcomes = {HTT, THT, TTH, TTT}
Total number of outcomes = 8
Number of favorable outcomes = 4
Required probability = 4 / 8 = 0.50

Question 2: Find the probability of getting a red card when a card is drawn from a well-shuffled pack of cards.

Solution:

Total number of outcomes = 52
Number of favorable outcomes = Number of red cards = 26
Required probability = 26 / 52 = 0.50

Question 3: A bag contains 6 white and 4 black balls. Two balls are drawn at random from the bag. Find the probability that both the balls are of the same color.

Solution:

Outcome will be favorable if the two balls drawn are of the same color. => Number of favorable outcomes = ⁶C₂ + ⁴C₂ = 21
Total number of outcomes = ¹⁰C₂ = 45
Therefore, required probability = 21 / 45 = 7 / 15

Question 4: An unbiased die is tossed. Find the probability of getting an even number.

Solution:

S = {1, 2, 3, 4, 5, 6}
Favorable outcomes = {2, 4, 6}
Required probability = 3 / 6 = 0.50

Question 5: From a bag containing red and blue balls, 10 each, 2 balls are drawn at random. Find the probability that one of them is red and the other is blue.

Solution:

Total number of outcomes = ²⁰C₂ = 190
Number of favorable outcomes = ¹⁰C₁ x ¹⁰C₁ = 100
Therefore, required probability = 100 / 190 = 10 / 19

Question 6: If a coin is thrown two times, what is the probability that at least one tail is obtained?

A) 3/4
B) 1/4
C) 1/3
D) 2/3
E) None of these

Answer: A

Solution:

Sample space = [TT, TH, HT, HH]
Total number of ways = 2 × 2 = 4.
Favourite Cases = 3
P (A) = 3/4

Question 7: There are 7 purple clips and 5 brown clips. Two clips are selected one by one without replacement. Find the probability that the first is brown and the second is purple.

A) 1/35
B) 35/132
C) 1/132
D) 35/144
E) None of these

Answer: B

Solution:

P (B) × P (P) = (5/12) x (7/11) = 35/132

Question 8: Find the probability of getting a sum of 8 when two dice are thrown.

A) 1/8
B) 1/5
C) 1/4
D) 5/36
E) 1/3

Answer: D

Solution:

Total number of ways = 6 × 6 = 36 ways.
Favorable cases = (2, 6) (6, 2) (3, 5) (5, 3) (4, 4) --- 5 ways.
P (A) = 5/36 = 5/36

Question 9: If two dice are rolled together then find the probability of getting at least one '3'.

A) 11/36
B) 1/12
C) 1/36
D) 13/25
E) 13/36

Answer: A

Solution:

Total number of ways = 6 × 6 = 36.
Probability of getting number ‘3' at least one time
= 1 – (Probability of getting no number 4)
= 1 – (5/6) × (5/6)
= 1 - 25/36
= 11/36

Question 10: A container contains 1 red, 3 black, 2 pink, and 4 violet gems. If a single gem is chosen at random from the container, then find the probability that it is violet or black.

A) 1/10
B) 3/10
C) 7/10
D) 9/10
E) None of these

Answer: C

Solution:

Total gems =( 1 + 3 + 2 + 4 ) = 10
probability of getting a violet gem = 4/10
The probability of getting a black gem = 3/10
Now, P ( Violet or Black) = P(violet) + P(Black)
= 4/10 + 3/10
= 7/10

Question 11: A jar contains 63 balls ( 1, 2, 3,......., 63). Two balls are picked at random from the jar one after one and without any replacement. What is the probability that the sum of both balls drawn is even?

A) 5/21
B) 3/23
C) 5/63
D) 19/63
E) None of these

Answer: E

Solution:

The sum of numbers can be even if we add either two even numbers or two odd numbers.
Number of even numbers from 1 to 63 = 31
Number of odd numbers from 1 to 63 = 32
Probability of getting two even numbers = (32/63) * (31/62) = 16/63
Probability of getting two odd numbers = (31/63) * (30/62) = 5/21
P(two even numbers OR two odd numbers) = 16/63 + 5/21 = 31/63

Question 12: There are 30 students in a class, 15 are boys and 15 are girls. In the final exam, 5 boys and 4 girls made an A grade. If a student is chosen at random from the class, what is the probability of choosing a girl or an 'A-grade student?

A) 1/4
B) 3/10
C) 1/3
D) 2/3
E) None of these

Answer: D

Solution:

Here, the total number of boys = 15 and the total number of girls = 15
Also, girls getting A grade = 4 and boys getting an A grade = 5
Probability of choosing a girl = 15/30
Probability of choosing A grade student= 9/30
Now, an A-grade student chosen can be a girl.
So the probability of choosing it = 4/30
Required probability of choosing a girl or an A-grade student
= 15/30 + 9/30 - 4/30
= 1/2 + 3/10 - 2/15
= 2/3

Question 13: If P(A) = 7/13, P(B) = 9/13 and P(A∩B) = 4/13, find the value of P(A|B).

A) 1/9
B) 2/9
C) 3/9
D) 4/9
E) None of these

Answer: D

Solution:

P(A|B) = P(A∩B)/P(B) = (4/13)/(9/13) = 4/9.

Question 14: A one-rupee coin and a two-rupee coin are tossed once, and then calculate a sample space.

A) [ HH, HT, TH, TT]
B) [ HH, TT]
C) [ TH, HT]
D) [HH, TH, TT]
E) None of these

Answer: A

Solution:

The outcomes are either Head (H) or tail(T).
Now,heads on both coins = (H, H) = HH
Tails on both coins = ( T, T) = TT
Probability of head on one rupee coin and Tail on the two rupee coins = (H, T) = HT
And Tail on one rupee coin and Head on the two rupee coin = (T, H) = TH
Thus, the sample space,S = [HH, HT, TH, TT]

Question 15: There are 20 tickets numbered 1 to 20. These tickets are mixed up and then a ticket is drawn at random. Find the probability that the ticket drawn has a number that is a multiple of 4 or 5.

A) 1/4
B) 2/13
C) 8/15
D) 9/20
E) None of these

Answer: E

Solution:

Here, S = {1, 2, 3, 4, ...., 19, 20} = 20
Multiples of 4: 4, 8, 12, 16, 20 (5 tickets)
Multiples of 5: 5, 10, 15, 20 (4 tickets)
Notice that ticket number 20 is a multiple of both 4 and 5, so we have counted it twice. Therefore, we need to subtract one from the total count.
Total number of tickets with numbers that are multiples of 4 or 5: 5 + 4 - 1 = 8
The total number of tickets is 20, so the probability of drawing a ticket with a number that is a multiple of 4 or 5 is:
P = 8/20 = 2/5 = 0.4
Therefore, the probability that the ticket drawn has a number that is a multiple of 4 or 5 is 0.4 or 40%.

Direction ( 16 - 18):
In a school the total number of students is 300, 95 students like chicken only, 120 students like fish only, 80 students like mutton only and 5 students do not like anything above. If randomly one student is chosen, find the probability that.

23) The student likes mutton.
24) He likes either chicken or mutton
25) He likes neither fish nor mutton.

Solution ( 16 - 18):

The total number of favorable outcomes = 300 (Since there are 300 students altogether).
The number of times a chicken liker is chosen = 95 (Since 95 students like chicken).
The number of times a fish liker is chosen = 120.
The number of times a mutton liker is chosen = 80.
The number of times a student is chosen who likes none of these = 5.

Question 19: Find the probability that the student likes mutton.

A) 3/10
B) 4/15
C) 1/10
D) 1/15
E) None of these

Answer: B

Solution:

Therefore, the probability of getting a student who likes mutton
= 80/300
= 4/15

Question 20: What is the probability that the student likes either chicken or mutton?

A) 7/12
B) 5/12
C) 3/4
D) 1/12
E) None of these

Answer: A

Solution:

The probability of getting a student who likes either chicken or mutton
= (95+80)/300
= 175/300
= 7/12

Question 21: Find the probability that the student likes neither fish nor mutton.

A) 1/2
B) 1/5
C) 1/3
D) 1/4
E) 1/6

Answer: C

Solution:

The probability of getting a student who likes neither fish nor mutton
= (300–120−80)/300
= 100/300
= 1/3

Direction ( 22-24):
A box contains 90 number plates numbered 1 to 90. If one number plate is drawn at random from the box then find out the probability that

26) The number is a two-digit number
27) The number is a perfect square
28) The number is a multiply of 5

Question 22: Find the probability that the number is a two-digit number.

A) 1/9
B) 1/10
C) 9/10
D) 7/10
E) None of these

Answer: C

Solution :

Total possible outcomes = 90 (Since the number plates are numbered from 1 to 90).
Number of favorable outcomes
= 90 - 9 = 81 ( here, except 1 to 9, other numbers are two-digit number.)
Thus required probability
= Number of Favourable Outcomes /Total Number of Possible Outcomes
= 81/90
= 9/10.

Question 23: What is the probability that the number is a perfect square?

A) 1/9
B) 1/10
C) 9/10
D) 1/7
E) None of these

Answer: B

Solution:

Total possible outcomes = 90.
Number of favorable outcomes = 9 [here 1, 4, 9, 16, 25, 36, 49, 64, and 81 are the perfect squares]
Thus the required probability = 9/90 =1/10

Question 24: Find the probability that the number is a multiple of 5.

A) 1/5
B) 1/6
C) 1/10
D) 1/8
E) 9/10

Answer: A

Solution:

Total possible outcomes = 90.
Number of favourable outcomes = 18 (here, 5 × 1, 5 × 2, 5 × 3, ...., 5 × 18 are multiple of 5).
Thus, the required probability= 18/90 =1/5