The random variable X has the following probability distribution

such that $E(x^2) = 2E(X)$, then the value of $b$ is:

$\frac{1}{8}$

The correct answer is Option (1) → $\frac{1}{8}$

Given probability distribution:

$X: 0, 1, 2, 3$

$P(X): a, a, b, b$

Sum of probabilities: $2a + 2b = 1 \Rightarrow a + b = \frac{1}{2}$

Given: $E(X^2) = 2E(X)$

Compute $E(X)$:

$E(X) = 0\cdot a + 1\cdot a + 2\cdot b + 3\cdot b = a + 5b$

Compute $E(X^2)$:

$E(X^2) = 0^2\cdot a + 1^2\cdot a + 2^2\cdot b + 3^2\cdot b = a + (4+9)b = a + 13b$

Set $E(X^2) = 2E(X)$:

$a + 13b = 2(a + 5b) \Rightarrow a + 13b = 2a + 10b \Rightarrow 3b - a = 0 \Rightarrow a = 3b$

Also, $a + b = \frac{1}{2} \Rightarrow 3b + b = 4b = \frac{1}{2} \Rightarrow b = \frac{1}{8}$

$\text{Answer: } b = \frac{1}{8}$