25 Probability Questions to Test Your Understanding (With Answers)
Hey there! If you’re looking to brush up on your probability skills, you’ve come to the right place. Whether you’re studying for an exam, prepping for an interview, or just love a good math puzzle, working through problems is the best way to learn. I remember when I first tackled probability—it felt like learning a new language. But with practice, those concepts like “favourable outcomes” and “sample spaces” start to click.
Below, I’ve put together 25 common probability questions, complete with detailed solutions. I’ve structured them to build from foundational ideas to slightly more complex ones, just like I would if I were tutoring you one-on-one. Grab a pen and paper, give them a try, and then check your reasoning against the explanations. Let’s dive in.
Practice Questions & Step-by-Step Solutions
Q1. The Classic Dice Roll
In a single throw of a fair die, what is the probability of getting a number greater than 4?
- (a) 1/6
- (b) 1/3
- (c) 1/2
- (d) 2/3
Answer: (b) 1/3
Explanation: A fair die has outcomes {1, 2, 3, 4, 5, 6}. The numbers greater than 4 are just 5 and 6. That’s 2 favourable outcomes out of 6 total possibilities. So, the probability is 2/6, which simplifies to 1/3.
Q2. Not That Colour!
A bag contains 5 red, 3 green and 2 blue balls. If one ball is drawn at random, what is the probability that it is not green?
- (a) 3/10
- (b) 7/10
- (c) 2/5
- (d) 1/2
Answer: (b) 7/10
Explanation: First, find the total number of balls: 5 + 3 + 2 = 10. The balls that are NOT green are the red and blue ones: 5 red + 2 blue = 7 balls. Therefore, the probability is 7/10.
Q3. Flipping Two Coins
Two coins are tossed simultaneously. What is the probability of getting exactly one head?
- (a) 1/4
- (b) 1/2
- (c) 3/4
- (d) 1
Answer: (b) 1/2
Explanation: List all possible outcomes when tossing two coins: HH, HT, TH, TT. These are the 4 outcomes in the sample space. The outcomes with exactly one head are HT and TH. That’s 2 favourable outcomes. Probability = 2/4 = 1/2.
Q4. Royalty from a Deck
From a deck of 52 cards, one card is drawn at random. The probability that the card is a king or a queen is:
- (a) 1/13
- (b) 2/13
- (c) 4/13
- (d) 8/52
Answer: (b) 2/13
Explanation: A standard deck has 4 kings and 4 queens, making 8 cards that are either a king or a queen. The total number of cards is 52. So, the probability is 8/52, which simplifies to 2/13.
Q5. The Complement Rule
If the probability of an event occurring is 0.35, what is the probability that it does not occur?
- (a) 0.35
- (b) 0.65
- (c) 0.5
- (d) 0.25
Answer: (b) 0.65
Explanation: This is a fundamental rule: the probability of an event not happening is 1 minus the probability that it does happen. So, P(not A) = 1 – P(A) = 1 – 0.35 = 0.65.
Q6. Playing Sports (Inclusion-Exclusion)
In a class of 30 students, 18 play cricket, 12 play football and 5 play both. What is the probability that a randomly selected student plays either cricket or football?
- (a) 0.5
- (b) 0.6
- (c) 0.833
- (d) 0.9
Answer: (c) 5/6 or ≈0.833
Explanation: We use the addition rule here to avoid double-counting the students who play both. P(Cricket or Football) = P(C) + P(F) – P(Both). That’s (18/30) + (12/30) – (5/30) = 25/30, which simplifies to 5/6 (about 0.833).
Q7. Drawing Without Replacement
A box contains 4 defective and 6 non‑defective bulbs. Two bulbs are drawn at random without replacement. What is the probability that both are defective?
- (a) 2/15
- (b) 1/15
- (c) 4/25
- (d) 6/25
Answer: (a) 2/15
Explanation: “Without replacement” means the second draw depends on the first. Probability the first is defective = 4/10. If that happens, there are 3 defective bulbs left out of 9 total. So, probability the second is defective = 3/9 = 1/3. Multiply these probabilities: (4/10) * (1/3) = 4/30 = 2/15.
Q8. Rolling for a Sum
A die is rolled twice. What is the probability that the sum of the numbers on the two faces is 7?
- (a) 1/6
- (b) 1/12
- (c) 1/36
- (d) 5/36
Answer: (a) 1/6
Explanation: Total possible outcomes when rolling a die twice: 6 * 6 = 36. The pairs that sum to 7 are: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). That’s 6 favourable pairs. Probability = 6/36 = 1/6.
Q9. Lottery Tickets
In a lottery, there are 10 tickets and only 1 winning ticket. If a person buys 2 tickets, what is the probability of winning at least one prize?
- (a) 1/5
- (b) 3/10
- (c) 19/45
- (d) 2/9
Answer: (a) 1/5
Explanation: Sometimes it’s easier to find the probability of the opposite event. The probability of winning nothing (i.e., both tickets are losers) is: (9/10) * (8/9) = 72/90 = 4/5. Therefore, the probability of winning at least once is 1 – 4/5 = 1/5.
Q10. Drawing With Replacement
A bag contains 7 white and 5 black balls. Two balls are drawn at random with replacement. What is the probability that both are black?
- (a) 25/144
- (b) 5/12
- (c) 35/144
- (d) 1/4
Answer: (a) 25/144
Explanation: “With replacement” means the bag is restored after the first draw, so each draw is independent. The probability of drawing a black ball on any single draw is 5/12. For two independent draws, we multiply: (5/12) * (5/12) = 25/144.
Q11. Understanding Odds
The odds in favour of an event are 3:2. What is the probability of the event?
- (a) 3/5
- (b) 2/5
- (c) 3/2
- (d) 5/3
Answer: (a) 3/5
Explanation: Odds in favour of 3:2 means for every 3 favourable outcomes, there are 2 unfavourable ones. So, total possible outcomes in this context are 3+2=5. The probability is simply favourable divided by total: 3/5.
Q12. Union of Events
If P(A)=0.4, P(B)=0.5 and P(A∩B)=0.2, then P(A∪B) is:
- (a) 0.7
- (b) 0.9
- (c) 0.3
- (d) 0.1
Answer: (a) 0.7
Explanation: This directly applies the general addition rule: P(A or B) = P(A) + P(B) – P(A and B). So, 0.4 + 0.5 – 0.2 = 0.7.
Q13. Three Coin Tosses
A fair coin is tossed three times. What is the probability of getting at least two heads?
- (a) 1/2
- (b) 3/8
- (c) 1/4
- (d) 1/8
Answer: (a) 1/2
Explanation: List all 8 possible sequences (HHH, HHT, HTH, HTT, THH, THT, TTH, TTT). “At least two heads” means 2 or 3 heads. The favourable outcomes are: HHH, HHT, HTH, THH. That’s 4 outcomes. Probability = 4/8 = 1/2.
Q14. Forming a Committee
From a group of 8 men and 6 women, a committee of 3 is to be formed. What is the probability that the committee consists of exactly 2 women?
- (a) 15/91
- (b) 30/91
- (c) 45/91
- (d) 60/91
Answer: (b) 30/91
Explanation: This is a combinations problem. Total ways to choose any 3 people from 14: C(14,3) = 364. Favourable ways: Choose 2 women from 6 [C(6,2)=15] AND choose 1 man from 8 [C(8,1)=8]. Multiply: 15 * 8 = 120. Probability = 120/364 = 30/91 after simplifying.
Q15. Independent Days
The probability that it will rain on a given day is 0.3. What is the probability that it will not rain on two consecutive days?
- (a) 0.09
- (b) 0.21
- (c) 0.49
- (d) 0.7
Answer: (c) 0.49
Explanation: Assuming weather on different days is independent. P(no rain on one day) = 1 – 0.3 = 0.7. For two independent days, we multiply the probabilities: 0.7 * 0.7 = 0.49.
Q16. Rolling a Double
In a single throw of two dice, what is the probability of getting a doublet (same number on both dice)?
- (a) 1/6
- (b) 1/12
- (c) 1/36
- (d) 1/3
Answer: (a) 1/6
Explanation: The doublets are (1,1), (2,2), (3,3), (4,4), (5,5), (6,6). That’s 6 favourable outcomes. With two dice, there are 36 total outcomes. Probability = 6/36 = 1/6.
Q17. One of Each Colour
A box contains 3 red, 4 blue and 5 green balls. If two balls are drawn at random without replacement, what is the probability that one is red and the other is blue?
- (a) 10/55
- (b) 6/55
- (c) 24/55
- (d) 8/55
Answer: (a) 2/11 or 10/55
Explanation: Total ways to pick 2 balls from 12: C(12,2) = 66. Favourable ways: pick 1 red (from 3) AND 1 blue (from 4). Number of ways = C(3,1) * C(4,1) = 3 * 4 = 12. Probability = 12/66 = 2/11 (which is equivalent to 10/55).
