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Practice Problems on Probability (Medium)

Last Updated : 10 Jun, 2025
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Probability is an important chapter for the students of Class 9, 10, 11, and 12. The Probability Questions, with their answers included in this article, will help you understand the basic concepts and formulas through the number of solved and unsolved questions. These questions cover concepts like Sample Space, Events, Coin Probability, etc. Solving these problems will improve your understanding and problem-solving skills in probability.

Check Tricks To Solve Probability Questions

Important Formulas on Probability

Probability Rules and Formulas

Probability of an Event

P(E) = Number of favorable outcomes​ / Total number of outcomes

Addition Rule

P(A ⋃ B) = P(A) + P(B) - P(A ⋂ B)

P(A ⋃ B) = P(A) + P(B) If A and B are mutually exclusive (disjoint).

Multiplication Rule

P(A ⋂ B) = P(A) · P(B | A)

P(A ⋂ B) = P(A) · P(B) If A and B are independent.

Conditional Probability

P(B | A) = P(A ⋂ B) / P(A)

Bayes' Theorem

P(A | B) = [P(B | A) · P(A)] / P(B)

Total Probability Rule

P(A) = P(A ⋂ B1) + P(A ⋂ B2) + ... + P(A ⋂ Bn) Where, B1, B2, ..., Bn are mutually exclusive and exhaustive events,

Solved Questions on Probability (Medium)

Question 1: What is the probability of flipping a coin three times and getting at least one head?

Solution:

Probability of getting no heads (all tails) = (1/2)³ = 1/8.
Probability of getting at least one head = 1 - Probability of getting no heads.
= 1 - 1/8 = 7/8 = 0.875 or 87.5%.

Question 2: A bag contains 5 red, 7 blue, and 8 green balls. Two balls are drawn without replacement. What is the probability that both balls are red?

Solution:

Probability of first ball being red = 5/20.
After drawing one red ball, there are 4 red balls left and 19 balls in total.
Probability of second ball being red = 4/19.
Total probability = (5/20) × (4/19) = 20/380 = 1/19 ≈ 0.0526 or 5.26%.

Question 3: Sumit is playing cricket with his friends, to decide who is going to bat is decided by tossing a coin, whichever wins the toss will bat first. Assuming Sumit and Mohit are captains of the two teams which have chosen heads and tails respectively. Find the chances of Sumit to bat first.

Solution:

We know, there are only two possible outcomes of the toss i.e. heads or tails.

Therefore, Sample space(s) = Total possible outcomes = {H, T}

 Sumit needs heads to win the toss, therefore there is only one favorable outcome.

 Probability of Sumit to win the toss = favorable outcome / total outcome 

= 1/2 = 0.5

Question 4: Two dice are tossed. Find the probability that the total score is a prime number.that

Solution:

Since, two dices are tossed therefore total no of combination = n(S) = (6 x 6) = 36 combinations.

Let us considered E be the event that the sum is a prime number.

All the favorable outcomes are (E) = {(1, 1), (1, 2), (1, 4), (1, 6), (2, 1),
                                                               (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), 
                                                              (4, 3), (5, 2), (5, 6), (6, 1), (6, 5)}

Therefore, n(E) = 15

Probability of score to be prime number = n(E)/n(S) = 15/36 = 5/12.

Question 5: In a lottery box, there are 10 prizes and 25 blanks. A slip is drawn at random from the lottery box. What is the probability of getting a prize?

Solution:

Given: Total number of prize = 10

Total number of blanks = 25

So, the total number of possible outcomes(i.e., n(S)) are = 10 + 25 = 35

Total number of prizes, n(E) = 10

According to the formula

P(E) = n(E)/n(S) = 10/35 = 2/7

Question 6: In a lottery, there are 100 tickets numbered from 1 to 100. What is the probability that a ticket drawn at random has a number that is a multiple of 5 or 7?

Solution:

Number of multiples of 5 between 1 and 100 = 20.
Number of multiples of 7 between 1 and 100 = 14.
Number of multiples of both 5 and 7 (35 and 70) = 2.
Total favorable outcomes = 20 + 14 - 2 = 32.
Probability = Number of favorable outcomes / Total tickets.
= 32 / 100 = 0.32 or 32%.

Question 7: In a class of 60 students, 30 are boys. If two students are selected at random, what is the probability that both are girls?

Solution:

Total girls = 60 - 30 = 30.
Probability of first student being a girl = 30/60.
After selecting one girl, remaining girls = 29, total students = 59.
Probability of second student being a girl = 29/59.
Total probability = (30/60) × (29/59).

Question 8: A fair coin is tossed 5 times. What is the probability of getting exactly 3 heads?

Solution:

The number of ways to get exactly 3 heads in 5 tosses can be calculated using the binomial formula: C(n, k) = n! / [k!(n-k)!], where C is the combination, n is the total number of events, and k is the number of successful events.
C(5, 3) = 5! / [3!(5-3)!] = 10.
Probability of getting a head in one toss = 1/2.
Probability = 10 × (1/2)3 × (1/2)2 = 10/32 = 0.3125 or 31.25%.

Question 9: A deck of 52 cards is shuffled. What is the probability of drawing either a heart or a queen, but not the queen of hearts?

Solution:

Total hearts = 13, including the queen of hearts.
Total queens = 4, including the queen of hearts.
Favorable outcomes = (13 hearts - 1 queen of hearts) + (4 queens - 1 queen of hearts) = 12 + 3 = 15.
Probability = Number of favorable outcomes / Total outcomes = 15 / 52 ≈ 0.2885 or 28.85%.

Question 10: In a lottery, there are 100 tickets numbered 1 to 100. What is the probability of drawing a number that is both a multiple of 5 and a multiple of 7?

Solution:

Multiples of 5 and 7 between 1 and 100 are 35 and 70.
Total favorable outcomes = 2.
Probability = Number of favorable outcomes / Total tickets = 2 / 100 = 0.02 or 2%.

Question 11: A bag contains 4 white, 5 red, and 6 blue balls. Three balls are drawn at random from the bag. The probability that all of them are red, is:

Solution:

Let S be the sample space.

Then, n(S) = Number of ways of drawing 3 balls out of 15

n(S) = 15C3 = 455

Let E = event of getting all the 3 red balls. 

n(E) = 5C3 = 10

P(E) = n(E)/n(S) = 10/455 = 2/91.

Question 12: In a class, there are 10 girls and 15 boys, what is the probability that 1 girl and 2 boys are selected?

Solution:

Let S be the sample space.

Then, n(S) = Number of ways of selecting 3 children out of 25

n(S) = 25C3

n(S) = 2300. 

Let  E= event of selecting 1 girl and 2 boys. 

n(E) = 10C1*15C2 = 1050

P(E) = n(E)/n(S) = 1050/2300 = 21/46. 

Practice More:


Easy-Level Probability Questions


Hard-Level Probability Questions

Binomial Distribution Practice

Conditional Probability Practice

Check - Probability Quiz

Practice Problems on Probability (Medium)

Question 1: What is the probability of flipping a coin four times and getting at least one tail?

Question 2: A bag contains 6 red, 5 blue, and 9 green balls. Two balls are drawn without replacement. What is the probability that both balls are blue?

Question 3: Two players are playing rock-paper-scissors. What is the probability that the first player wins by choosing rock if the second player chooses randomly?

Question 4: Two dice are rolled. Find the probability that the total score is an even number.

Question 5: A box contains 8 prizes and 12 blanks. A slip is drawn at random from the box. What is the probability of getting a blank?

Question 6: In a lottery, there are 200 tickets numbered from 1 to 200. What is the probability that a ticket drawn at random has a number that is a multiple of 4 or 6?

Question 7: In a class of 50 students, 20 are boys. If two students are selected at random, what is the probability that both are boys?

Question 8: A fair coin is tossed 6 times. What is the probability of getting exactly 4 heads?

Question 9: A deck of 52 cards is shuffled. What is the probability of drawing either a spade or a king, but not the king of spades?

Question 10: In a raffle, there are 150 tickets numbered from 1 to 150. What is the probability of drawing a number that is both a multiple of 3 and a multiple of 5?

Answer key

  1. 15/16 ≈ 0.9375 or 93.75%.
  2. 1/19 ≈ 0.0526 or 5.26%.
  3. 1/3 ≈ 33.33% or 33.33%.
  4. 1/2 = 0.5 or 50%.
  5. 3/5 = 0.63/5 or 60%.
  6. 67/200 = 0.335 or 33.5%.
  7. 380/2450 ≈ 0.1551 or 15.51%.
  8. 15/64 ≈ 0.2344 or 23.44%.
  9. 15/52 ≈ 0.2885 or 28.85%.
  10. 1/15 ≈ 0.0667 or 6.67%.

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