JEE (Main) · Mathematics, Class 12
Probability: JEE Main PYQ, 2020–2025
Probability (Mathematics, Class 12) has been asked 135 times in the 111 JEE (Main) shift papers from 2020–2025: about 1.2 questions a paper, 4% of all Mathematics questions. It is the 13th most-asked of 19 Mathematics chapters. The 2025 shifts had 18 from it in all. Last checked 28 October 2026.
Questions per year, all shifts
Year by year.
| Year | Questions |
|---|---|
| 2020 | 15in 14 papers |
| 2021 | 32in 26 papers |
| 2022 | 33in 24 papers |
| 2023 | 13in 12 papers |
| 2024 | 24in 18 papers |
| 2025 | 18in 17 papers |
JEE (Main) is sat in many shifts a year, each with its own paper; the counts add up every shift. 2021–2024 papers had 30 questions a subject, the others 25.
Inside the chapter
Topics asked most.
- Probability distribution of a random variable46
- Probability of an event40
- Addition and multiplication theorems32
- Bayes' theorem17
Try it
10 Probability questions to practise.
Our own questions, written in the JEE Main pattern and checked, each with an explanation. Pick an option to see the answer.
Q1 · Mathematics · Probability
A raffle has 100 tickets. One ticket wins a prize of ₹1000, three tickets each win a prize of ₹200, and the remaining tickets win nothing. If X is the prize amount (in ₹) won by a randomly chosen ticket, find E(X).
Answer: (A)
The distribution of X is P(X = 1000) = 1/100, P(X = 200) = 3/100, P(X = 0) = 96/100. The expected value is E(X) = Σ xᵢpᵢ = 1000 × 1/100 + 200 × 3/100 + 0 × 96/100 = 10 + 6 + 0 = 16. So the expected prize per ticket is ₹16.
Q2 · Mathematics · Probability
Four cards are drawn at random, without replacement, from a well-shuffled standard pack of 52 cards. What is the probability that one card is drawn from each of the four suits?
Answer: (A)
Treat the hand as an unordered set of 4 cards, so there are C(52,4) = 270725 equally likely hands. A hand with all four suits has one card from each suit, which can be chosen in 13 × 13 × 13 × 13 = 28561 ways. So P = 28561/270725 = 2197/20825, which is about 0.105.
Q3 · Mathematics · Probability
A and B are events of a sample space with P(A) = 3/4 and P(B) = 5/8. Which of the following is the exact range of possible values of the conditional probability P(A | B)?
Answer: (A)
By the addition theorem, P(A ∩ B) = P(A) + P(B) − P(A ∪ B). Since P(A ∪ B) ≤ 1, P(A ∩ B) ≥ 3/4 + 5/8 − 1 = 3/8. Also A ∩ B ⊂ B, so P(A ∩ B) ≤ P(B) = 5/8, and this is smaller than P(A) = 3/4, so it is the binding upper bound. Both extremes can occur: P(A ∩ B) = 5/8 when B ⊂ A, and P(A ∩ B) = 3/8 when A ∪ B is the whole sample space. Then P(A | B) = P(A ∩ B)/(5/8) lies between (3/8)/(5/8) = 3/5 and (5/8)/(5/8) = 1.
Q4 · Mathematics · Probability
For two events A and B, P(A) = 0.35, P(B) = 0.45 and P(A ∩ B) = 0.10. What is the probability that neither A nor B occurs?
Answer: (A)
Use the addition theorem to get P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 0.35 + 0.45 − 0.10 = 0.70. 'Neither A nor B' is the complement of A ∪ B, so its probability is 1 − 0.70 = 0.30.
Q5 · Mathematics · Probability
All the letters of the word MISSISSIPPI are arranged in a row at random, and all distinct arrangements are equally likely. What is the probability that the four S's occupy four consecutive positions?
Answer: (A)
Only the positions of the S's matter, and each set of 4 positions out of 11 is equally likely, so there are C(11,4) = 330 outcomes. The S's are consecutive if they start at position 1, 2, ..., 8, which gives 8 favourable outcomes. So P = 8/330 = 4/165. Counting arrangements gives the same result: 8!/(4!·2!) = 840 favourable out of 11!/(4!·4!·2!) = 34650, and 840/34650 = 4/165.
Q6 · Mathematics · Probability
In a population, 2% of people have a certain disease. A screening test gives a positive result for 90% of people who have the disease and for 5% of people who do not. A person is tested twice. Given the person's disease status, the two test results are independent. Both tests come out positive. What is the probability that the person has the disease?
Answer: (A)
Use Bayes' theorem with the two test results combined into one event. Given the disease status, the tests are independent, so P(++|D) = 0.9² = 0.81 and P(++|Dᶜ) = 0.05² = 0.0025. Then P(D|++) = (0.02·0.81)/(0.02·0.81 + 0.98·0.0025) = 0.0162/0.01865 = 324/373. Applying the prior only once and squaring both likelihoods is the key step. The result is not the single-test value 18/67, and it is not found by dropping the prior.
Q7 · Mathematics · Probability
A bag contains 3 red and 2 blue balls. Two balls are drawn one after the other without replacement, and X denotes the number of red balls drawn. Find P(X = 2).
Answer: (A)
For X = 2, both balls must be red. The first ball is red with probability 3/5. Given that, the bag has 2 red balls among 4, so the second is red with probability 2/4. Therefore P(X = 2) = 3/5 × 1/2 = 3/10.
Q8 · Mathematics · Probability
For events A and B, P(A ∩ B) = 1/12, P(A | B) = 1/2 and P(B | A) = 1/3. What is P(A ∪ B)?
Answer: (A)
From P(A | B) = P(A ∩ B)/P(B), P(B) = (1/12)/(1/2) = 1/6. From P(B | A) = P(A ∩ B)/P(A), P(A) = (1/12)/(1/3) = 1/4. The addition theorem gives P(A ∪ B) = 1/4 + 1/6 − 1/12 = 3/12 + 2/12 − 1/12 = 4/12 = 1/3.
Q9 · Mathematics · Probability
An urn contains 2 white and 3 black balls. Balls are drawn one at a time without replacement until the first white ball appears. If X is the number of draws made, find Var(X).
Answer: (A)
P(X = 1) = 2/5. P(X = 2) = (3/5)(2/4) = 3/10. P(X = 3) = (3/5)(2/4)(2/3) = 1/5. P(X = 4) = (3/5)(2/4)(1/3)(2/2) = 1/10. These sum to 1. E(X) = 2/5 + 6/10 + 3/5 + 4/10 = 2. E(X²) = 2/5 + 12/10 + 9/5 + 16/10 = 5. Var(X) = 5 − 2² = 1.
Q10 · Mathematics · Probability
Two fair six-faced dice are thrown together. What is the probability that the sum of the numbers on their upper faces is 9?
Answer: (A)
When two dice are thrown, the sample space has 6 × 6 = 36 equally likely ordered pairs. List the pairs with sum 9: (3,6), (4,5), (5,4), (6,3). That is 4 favourable outcomes. P = 4/36 = 1/9. The common errors are treating (3,6) and (6,3) as one outcome, or dropping a pair from the list.
Free with an account
All 135 real JEE Main questions from Probability.
The questions from the actual papers, with the official answer and an explanation for each — chapter by chapter, or as full timed papers.
Solve them freeQuestions
About Probability in JEE Main.
How many questions come from Probability in JEE Main?
Probability (Mathematics, Class 12) has been asked 135 times in the 111 JEE (Main) shift papers from 2020–2025: about 1.2 questions a paper, 4% of all Mathematics questions. It is the 13th most-asked of 19 Mathematics chapters. The 2025 shifts had 18 from it in all.
Which topics of Probability are asked most in JEE Main?
Probability distribution of a random variable (46), Probability of an event (40), Addition and multiplication theorems (32), Bayes' theorem (17), counted over the papers from 2020–2025.
Is Probability important for JEE Main 2027?
It is in the syllabus in force, so it can be asked. By past papers it ranks 13 of 19 in Mathematics; use that to decide how much time it gets, not whether to study it.
Where can I solve Probability JEE Main previous year questions?
On Nikaalle, free: sign up and practise every past JEE Main question from Probability with an explanation for each, or sit the full papers timed. The ten questions on this page are our own, written in the same pattern.
Your JEE Main 2027 plan, chapter by chapter.
A week-wise plan that gives the heavy chapters their time. Free, from a non-profit trust.