Probability: Advanced JEE Main PYQs PDF, 2020–2026

Probability: Advanced questions per year

Topic-wise weightage by year

How many Probability: Advanced questions each topic had in every JEE Main year (all shifts), counting every question asked, including ones not in the PDF. Darker cells mean more questions.

Swipe the table sideways for older years.

Newest questions (2026)

All 10 JEE Main 2026 Probability: Advanced questions in the free PDF, latest shift first.

1. JEE Main 2026 · 8 Apr · Shift 2 · Bayes' Theorem

   A candidate has to go to the examination centre to appear in an examination. The candidate uses only one means of transportation for the entire distance out of bus, scooter and car. The probabilities of the candidate going by bus, scooter and car, respectively, are $\frac{2}{5},\frac{1}{5}$ and $\frac{2}{5}$. The probabilities that the candidate reaches late at the examination centre are $\frac{1}{5},\frac{1}{3}$ and $\frac{1}{4}$ if the candidate uses bus, scooter and car, respectively. Given that the candidate reached late at the examination centre, the probability that the candidate travelled by bus is :

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2. JEE Main 2026 · 6 Apr · Shift 1 · Bayes' Theorem

   A bag contains $({\mathrm{N}}+1)$ coins -N fair coins, and one coin with 'Head' on both sides. A coin is selected at random and tossed. If the probability of getting 'Head' is $\frac{9}{16}$, then N is equal to:

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3. JEE Main 2026 · 5 Apr · Shift 2 · Bayes' Theorem

   The probabilities that players A and B of a team are selected for the captaincy for a tournament are 0.6 and 0.4 , respectively. If $A$ is selected the captain, the probability that the team wins the tournament is 0.8 and if B is selected the captain, the probability that the team wins the tournament is 0.7 . Then the probability, that the team wins the tournament, is :

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4. JEE Main 2026 · 5 Apr · Shift 1 · Bayes' Theorem

   A letter is known to have arrived by post either from KANPUR or from ANANTPUR. On the envelope just two consecutive letters AN are visible. The probability, that the letter came from ANANTPUR, is:

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5. JEE Main 2026 · 28 Jan · Shift 1 · Bayes' Theorem

   A bag contains 10 balls out of which $k$ are red and $(10-k)$ are black, where $0\leq k\leq 10$. If three balls are drawn at random without replacement and all of them are found to be black, then the probability that the bag contains 1 red and 9 black balls is:

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6. JEE Main 2026 · 28 Jan · Shift 2 · Random Variables and Distributions

   The probability distribution of a random variable X is given below :

   X	4k	$\frac{30}{7}k$	$\frac{32}{7}k$	$\frac{34}{7}k$	$\frac{36}{7}k$	$\frac{38}{7}k$	$\frac{40}{7}k$	6k
   P(X)	$\frac{2}{15}$	$\frac{1}{15}$	$\frac{2}{15}$	$\frac{1}{5}$	$\frac{1}{15}$	$\frac{2}{15}$	$\frac{1}{5}$	$\frac{1}{15}$

   If E(X) = $\frac{263}{15}$, then P(X < 20) is equal to :

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7. JEE Main 2026 · 24 Jan · Shift 1 · Binomial Distribution

   From a lot containing 10 defective and 90 non-defective bulbs, 8 bulbs are selected one by one with replacement. Then the probability of getting at least 7 defective bulbs is

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8. JEE Main 2026 · 23 Jan · Shift 2 · Bayes' Theorem

   Bag A contains 9 white and 8 black balls, while bag B contains 6 white and 4 black balls. One ball is randomly picked up from the bag B and mixed up with the balls in the bag A . Then a ball is randomly drawn from the bag A . If the probability, that the ball drawn is white, is $\frac{{\mathrm{p}}}{{\mathrm{q}}},\gcd ({\mathrm{p}},{\mathrm{q}})=1$, then ${\mathrm{p}}+{\mathrm{q}}$ is equal to

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9. JEE Main 2026 · 22 Jan · Shift 1 · Random Variables and Distributions

   If a random variable $x$ has the probability distribution

   $\begin{array}{ccccccccc}x & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ {\mathrm{P}}(x) & 0 & 2{\mathrm{k}} & {\mathrm{k}} & 3{\mathrm{k}} & 2{\mathrm{k}}^{2} & 2{\mathrm{k}} & {\mathrm{k}}^{2}+{\mathrm{k}} & 7{\mathrm{k}}^{2}\end{array}$

   $\text{ then}{\mathrm{P}}(3<x\leq 6)\text{ is equal to}$

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10. JEE Main 2026 · 21 Jan · Shift 2 · Random Variables and Distributions

    A random variable X takes values 0, 1, 2, 3 with probabilities $\frac{2a+1}{30}$, $\frac{8a-1}{30}$, $\frac{4a+1}{30}$, $b$ respectively, where $a,b\in {\mathbb{R}}$.
    
    Let $\mu$ and $\sigma$ respectively be the mean and standard deviation of $X$ such that $\sigma^{2}+\mu^{2}=2$.

    Then $\frac{a}{b}$ is equal to:

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