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CUET · MATHS · PYQ PAPER 2025

The random variable X has the following probability distribution
X0123
P(X)aabb
such that E\(\left(x^2\right)\) = 2E(x), then the value of b is:

  1. A \(\frac{1}{8}\)
  2. B \(\frac{1}{4}\)
  3. C \(\frac{1}{2}\)
  4. D \(\frac{3}{8}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{1}{8}\)

Step-by-step Solution

Detailed explanation

\(\sum P(X) = 1 \implies 2a + 2b = 1\) \(E(X) = 0a + 1a + 2b + 3b = a + 5b\) \(E(X^2) = 0^2a + 1^2a + 2^2b + 3^2b = a + 13b\) \(E(X^2) = 2E(X) \implies a + 13b = 2(a + 5b)\) \(a + 13b = 2a + 10b \implies 3b = a\) \(2(3b) + 2b = 1 \implies 8b = 1 \implies b = \frac{1}{8}\)
From CUET
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