The random variable X has the following probability distribution

X	0	1	2	3
P(X)	a	a	b	b

such that E(x²) = 2E(x), then the value of b is:

The sum of all probabilities must equal 1.

From the table:

• P(X=0) = a
• P(X=1) = a
• P(X=2) = b
• P(X=3) = b

$a + a + b + b = 1$

$2a + 2b = 1$

$a + b = \frac{1}{2}$ ... (Equation 1)

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The expected value is calculated as:

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

$E(X) = 0 + a + 2b + 3b$

$E(X) = a + 5b$

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Similarly:

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

$E(X^2) = 0 + a + 4b + 9b$

$E(X^2) = a + 13b$

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Using the given condition $E(X^2) = 2E(X)$:

$a + 13b = 2(a + 5b)$

$a + 13b = 2a + 10b$

$13b - 10b = 2a - a$

$3b = a$ ... (Equation 2)

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Substituting Equation 2 into Equation 1:

$3b + b = \frac{1}{2}$

$4b = \frac{1}{2}$

$b = \frac{1}{8}$

Therefore, the value of $b = \frac{1}{8}$.