Probability JEE Advanced previous year questions with solutions

5 solved JEE Advanced questions on Probability, free to read — no sign-in needed. The full chapter has 55 questions; sign in to attempt the remaining 50 in the exam simulator.

1. Q1JEE Advanced Adv 2026 (Paper 1)
   Suppose that Box I contains $6$ red balls and $9$ green balls, and Box II contains $8$ red balls and $12$ green balls. All the balls of Box I and Box II are mixed together and a ball is chosen at random from them. Let $E_1$ be the event that the ball chosen belonged to Box I and let $E_2$ be the event that the ball chosen belonged to Box II. Let $F_1$ be the event that the ball chosen is red and let $F_2$ be the event that the ball chosen is green. Then which of the following statements is (are) TRUE?
   1. A.The events $E_1$ and $F_1$ are independent
   2. B.The events $E_2$ and $F_2$ are dependent
   3. C.The conditional probability $P(F_1 \mid E_1)$ is equal to the conditional probability $P(F_1 \mid E_2)$
   4. D.The conditional probability $P(F_1 \mid E_1)$ is greater than the conditional probability $P(F_2 \mid E_2)$
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   Answer: A,C

   Total balls in Box I $= 6 + 9 = 15$ Total balls in Box II $= 8 + 12 = 20$ Total balls in the mixture $= 15 + 20 = 35$ Total red balls $= 6 + 8 = 14$ Total green balls $= 9 + 12 = 21$ The probabilities of the given events are: $P(E_1) = \dfrac{15}{35} = \dfrac{3}{7}$ $P(E_2) = \dfrac{20}{35} = \dfrac{4}{7}$ $P(F_1) = \dfrac{14}{35} = \dfrac{2}{5}$ $P(F_2) = \dfrac{21}{35} = \dfrac{3}{5}$ The joint probabilities are: $P(E_1 \cap F_1) = \dfrac{6}{35}$ $P(E_2 \cap F_2) = \dfrac{12}{35}$ For statement (A): $P(E_1) P(F_1) = \dfrac{3}{7} \times \dfrac{2}{5} = \dfrac{6}{35} = P(E_1 \cap F_1)$ Thus, $E_1$ and $F_1$ are independent. Statement (A) is TRUE. For statement (B): $P(E_2) P(F_2) = \dfrac{4}{7} \times \dfrac{3}{5} = \dfrac{12}{35} = P(E_2 \cap F_2)$ Thus, $E_2$ and $F_2$ are independent. Statement (B) is FALSE. For statement (C): $P(F_1 \mid E_1) = \dfrac{P(E_1 \cap F_1)}{P(E_1)} = \dfrac{6/35}{15/35} = \dfrac{2}{5}$ $P(F_1 \mid E_2) = \dfrac{P(E_2 \cap F_1)}{P(E_2)} = \dfrac{8/35}{20/35} = \dfrac{2}{5}$ Since $P(F_1 \mid E_1) = P(F_1 \mid E_2)$, statement (C) is TRUE. For statement (D): $P(F_2 \mid E_2) = \dfrac{P(E_2 \cap F_2)}{P(E_2)} = \dfrac{12/35}{20/35} = \dfrac{3}{5}$ Since $\dfrac{2}{5} \lt \dfrac{3}{5}$, $P(F_1 \mid E_1)$ is not greater than $P(F_2 \mid E_2)$. Statement (D) is FALSE. Answer: The events $E_1$ and $F_1$ are independent; The conditional probability $P(F_1 \mid E_1)$ is equal to the conditional probability $P(F_1 \mid E_2)$
2. Q2JEE Advanced Adv 2025 (Paper 1)
   Three students $S_1, S_2$ and $S_1$ are given a problem to solve. Consider the following events : $U:$ At least one of $S_1, S_2$ and $S_3$ can solve the problem, $V: S_1$ can solve the problem, given that neither $\mathrm{S}_2$ nor $\mathrm{S}_3$ can solve the problem, $W: S_2$ can solve the problem and $S_3$ cannot solve the problem, $T: S_3$ can solve the problem. For any event $E$, let $P(E)$ denote the probability of $E$. If $P(U)=\frac{1}{2}, P(V)=\frac{1}{10}$ and $P(W)=\frac{1}{12}$, then $P(T)$ is equal to
   1. A.$\frac{13}{36}$
   2. B.$\frac{1}{3}$
   3. C.$\frac{19}{60}$
   4. D.$\frac{1}{4}$
   Show answer & solution

   Answer: (A)

   $\begin{aligned} & \mathrm{P}(\mathrm{U})=1-\mathrm{P}\left(\mathrm{~S}_1^{\prime} \cap \mathrm{S}_2^{\prime} \cap \mathrm{S}_3^{\prime}\right)=\frac{1}{2} \\ & \Rightarrow \mathrm{P}\left(\mathrm{~S}_1^{\prime} \cap \mathrm{S}_2^{\prime} \cap \mathrm{S}_3^{\prime}\right)=\frac{1}{2} ; \mathrm{P}\left(\mathrm{~S}_1^{\prime}\right) \cdot \mathrm{P}\left(\mathrm{~S}_2^{\prime}\right) \cdot \mathrm{P}\left(\mathrm{~S}_3^{\prime}\right)=\frac{1}{2} \\ & \Rightarrow\left(1-\mathrm{P}\left(\mathrm{~S}_1\right)\right)\left(1-\mathrm{P}\left(\mathrm{~S}_2\right)\right)\left(1-\mathrm{P}\left(\mathrm{~S}_3\right)\right)=\frac{1}{2} \quad \ldots(1) \end{aligned}$ $\begin{aligned} & \mathrm{P}(\mathrm{V})=\frac{\mathrm{P}\left(\mathrm{S}_1 \cap \mathrm{~S}_2^{\prime} \cap \mathrm{S}_3^{\prime}\right)}{\mathrm{P}\left(\mathrm{S}_2^{\prime} \cap \mathrm{S}_3^{\prime}\right)}=\frac{1}{10} \\ & \Rightarrow \mathrm{P}\left(\mathrm{S}_1\right) \cdot \mathrm{P}\left(\mathrm{S}_2^{\prime}\right) \mathrm{P}\left(\mathrm{S}_3^{\prime}\right)=\frac{1}{10} \mathrm{P}\left(\mathrm{S}_2^{\prime}\right) \cdot \mathrm{P}\left(\mathrm{S}_3^{\prime}\right) \\ & \Rightarrow \mathrm{P}\left(\mathrm{S}_1\right)=\frac{1}{10}\end{aligned}$ $\begin{aligned} & \mathrm{P}(\mathrm{~W})=\mathrm{P}\left(\mathrm{~S}_2 \cap \mathrm{~S}_3^{\prime}\right)=\frac{1}{12} \\ & \mathrm{P}\left(\mathrm{~S}_2\right) \cdot \mathrm{P}\left(\mathrm{~S}_3^{\prime}\right)=\frac{1}{12} \\ & \mathrm{P}\left(\mathrm{~S}_2\right)\left(1-\mathrm{P}\left(\mathrm{~S}_3\right)\right)=\frac{1}{12}...(2) \end{aligned}$ Eq. (1) $\left(1-\frac{1}{10}\right)\left(1-\mathrm{P}\left(\mathrm{~S}_2\right)\right)\left(1-\mathrm{P}\left(\mathrm{~S}_3\right)\right)=\frac{1}{2}$ $\left(1-\mathrm{P}\left(\mathrm{~S}_2\right)\right)\left(1-\mathrm{P}\left(\mathrm{~S}_3\right)\right)=\frac{5}{9}....(3)$ $\begin{aligned} & \frac{\text { Eq. }(2)}{\text { Eq. }(3)} \Rightarrow \frac{\mathrm{P}\left(\mathrm{~S}_2\right)}{1-\mathrm{P}\left(\mathrm{~S}_2\right)}=\frac{1}{12} \times \frac{9}{5} \\ & \mathrm{P}\left(\mathrm{~S}_2\right)=\frac{3}{23} \end{aligned}$ Put in Eq. (2) $\begin{aligned} & \frac{3}{23}\left(1-\mathrm{P}\left(\mathrm{~S}_3\right)\right)=\frac{1}{12} \\ & 1-\mathrm{P}\left(\mathrm{~S}_3\right)=\frac{23}{36} \\ & \mathrm{P}\left(\mathrm{~S}_3\right)=\frac{13}{36} \\ & \mathrm{P}(\mathrm{~T})=\frac{13}{36} \end{aligned}$
3. Q3JEE Advanced Adv 2024 (Paper 1)
   A student appears for a quiz consisting of only true-false type questions and answers all the questions. The student knows the answers of some questions and guesses the answers for the remaining questions. Whenever the student knows the answer of a question, he gives the correct answer. Assume that the probability of the student giving the correct answer for a question, given that he has guessed it, is $\frac{1}{2}$. Also assume that the probability of the answer for a question being guessed, given that the student's answer is correct, is $\frac{1}{6}$. Then the probability that the student knows the answer of a randomly chosen question is
   1. A.$\frac{1}{12}$
   2. B.$\frac{1}{7}$
   3. C.$\frac{5}{7}$
   4. D.$\frac{5}{12}$
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   Answer: (C)

   $\mathrm{C} \rightarrow$ Correct $\mathrm{G} \rightarrow$ Guess $\mathrm{K} \rightarrow$ Knows $\begin{array}{ll}\mathrm{P}\left(\frac{\mathrm{C}}{\mathrm{G}}\right)=\frac{1}{2} \quad, \quad \mathrm{P}\left(\frac{\mathrm{C}}{\mathrm{K}}\right)=1 \\\mathrm{P}\left(\frac{\mathrm{G}}{\mathrm{C}}\right)=\frac{1}{6}\end{array}$ $\begin{aligned} & \text { Let required probability }=\mathrm{x} \\ & \therefore \mathrm{P}\left(\frac{\mathrm{G}}{\mathrm{C}}\right)=\frac{(1-\mathrm{x}) \mathrm{P}\left(\frac{\mathrm{C}}{\mathrm{G}}\right)}{(1-\mathrm{x}) \mathrm{P}\left(\frac{\mathrm{C}}{\mathrm{G}}\right)+\mathrm{x} \cdot \mathrm{P}\left(\frac{\mathrm{C}}{\mathrm{K}}\right)} \\ & \frac{1}{6}=\frac{(1-\mathrm{x})\left(\frac{1}{2}\right)}{(1-\mathrm{x})\left(\frac{1}{2}\right)+(\mathrm{x})(1)} \\ & \Rightarrow \mathrm{x}=\frac{5}{7} \Rightarrow \text { Option (3) is correct. }\end{aligned}$
4. Q4JEE Advanced Adv 2023 (Paper 2)
   Consider an experiment of tossing a coin repeatedly until the outcomes of two consecutive tosses are same. If the probability of a random toss resulting in head is $\dfrac{1}{3}$, then the probability that the experiment stops with head is
   1. A.$\dfrac{1}{3}$
   2. B.$\dfrac{5}{21}$
   3. C.$\dfrac{4}{21}$
   4. D.$\dfrac{2}{7}$
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   Answer: (B)

   Given, $P(H)=\dfrac{1}{3}$ So, $P(T)=1-\dfrac{1}{3}=\dfrac{2}{3}$ Now probability of the event is given by, $P(E)=P(HH)+P(THH)+P(HTHH)+P(THTHH)+P(HTHTHH)+P(THTHTHH)+\ldots$ $\Rightarrow P(E)=\dfrac{1}{3}\times \dfrac{1}{3}+\dfrac{2}{3}\times \dfrac{1}{3}\times \dfrac{1}{3}+\dfrac{1}{3}\times \dfrac{2}{3}\times \dfrac{1}{3}\times \dfrac{1}{3}+\dfrac{4}{{3}^{5}}+\dfrac{4}{{3}^{6}}+\dfrac{8}{{3}^{7}}+\dfrac{8}{{3}^{8}}+\ldots$ $\Rightarrow P(E)=\dfrac{1}{{3}^{2}}+\dfrac{2}{{3}^{3}}+\dfrac{2}{{3}^{4}}+\dfrac{4}{{3}^{5}}+\dfrac{4}{{3}^{6}}+\dfrac{8}{{3}^{7}}+\dfrac{8}{{3}^{8}}+\ldots$ $\Rightarrow P(E)=\left(\dfrac{1}{{3}^{2}}+\dfrac{2}{{3}^{4}}+\dfrac{4}{{3}^{6}}+\ldots .\right)+\left(\dfrac{2}{{3}^{3}}+\dfrac{4}{{3}^{5}}+\dfrac{8}{{3}^{7}}+\ldots .\right)$ $\Rightarrow P(E)=\left(\dfrac{\dfrac{1}{{3}^{2}}}{\left(1-\dfrac{2}{{3}^{2}}\right)}\right)+\left(\dfrac{\dfrac{2}{{3}^{3}}}{\left(1-\dfrac{2}{{3}^{2}}\right)}\right)$ $\Rightarrow P(E)=\dfrac{1}{7}+\dfrac{2}{21}=\dfrac{5}{21}$
5. Q5JEE Advanced Adv 2021 (Paper 2)
   A number is chosen at random from the set $\left\{1,2,3,\ldots ,2000\right\}$. Let $p$ be the probability that the chosen number is a multiple of 3 or $7$. Then the value of $500p$ is ____.
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   Answer: 214

   Multiple of $3=3,6,9\ldots .1998$ are $666$ Multiple of $7=7,14\ldots .1995$ are $285$ Multiple of $21=21,42\ldots 1995$ are $95$ $p=\dfrac{666+285-95}{2000}=\dfrac{856}{2000}=\dfrac{214}{500}=p$ $500p=214$