Probability JEE Main PYQs (2020–2026)

Solved Probability PYQs

3 recent questions with answers. The full set of 173 is free to practise.

1. JEE Main 2026MediumConditional Probability & Multiplication Rule
   Q1

   A bag contains 6 blue and 6 green balls. Pairs of balls are drawn without replacement until the bag is empty. The probability that each drawn pair consists of one blue and one green ball is :

   • A$\frac{63}{925}$
   • B$\frac{17}{231}$
   • C$\frac{16}{231}$
   • D$\frac{64}{925}$

   Show answer & solution

   Correct answer: (C)

   We want every pair drawn to have one blue and one green. Since there are 6 blue and 6 green balls, we will draw a total of 6 pairs.
   
   For each pair to be mixed (one of each colour), we must match the 6 blue balls with the 6 green balls in pairs. The number of such “good” pairings is counted by choosing and matching one ball of one colour with one ball of the other colour.
   
   The total number of ways to split 12 balls into 6 pairs (without caring about the order of drawing inside a pair) is counted separately. The required probability is:
   
   Probability $=\frac{\lfloor 6\times \lfloor 6}{\left(\frac{12}{(\lfloor 2{)}^6\lfloor 6}\right)\times \lfloor 6}=\frac{16}{231}$
   
   So, the probability that each drawn pair consists of one blue and one green ball is $\frac{16}{231}$ .
2. JEE Main 2026EasyLaw of Total Probability & Bayes' Theorem
   Q2

   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 :

   • A0.74
   • B0.76
   • C0.72
   • D0.78

   Show answer & solution

   Correct answer: (B)

   Let
   
   $A$ = event that player A is selected as captain
   
   $B$ = event that player B is selected as captain
   
   $W$ = event that the team wins the tournament
   
   Given:
   
   $$P(A)=0.6,P(B)=0.4$$
   
   Also,
   
   $$P(W|A)=0.8,P(W|B)=0.7$$
   
   We have to find $P(W)$ .
   
   Using the formula of total probability:
   
   $$P(W)=P(A)P(W|A)+P(B)P(W|B)$$
   
   Substituting the values:
   
   $$P(W)=0.6\times 0.8+0.4\times 0.7$$
   
   $$P(W)=0.48+0.28$$
   
   $$P(W)=0.76$$
   
   So, the required probability is
   
   $$\boxed{0.76}$$
   
   Hence, the correct option is Option B.
3. JEE Main 2026EasyBinomial Distribution (Bernoulli Trials)
   Q3

   A man throws a fair coin repeatedly. He gets 10 points for each head he throws and 5 points for each tail he throws. If the probability that he gets exactly 30 points is $\frac{m}{n}$ , gcd $(m,n)=1$ , then m + n is equal to :

   • A53
   • B55
   • C107
   • D105

   Show answer & solution

   Correct answer: (C)

   $$\begin{array}{rl}& P=(6\text{ tail })+(4\text{ tail }+1\text{ head })+(2\text{ tail }+2\text{ head })+(3\text{ head })\\ & P={\left(\frac{1}{2}\right)}^6+\frac{5!}{1!4!}{\left(\frac{1}{2}\right)}^5+\frac{4!}{2!2!}{\left(\frac{1}{2}\right)}^4+{\left(\frac{1}{2}\right)}^3\\ & P=\frac{1}{64}+\frac{5}{32}+\frac{3}{8}+\frac{1}{8}\\ & P=\frac{1+10+24+8}{64}=\frac{43}{64}=\frac{m}{n}\\ & m+n=107\end{array}$$

Topic-wise question count

Where JEE Main has asked from Probability, most frequent first.

• Classical Probability, Sample Space & Addition Theorem70
• Law of Total Probability & Bayes' Theorem26
• Binomial Distribution (Bernoulli Trials)25
• Mean and Variance of a Random Variable19
• Conditional Probability & Multiplication Rule14
• Independent Events11
• Random Variables & Probability Distributions8

Year-wise Probability questions in JEE Main

Number of questions from this chapter in each year's paper(s).