Probability Solved Questions and Answers (original) (raw)
Last Updated : 21 Apr, 2026
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 head 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 3)
= 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 = 32Probability of getting two even numbers = (32/63) * (31/62) = **16/63
Probability of getting **two odd numbers = (31/63) * (30/62) = **5/21P(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 = 15/30Probability of choosing A grade student= 9/30
Now, an A-grade student chosen can be a girl.
So the probability of choosing it = 4/30Required 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) = TTProbability 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) = THThus, 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.
- The student likes mutton.
- He likes either chicken or mutton
- 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
- The number is a two-digit number
- The number is a perfect square
- 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