Probability JEE Main previous year questions with solutions

5 solved JEE Main questions on Probability, free to read — no sign-in needed. The full chapter has 212 questions; sign in to attempt the remaining 207 in the exam simulator.

1. Q1JEE Main 2026 (05 Apr, Shift 1)Baye's 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:
   1. A.$\dfrac{7}{10}$
   2. B.$\dfrac{10}{17}$
   3. C.$\dfrac{12}{19}$
   4. D.$\dfrac{7}{19}$
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   Answer: (B)

   Let $E_1$ be the event that the letter came from KANPUR and $E_2$ be the event that the letter came from ANANTPUR. Since the letter is equally likely to come from either city, $P(E_1) = P(E_2) = \dfrac{1}{2}$. Let $A$ be the event that the two consecutive visible letters are AN. In the word KANPUR, there are 6 letters, so there are 5 pairs of consecutive letters: KA, AN, NP, PU, UR. The pair AN appears exactly once. Thus, $P(A|E_1) = \dfrac{1}{5}$. In the word ANANTPUR, there are 8 letters, so there are 7 pairs of consecutive letters: AN, NA, AN, NT, TP, PU, UR. The pair AN appears exactly twice. Thus, $P(A|E_2) = \dfrac{2}{7}$. By Bayes' theorem, the probability that the letter came from ANANTPUR given that AN is visible is: $P(E_2|A) = \dfrac{P(E_2) P(A|E_2)}{P(E_1) P(A|E_1) + P(E_2) P(A|E_2)}$ $P(E_2|A) = \dfrac{\dfrac{1}{2} \times \dfrac{2}{7}}{\dfrac{1}{2} \times \dfrac{1}{5} + \dfrac{1}{2} \times \dfrac{2}{7}}$ $P(E_2|A) = \dfrac{\dfrac{2}{7}}{\dfrac{1}{5} + \dfrac{2}{7}}$ $P(E_2|A) = \dfrac{\dfrac{2}{7}}{\dfrac{17}{35}} = \dfrac{2}{7} \times \dfrac{35}{17} = \dfrac{10}{17}$ Answer: $\dfrac{10}{17}$
2. Q2JEE Main 2025 (08 Apr, Shift 2)Classical Definition of Probability
   If $A$ and $B$ are two events such that $P(A)=0.7$, $\mathrm{P}(\mathrm{B})=0.4$ and $\mathrm{P}(\mathrm{A} \cap \overline{\mathrm{B}})=0.5$, where $\overline{\mathrm{B}}$ denotes the complement of $B$, then $P(B \mid(A \cup \bar{B}))$ is equal:-
   1. A.$\frac{1}{4}$
   2. B.$\frac{1}{2}$
   3. C.$\frac{1}{6}$
   4. D.$\frac{1}{3}$
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   Answer: (A)

   $\begin{aligned} & \mathrm{P}(\mathrm{A})=\frac{7}{10}, \mathrm{P}(\mathrm{B})=\frac{4}{10} \\ & \mathrm{P}(\mathrm{A} \cup \overline{\mathrm{B}})=\frac{5}{10} \\ & \mathrm{P}\left(\frac{\mathrm{B}}{\mathrm{A} \cup \overline{\mathrm{B}}}\right)=\frac{\mathrm{P}(\mathrm{B} \cap(\mathrm{A} \cup \overline{\mathrm{B}}))}{\mathrm{P}(\mathrm{A} \cup \overline{\mathrm{B}})} \\ & =\frac{\mathrm{P}((\mathrm{B} \cap \overline{\mathrm{B}}) \cup(\mathrm{B} \cap \mathrm{A}))}{\mathrm{P}(\mathrm{A} \cup \overline{\mathrm{B}})}=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{A} \cup \overline{\mathrm{B}})}\end{aligned}$ $\begin{aligned} & =\frac{\mathrm{P}(\mathrm{A})-\mathrm{P}(\mathrm{A} \cap \overline{\mathrm{B}})}{\mathrm{P}(\mathrm{A})+\mathrm{P}(\overline{\mathrm{B}})-\mathrm{P}(\mathrm{A} \cap \overline{\mathrm{B}})}=\frac{\frac{7}{10}-\frac{5}{10}}{\frac{7}{10}+\left(1-\frac{4}{10}\right)-\frac{5}{10}} \\ & =\frac{2}{8}=\frac{1}{4}\end{aligned}$
3. Q3JEE Main 2024 (08 Apr, Shift 1)Addition and Subtraction Theorems
   Let the sum of two positive integers be 24 . If the probability, that their product is not less than $\frac{3}{4}$ times their greatest possible product, is $\frac{m}{n}$, where $\operatorname{gcd}(m, n)=1$, then $n-m$ equals
   1. A.10
   2. B.9
   3. C.11
   4. D.8
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   Answer: (A)

   $\begin{aligned} & x+y=24, x, y \in N \\ & A M\gt G M \Rightarrow x y \leq 144 \\ & x y \geq 108\end{aligned}$ Favorable pairs of $(x, y)$ are $\begin{aligned} & (13,11),(12,12),(14,10),(15,9),(16,8), \\ & (17,7),(18,6),(6,18),(7,17),(8,16),(9,15), \\ & (10,14),(11,13)\end{aligned}$ i.e. 13 cases Total choices for $\mathrm{x}+\mathrm{y}=24$ is 23 $\begin{aligned} & \text { Probability }=\frac{13}{23}=\frac{m}{n} \\ & n-m=10\end{aligned}$
4. Q4JEE Main 2023 (30 Jan, Shift 2)Conditional Probability and Multiplication Theorem
   <p>A bag contains six balls of different colours. Two balls are drawn in succession with replacement. The probability that both the balls are of the same colour is $p$. Next four balls are drawn in succession with replacement and the probability that exactly three balls are of the same colours is $q$. If $p:q=m$ $:n$, where $m$ and $n$ are co-prime, then $m+n$ is equal to</p>
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   Answer: 14

   <p>Given, A bag contains six balls of different colours, two balls are drawn in succession with replacement, The probability that both the balls are of the same colour will be $p=\dfrac{6}{36}$ Now Next four balls are drawn in succession with replacement and the probability that exactly three balls are of the same colours is $q$, So, probability will be $q=\dfrac{C16\times C15\times C14}{{6}^{4}}=\dfrac{120}{1296}=\dfrac{5}{54}$ Now the ratio of p,q will be $\dfrac{p}{q}=\dfrac{9}{5}$ Now, on comparing with $\dfrac{p}{q}=\dfrac{m}{n}\Rightarrow m+n=14$</p>
5. Q5JEE Main 2022 (28 Jul, Shift 2)Random Variable and its Probability Distribution
   A bag contains $4$ white and $6$ black balls. Three balls are drawn at random from the bag. Let $X$ be the number of white balls, among the drawn balls. If ${\sigma }^{2}$ is the variance of $X$, then $100{\sigma }^{2}$ is equal to
   Show answer & solution

   Answer: 56

   Given, $X=$ Number of white ball drawn, And bag contains $4$ white and $6$ black balls, total $3$ balls are to be drawn, So, $P\left(X=0\right)=\dfrac{C36}{C310}=\dfrac{1}{6}$ $P\left(X=1\right)=\dfrac{C26\times C14}{C310}=\dfrac{1}{2}$ $P\left(X=2\right)=\dfrac{C16\times C24}{C310}=\dfrac{3}{10}$ and $P\left(X=3\right)=\dfrac{C06\times C34}{C310}=\dfrac{1}{30}$ $X$ $0$ $1$ $2$ $3$ $P\left(X\right)$ $\dfrac{1}{6}$ $\dfrac{1}{2}$ $\dfrac{3}{10}$ $\dfrac{1}{30}$ So, variance $={\sigma }^{2}=\sum {P}_{i}{X}_{i}^{2}-{\left(\sum {P}_{i}{X}_{i}\right)}^{2}$ $\Rightarrow {\sigma }^{2}=\dfrac{1}{2}+\dfrac{12}{10}+\dfrac{3}{10}-{\left(\dfrac{1}{2}+\dfrac{6}{10}+\dfrac{1}{10}\right)}^{2}$ $\Rightarrow {\sigma }^{2}=\dfrac{56}{100}$ $\Rightarrow 100{\sigma }^{2}=56$