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Probability — JEE Main Maths PYQ MCQs with Solutions
Free JEE Main Maths PYQ Probability MCQs with step-by-step solutions covering Random Experiment & Events, Classical & Axiomatic Probability, Conditional Probability & Bayes' Theorem, Probability Distributions. Practise online on Prepizo — no login needed.
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Sample questions with solutions
Q1 — Random Experiment & Events · easy · theory
A coin is tossed four times. The number of possible outcomes in the sample space is:
A. $2$
B. $8$
C. $16$ ✓ Correct
D. $32$
Solution: When a coin is tossed, there are 2 possible outcomes (H or T). For 4 tosses, the total number of outcomes = $2^4 = 16$. Each outcome is a sequence like HHHH, HHHT, HHTH, ..., TTTT.
Q2 — Random Experiment & Events · easy · theory
Two dice are rolled simultaneously. How many elements are in the sample space?
A. $12$
B. $24$
C. $36$ ✓ Correct
D. $48$
Solution: Each die has 6 faces. When two dice are rolled, the first die can show any of 6 outcomes and the second die can show any of 6 outcomes. Total outcomes = $6 \times 6 = 36$.
Q3 — Random Experiment & Events · easy · theory
In a random experiment, the sample space is $S = \{1, 2, 3, 4, 5, 6\}$. Which of the following is a valid event?
A. $\{1, 2, 7\}$
B. $\{2, 4, 6\}$ ✓ Correct
C. $\{0, 1, 2\}$
D. $\{1, 2, 3, 4, 5, 6, 7\}$
Solution: An event must be a subset of the sample space. $\{2, 4, 6\} \subseteq S$ is a valid event. The other options contain elements not in $S$ or are supersets of $S$.
Q4 — Random Experiment & Events · easy · theory
Two events $A$ and $B$ in a sample space $S$ are called mutually exclusive if:
A. $A \cup B = \emptyset$
B. $A \cap B = \emptyset$ ✓ Correct
C. $A \subseteq B$
D. $A^c = B$
Solution: Two events are mutually exclusive (or disjoint) if they cannot occur simultaneously, i.e., they have no outcomes in common. This means $A \cap B = \emptyset$ (their intersection is empty).
Q5 — Random Experiment & Events · easy · theory
Let $S$ be a sample space. If $A$ is an event, then $A \cup A^c$ equals:
A. $A$
B. $A^c$
C. $S$ ✓ Correct
D. $\emptyset$
Solution: For any event $A$ in sample space $S$, the complement $A^c$ contains all outcomes not in $A$. Therefore, $A \cup A^c$ contains all outcomes in $A$ plus all outcomes not in $A$, which is the entire sample space $S$.
Q6 — Random Experiment & Events · easy · theory
A die is rolled. Let $A = \{2, 4, 6\}$ (even numbers) and $B = \{1, 3, 5\}$ (odd numbers). Which statement is true?
A. $A$ and $B$ are neither mutually exclusive nor exhaustive
B. $A$ and $B$ are mutually exclusive but not exhaustive
C. $A$ and $B$ are mutually exclusive and exhaustive ✓ Correct
D. $A$ and $B$ are exhaustive but not mutually exclusive
Solution: For events on rolling a die with $S = \{1, 2, 3, 4, 5, 6\}$: Mutually exclusive: $A \cap B = \emptyset$ ✓ (no outcome is both even and odd). Exhaustive: $A \cup B = \{1, 2, 3, 4, 5, 6\} = S$ ✓ (every outcome is either even or odd).
Q7 — Random Experiment & Events · easy · theory
Let $S = \{a, b, c, d, e\}$. The number of events that contain exactly 3 elements from $S$ is:
A. $3$
B. $5$
C. $10$ ✓ Correct
D. $15$
Solution: An event is any subset of $S$. The number of 3-element subsets of a 5-element set is $\binom{5}{3} = \frac{5!}{3! \cdot 2!} = 10$.
Q8 — Classical & Axiomatic Probability · easy · theory
A card is drawn from a standard deck of 52 cards. The probability of drawing either a spade or a king is:
A. 4/13 ✓ Correct
B. 1/4
C. 13/52
D. 17/52
Solution: Let A = event of drawing a spade, B = event of drawing a king. P(A) = 13/52, P(B) = 4/52, P(A∩B) = 1/52 (king of spades). By addition theorem: P(A∪B) = 13/52 + 4/52 - 1/52 = 16/52 = 4/13.
Q9 — Classical & Axiomatic Probability · easy · theory
Two dice are rolled simultaneously. The probability that the sum is either 7 or 11 is:
A. 2/9 ✓ Correct
B. 1/6
C. 5/36
D. 1/4
Solution: Sum = 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) → 6 outcomes. Sum = 11: (5,6), (6,5) → 2 outcomes. These are mutually exclusive events. P(7 or 11) = (6 + 2)/36 = 8/36 = 2/9.
Q10 — Classical & Axiomatic Probability · easy · theory
If P(A) = 0.3, P(B) = 0.4, and A and B are independent events, then P(A∪B) equals:
A. 0.58 ✓ Correct
B. 0.70
C. 0.12
D. 0.67
Solution: For independent events, P(A∩B) = P(A) × P(B) = 0.3 × 0.4 = 0.12. Using addition theorem: P(A∪B) = P(A) + P(B) - P(A∩B) = 0.3 + 0.4 - 0.12 = 0.58.
Q11 — Classical & Axiomatic Probability · easy · theory
Three coins are tossed simultaneously. What is the probability of getting at least one head?
A. 7/8 ✓ Correct
B. 3/8
C. 1/2
D. 5/8
Solution: Using complement: P(at least one head) = 1 - P(no heads) = 1 - P(all tails) = 1 - (1/2)³ = 1 - 1/8 = 7/8.
Q12 — Classical & Axiomatic Probability · easy · theory
In a class of 30 students, 12 play cricket, 15 play football, and 6 play both. If a student is selected at random, the probability that the student plays cricket or football is:
A. 7/10 ✓ Correct
B. 3/5
C. 1/2
D. 4/5
Solution: Let C = plays cricket, F = plays football. |C| = 12, |F| = 15, |C∩F| = 6. P(C∪F) = (12 + 15 - 6)/30 = 21/30 = 7/10.
Q13 — Classical & Axiomatic Probability · easy · theory
If P(A) = 1/3 and P(B) = 1/4, where A and B are mutually exclusive events, then P(A∪B) is:
A. 7/12 ✓ Correct
B. 1/12
C. 1/7
D. 5/12
Solution: For mutually exclusive events, P(A∩B) = 0. Therefore, P(A∪B) = P(A) + P(B) = 1/3 + 1/4 = 4/12 + 3/12 = 7/12.
Q14 — Classical & Axiomatic Probability · easy · theory
The probability that a randomly selected student from a class passes Math is 0.7 and passes English is 0.8. Assuming independence, the probability of passing both is:
A. 0.56 ✓ Correct
B. 0.75
C. 0.87
D. 1.5
Solution: For independent events, P(Math ∩ English) = P(Math) × P(English) = 0.7 × 0.8 = 0.56.
Q15 — Classical & Axiomatic Probability · easy · theory
A fair coin is tossed 3 times. What is the probability of getting exactly 2 heads?
A. 3/8 ✓ Correct
B. 1/4
C. 1/2
D. 5/8
Solution: Total outcomes = 2³ = 8. Favorable outcomes (exactly 2 heads): HHT, HTH, THH → 3 outcomes. Probability = 3/8.
Q16 — Classical & Axiomatic Probability · easy · theory
A die is rolled twice. The probability of getting a sum of 10 is:
A. 1/12 ✓ Correct
B. 1/6
C. 1/9
D. 1/18
Solution: Sum = 10 can occur in: (4,6), (5,5), (6,4) → 3 favorable outcomes. Total outcomes = 36. Probability = 3/36 = 1/12.
Q17 — Classical & Axiomatic Probability · easy · theory
The axioms of probability state that for any event A:
A. 0 ≤ P(A) ≤ 1 ✓ Correct
B. P(A) > 0
C. P(A) < 1
D. -1 ≤ P(A) ≤ 1
Solution: One of the fundamental axioms of probability is that the probability of any event is between 0 and 1, inclusive. P(impossible event) = 0 and P(certain event) = 1.
Q18 — Classical & Axiomatic Probability · easy · theory
If the sample space S contains n equally likely outcomes and event A contains m outcomes, then P(A) is defined as:
A. m/n ✓ Correct
B. n/m
C. m-n
D. (m+n)/2
Solution: In classical probability, when all outcomes in the sample space are equally likely, P(A) = (number of favorable outcomes) / (total number of outcomes) = m/n.
Q19 — Classical & Axiomatic Probability · easy · theory
For any two events A and B, the addition theorem states that P(A∪B) equals:
A. P(A) + P(B) - P(A∩B) ✓ Correct
B. P(A) + P(B)
C. P(A) - P(B)
D. P(A) × P(B)
Solution: The addition theorem for probability states: P(A∪B) = P(A) + P(B) - P(A∩B). This avoids double-counting the intersection when A and B overlap.
Q20 — Conditional Probability & Bayes' Theorem · easy · theory
A bag contains 4 red and 6 blue balls. Two balls are drawn without replacement. What is the probability that the second ball is blue given that the first ball drawn is red?
A. $\dfrac{3}{5}$
B. $\dfrac{6}{9}$ ✓ Correct
C. $\dfrac{5}{9}$
D. $\dfrac{4}{9}$
Solution: After drawing a red ball, 9 balls remain (3 red, 6 blue). Using conditional probability: $P(B_2|R_1) = \dfrac{6}{9} = \dfrac{2}{3}$.
Q21 — Conditional Probability & Bayes' Theorem · easy · theory
If $P(A) = \dfrac{1}{3}$ and $P(B) = \dfrac{1}{4}$, and $A$ and $B$ are independent events, then $P(A \cap B)$ equals:
A. $\dfrac{1}{12}$ ✓ Correct
B. $\dfrac{1}{7}$
C. $\dfrac{7}{12}$
D. $\dfrac{1}{2}$
Solution: For independent events: $P(A \cap B) = P(A) \times P(B) = \dfrac{1}{3} \times \dfrac{1}{4} = \dfrac{1}{12}$.
Q22 — Conditional Probability & Bayes' Theorem · easy · theory
Two events $A$ and $B$ are such that $P(A) = 0.6$, $P(B) = 0.3$, and $P(A \cap B) = 0.2$. Find $P(A|B)$:
A. $\dfrac{2}{3}$ ✓ Correct
B. $\dfrac{1}{3}$
C. $\dfrac{2}{5}$
D. $\dfrac{3}{5}$
Solution: Using the definition: $P(A|B) = \dfrac{P(A \cap B)}{P(B)} = \dfrac{0.2}{0.3} = \dfrac{2}{3}$.
Q23 — Conditional Probability & Bayes' Theorem · easy · theory
If $P(A|B) = 0.5$ and $P(B) = 0.4$, and $A$ and $B$ are not independent, then $P(A \cap B)$ equals:
A. $0.2$ ✓ Correct
B. $0.9$
C. $0.1$
D. $0.25$
Solution: From the definition: $P(A|B) = \dfrac{P(A \cap B)}{P(B)}$, so $P(A \cap B) = P(A|B) \times P(B) = 0.5 \times 0.4 = 0.2$.
Q24 — Conditional Probability & Bayes' Theorem · easy · theory
A card is drawn from a deck. Event $A$ is 'card is a spade' and event $B$ is 'card is a queen'. Find $P(A|B)$:
A. $\dfrac{1}{4}$ ✓ Correct
B. $\dfrac{1}{13}$
C. $\dfrac{4}{13}$
D. $\dfrac{1}{52}$
Solution: Given that the card is a queen (event $B$), there are 4 possible cards. Only 1 of these 4 is the queen of spades. Therefore, $P(A|B) = \dfrac{1}{4}$.
Q25 — Conditional Probability & Bayes' Theorem · easy · theory
Two students solve a problem. The probability that student A solves it is $\dfrac{2}{3}$ and for student B is $\dfrac{3}{4}$. If they solve independently, what is the probability that the problem is solved by at least one of them?
A. $\dfrac{11}{12}$ ✓ Correct
B. $\dfrac{1}{12}$
C. $\dfrac{5}{12}$
D. $\dfrac{17}{12}$
Solution: Probability at least one solves = $1 -$ P(neither solves) $= 1 - (1 - \dfrac{2}{3})(1 - \dfrac{3}{4}) = 1 - \dfrac{1}{3} \times \dfrac{1}{4} = 1 - \dfrac{1}{12} = \dfrac{11}{12}$.
Q26 — Conditional Probability & Bayes' Theorem · easy · theory
A bag has 5 red balls and 3 blue balls. Two balls are drawn without replacement. What is the probability that the second ball is red given that the first is blue?
A. $\dfrac{5}{7}$ ✓ Correct
B. $\dfrac{5}{8}$
C. $\dfrac{4}{7}$
D. $\dfrac{1}{2}$
Solution: After drawing a blue ball, 7 balls remain: 5 red and 2 blue. $P(\text{Red}_2|\text{Blue}_1) = \dfrac{5}{7}$.
Q27 — Conditional Probability & Bayes' Theorem · easy · theory
The probability that a student passes mathematics is $\dfrac{2}{5}$ and the probability that he passes English is $\dfrac{3}{5}$. If passing these subjects are independent, the probability that he fails both is:
A. $\dfrac{6}{25}$ ✓ Correct
B. $\dfrac{19}{25}$
C. $\dfrac{1}{25}$
D. $\dfrac{5}{25}$
Solution: Probability of failing math = $1 - \dfrac{2}{5} = \dfrac{3}{5}$. Probability of failing English = $1 - \dfrac{3}{5} = \dfrac{2}{5}$. Since independent: $P(\text{fail both}) = \dfrac{3}{5} \times \dfrac{2}{5} = \dfrac{6}{25}$.
Q28 — Conditional Probability & Bayes' Theorem · easy · theory
If $P(A \cap B) = 0.15$, $P(B) = 0.3$, then $P(A|B)$ equals:
A. $0.5$ ✓ Correct
B. $0.45$
C. $0.15$
D. $0.2$
Solution: $P(A|B) = \dfrac{P(A \cap B)}{P(B)} = \dfrac{0.15}{0.3} = 0.5$.
Q29 — Conditional Probability & Bayes' Theorem · easy · theory
A test has two independent parts. The probability of passing part 1 is $p$ and part 2 is $q$. If the probability of passing at least one part is $\dfrac{3}{4}$, then:
A. $pq = \dfrac{1}{4}$
B. $p + q - pq = \dfrac{3}{4}$ ✓ Correct
C. $p + q = \dfrac{3}{4}$
D. $pq = \dfrac{3}{4}$
Solution: For independent events: $P(\text{at least one}) = P(A \cup B) = P(A) + P(B) - P(A)P(B) = p + q - pq = \dfrac{3}{4}$.
Q30 — Conditional Probability & Bayes' Theorem · easy · theory
Events $A$ and $B$ are independent. $P(A) = \dfrac{3}{4}$ and $P(B) = \dfrac{1}{3}$. What is $P(A \cup B)$?
A. $\dfrac{11}{12}$ ✓ Correct
B. $\dfrac{13}{12}$
C. $\dfrac{7}{12}$
D. $\dfrac{1}{4}$
Solution: $P(A \cup B) = P(A) + P(B) - P(A)P(B) = \dfrac{3}{4} + \dfrac{1}{3} - \dfrac{3}{4} \times \dfrac{1}{3} = \dfrac{9}{12} + \dfrac{4}{12} - \dfrac{3}{12} = \dfrac{10}{12}$. [Recheck: $\dfrac{3}{4} + \dfrac{1}{3} = \dfrac{13}{12}$, minus $\dfrac{1}{4} = \dfrac{11}{12}$.]