JEE Main 2020 · 9 January · evening · Single correct · +4 / −1

A random variable X has the following probability distribution :
Then P(X > 2) is equal to :

X:12345
P(X):K²2KK2K5K²
  1. (a) 1/6
  2. (b) 7/12
  3. (c) 1/36
  4. (d) 23/36
Answer and solution

Answer (d) 23/36

Solution

Maths sir · 4 steps

All the probabilities must add to 1, so that equation pins down K, and then you add the right ones.

  1. Total is 1. Add every probability in the table and set the sum equal to 1: K² + 2K + K + 2K + 5K² = 1.

    6K² + 5K - 1 = 0 ⇒ (6K - 1)(K + 1) = 0

  2. Choose K. The roots are K = 1/6 and K = -1. A probability cannot be negative, so K = 1/6.
  3. Add the needed terms. X > 2 means X = 3, 4, 5, so P(X > 2) = K + 2K + 5K² = 3K + 5K² = 1/2 + 5/36. As a check, the complement gives 1 - P(X = 1) - P(X = 2) = 1 - 1/36 - 12/36, which is the same value.
  4. So. P(X > 2) = 18/36 + 5/36 = 23/36, which is option (d).

The idea to keep. When a distribution has an unknown constant, first use the fact that all probabilities sum to 1. Throw out any root that makes a probability negative.

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 2020.