Probability questions with answers and concepts

Learn probability with concept notes and worked examples, then practise focused questions with explained answers. Build accuracy for placement and competitive aptitude tests before trying timed mode.

130 practice questionsPage updated 11 Nov 2026Review metadata pendingPractice questions

Practise this concept

Probability is a branch of mathematics that deals with the likelihood of events occurring. It quantifies uncertainty by assigning a numerical value between 0 and 1 to each possible outcome, where 0 means impossible and 1 means certain. Probability is essential in competitive exams for solving problems involving cards, dice, coins, and real-world scenarios.

Key formulas and rules

9 rules

Classical definition: $P (E) = \frac{n ( E )}{n ( S )} = \frac{Number of favourable outcomes}{Total number of possible outcomes}$

Complementary rule: $P (E) + P (E^{'}) = 1 ⟹ P (E) = 1 - P (not E)$

Addition Rule (general): $P (A \cup B) = P (A) + P (B) - P (A \cap B)$

Addition Rule (mutually exclusive): $P (A \cup B) = P (A) + P (B)$ (since $P (A \cap B) = 0$ )

Conditional Probability: $P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}$ (where $P (B) > 0$ )

Multiplication Rule: $P (A \cap B) = P (A) \times P (B ∣ A) = P (B) \times P (A ∣ B)$

Independent Events: $P (A \cap B) = P (A) \times P (B) ⟺ P (A ∣ B) = P (A)$

Odds in favour of $E$ : $n (E) : n (E^{'}) = P (E) : [1 - P (E)]$

Odds against $E$ : $n (E^{'}) : n (E) = [1 - P (E)] : P (E)$

Key concepts

Classical (Theoretical) Probability

This is the basic definition where probability is calculated as the ratio of favourable outcomes to total possible outcomes, assuming all outcomes are equally likely. For example, P(getting a head) = 1/2 when tossing a fair coin. This approach works well for simple, well-defined experiments like coin tosses, dice rolls, and drawing cards from a well-shuffled deck.

Complementary Events

The complement of an event E (denoted as E' or not E) consists of all outcomes where E does not occur. Since either E happens or it doesn't, P(E) + P(E') = 1. This is useful when calculating P(E) directly is difficult but calculating P(E') is easier. For example, P(at least one head in 3 tosses) = 1 - P(no heads) = 1 - 1/8 = 7/8.

Mutually Exclusive Events

Two events are mutually exclusive (or disjoint) if they cannot occur simultaneously. For example, when rolling a die, getting a 2 and getting a 5 are mutually exclusive. For mutually exclusive events A and B: P(A union B) = P(A) + P(B). This is the addition rule for mutually exclusive events. Note that complementary events are always mutually exclusive, but not all mutually exclusive events are complementary.

Independent Events

Two events are independent if the occurrence of one does not affect the probability of the other. For example, successive coin tosses are independent. For independent events A and B: P(A intersect B) = P(A) x P(B). This is the multiplication rule for independent events. Be careful not to confuse independence with mutual exclusivity - mutually exclusive events with non-zero probability cannot be independent.

Conditional Probability

Conditional probability P(A|B) represents the probability of event A occurring given that event B has already occurred. The formula is P(A|B) = P(A intersect B) / P(B), where P(B) > 0. This concept is crucial for problems involving drawing without replacement, where the probability changes after each draw. For example, P(second card is ace | first card was ace) = 3/51 when drawing from a standard deck.

Cards, Dice, and Coins

Standard deck: 52 cards with 4 suits (spades ♠, hearts ♥, diamonds ♦, clubs ♣), each with 13 cards (A, 2-10, J, Q, K). Face cards: Jack, Queen, King (12 total). Honour cards: Ace, 10, J, Q, K (20 total). Standard die: 6 faces numbered 1-6. For coins, remember that each toss has 2 outcomes. When multiple coins are tossed, total outcomes = $2^{n}$ . For multiple dice, total outcomes = $6^{n}$ .

Check the idea while it is fresh.Open a focused exercise without losing this article.Practise this concept

Worked examples

Example 1

A single card is drawn from a well-shuffled standard pack of 52 playing cards. What is the probability of drawing a King or a Queen?

Solution

Step 1: Total sample space cardinality.

n (S) = 52

Step 2: Favourable outcomes (mutually exclusive events).
- Number of Kings = 4
- Number of Queens = 4

n (E) = 4 + 4 = 8

Step 3: Compute probability.

P (E) = \frac{n ( E )}{n ( S )} = \frac{8}{52} = \frac{2}{13}

Answer: $\frac{2}{13}$ .

Example 2

Two standard six-sided dice are thrown simultaneously. What is the probability that the sum of the numbers showing is 7?

Solution

Step 1: Total possible outcomes.

n (S) = 6 \times 6 = 36

Step 2: Favourable outcomes summing to 7.

E = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)} ⟹ n (E) = 6

Step 3: Calculate probability.

P (Sum = 7) = \frac{6}{36} = \frac{1}{6}

Answer: $\frac{1}{6}$ .

Example 3

An urn contains 5 red balls and 7 blue balls. Two balls are drawn successively without replacement. What is the probability that both balls drawn are red?

Solution

Step 1: Probability of drawing red on first draw.

P (R_{1}) = \frac{5}{5 + 7} = \frac{5}{12}

Step 2: Conditional probability of drawing red on second draw.

P (R_{2} ∣ R_{1}) = \frac{4}{11}

Step 3: Joint probability.

P (R_{1} \cap R_{2}) = \frac{5}{12} \times \frac{4}{11} = \frac{20}{132} = \frac{5}{33}

Answer: $\frac{5}{33}$ .

Example 4

A fair die is rolled twice. What is the probability of obtaining at least one 6?

Solution

Step 1: Use complementary probability.

P (\geq 1 six) = 1 - P (no six on either roll)

Step 2: Probability of no six on a single roll is $\frac{5}{6}$ .

P (no six) = (\frac{5}{6})^{2} = \frac{25}{36}

Step 3: Compute complement.

P (\geq 1 six) = 1 - \frac{25}{36} = \frac{11}{36}

Answer: $\frac{11}{36}$ .

Representative solved questions

See the kind of question in this topic before opening the full practice set.

Question 1

A bag contains 6 red balls and some blue balls. If the probability of drawing a blue ball is 2/3, how many blue balls are there?

Answer: C. 12

Explanation

Let blue balls = x
Total balls = 6 + x
Probability of blue = x/(6+x) = 2/3
$3 x = 12 + 2 x$
$x = 12$

Sources and review notes

This is an AISEA-authored practice question.

Review status: accepted · Reviewed 2026-08-13 · structure and answer-key checks, editorial quality checks, duplicate screening

Question 2

A box contains 5 green, 4 yellow, and 3 white balls. If 2 balls are drawn at random, what is the probability that they are of different colours?

29/66
23/66
19/66
47/66

Answer: D. 47/66

Explanation

Total balls = 12
Ways to draw 2 balls: $C (12, 2) = 66$
Ways to draw same colour: $C (5, 2) + C (4, 2) + C (3, 2) = 10 + 6 + 3 = 19$
Ways to draw different colours = 66 - 19 = 47
Probability = 47/66

Sources and review notes

This is an AISEA-authored practice question.

Review status: accepted · Reviewed 2026-08-13 · structure and answer-key checks, editorial quality checks, duplicate screening

Question 3

A number is selected from numbers 1 to 25. What is the probability that it is a prime number?

9/25
8/25
7/25
10/25

Answer: A. 9/25

Explanation

Prime numbers from 1 to 25: 2, 3, 5, 7, 11, 13, 17, 19, 23 (9 numbers)
Total numbers = 25
Probability = 9/25

Sources and review notes

This is an AISEA-authored practice question.

Review status: accepted · Reviewed 2026-08-13 · structure and answer-key checks, editorial quality checks, duplicate screening

Common mistakes and useful habits

Always verify if events are mutually exclusive before using simple addition. Events like 'king' and 'spade' are NOT mutually exclusive (king of spades exists).
For 'at least one' type problems, use complementary probability: P(at least one) = 1 - P(none). This often simplifies calculations significantly.
When dealing with cards, remember: P(face card) = 12/52 = 3/13, P(honour card) = 20/52 = 5/13, P(card of a specific suit) = 13/52 = 1/4.
For problems with replacement, probabilities remain constant. For without replacement, probabilities change and conditional probability applies.
With dice problems, create a systematic list or table to ensure you count all favourable outcomes correctly, especially for sum-based questions.
When tossing n coins simultaneously or one coin n times, the probability calculation is identical: each has 2^n equally likely outcomes.

Ready to test your understanding?

Work through 130 questions with explanations after each answer.

Start practice

Next topic: Area (Mensuration)