Probability: JEE Main Previous Year Questions (2021-2026)

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JEE Main Previous Year Questions (20 1 2 -2026): 
Probability 
 
(January 2026) 
 
Q1: The probability distribution of a random variable X is given below : 
X 4k 30/7 k 32/7 k 34/7 k 36/7 k 38/7 k 40/7 k 6k 
P(X) 2/15 1/15 2/15 1/5 1/15 2/15 1/5 1/15 
If E(X) = 263/15, then P(X < 20) is equal to:  
(a) 3/5  
(b) 14/15  
(c) 8/15  
(d) 11/15 
Ans: (d) 
Sol: 
The expected value of a discrete random variable is given by the formula :  
E(X) = ?X
i
 · P (X
i
)  
Using the data from the table: 
 
Take k/15 factor common 
 
 
 
Given that E(X) = 263/15: 
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JEE Main Previous Year Questions (20 1 2 -2026): 
Probability 
 
(January 2026) 
 
Q1: The probability distribution of a random variable X is given below : 
X 4k 30/7 k 32/7 k 34/7 k 36/7 k 38/7 k 40/7 k 6k 
P(X) 2/15 1/15 2/15 1/5 1/15 2/15 1/5 1/15 
If E(X) = 263/15, then P(X < 20) is equal to:  
(a) 3/5  
(b) 14/15  
(c) 8/15  
(d) 11/15 
Ans: (d) 
Sol: 
The expected value of a discrete random variable is given by the formula :  
E(X) = ?X
i
 · P (X
i
)  
Using the data from the table: 
 
Take k/15 factor common 
 
 
 
Given that E(X) = 263/15: 
 
 
 
Calculate P (X < 20):  
Sum the probabilities for X = 14, 15, 16, 17, 18, and 19: 
P(X < 20) = P(X = 14) + P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) 
 
 
Q2: A bag contains 10 balls out of which k are red and (10 - k) are black, where 0 = k = 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:  
(a) 7/110  
(b) 7/11  
(c) 7/55  
(d) 14/55