JEE Main 2022 · 24 June · evening · Single correct · +4 / −1

A random variable X has the following probability distribution :
The value of P(1 < X < 4 | X ≤ 2) is equal to :

X01234
P(X)k2k4k6k8k
  1. (a) 4/7
  2. (b) 2/3
  3. (c) 3/7
  4. (d) 4/5
Answer and solution

Answer (a) 4/7

Solution

Maths sir · 4 steps

Given X ≤ 2, only X = 0, 1, 2 can happen, and among those only X = 2 lies strictly between 1 and 4.

  1. Find k. All the probabilities add up to 1, so k + 2k + 4k + 6k + 8k = 21k = 1. That gives k = 1/21, although k will cancel at the end anyway.

    P(A | B) = P(A ∩ B)/P(B)

  2. Denominator. B is X ≤ 2, which means X = 0, 1, 2. So P(B) = k + 2k + 4k = 7k.
  3. Numerator. A is 1 < X < 4, which means X = 2, 3. Of these, only X = 2 also satisfies X ≤ 2, so P(A ∩ B) = P(X = 2) = 4k.
  4. So. P(1 < X < 4 | X ≤ 2) = 4k/7k = 4/7, which is option (a).

The idea to keep. For a conditional probability, first shrink the sample space to the given condition, then keep only the values of the event that survive inside it. Read strict and non-strict inequalities carefully.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Probability — Advanced

Asked about 100 times in JEE Main since 2020. This question is from 2022.