The random variable X can take values 0, 1, 2. If $P(X=0)=P(X=1)=\alpha$, and $E(X^2)=E(X)$, then which of the following are correct?

(A) $E(X) = 2-3\alpha$

(B) $E(X^2) = 4+7\alpha$

(C) $\alpha = \frac{1}{2}$

(D) $\alpha = \frac{1}{5}$

Choose the correct answer from the options given below:

Given: X can take values 0, 1, 2 with $P(X=0) = P(X=1) = \alpha$ and $E(X^2) = E(X)$.

Since all probabilities must sum to 1:

$P(X=0) + P(X=1) + P(X=2) = 1$

$\alpha + \alpha + P(X=2) = 1$

$P(X=2) = 1 - 2\alpha$

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Calculate $E(X)$:

$E(X) = 0 \cdot P(X=0) + 1 \cdot P(X=1) + 2 \cdot P(X=2)$

$E(X) = 0 \cdot \alpha + 1 \cdot \alpha + 2 \cdot (1-2\alpha)$

$E(X) = \alpha + 2 - 4\alpha$

$E(X) = 2 - 3\alpha$

Statement (A) is correct.

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Calculate $E(X^2)$:

$E(X^2) = 0^2 \cdot P(X=0) + 1^2 \cdot P(X=1) + 2^2 \cdot P(X=2)$

$E(X^2) = 0 \cdot \alpha + 1 \cdot \alpha + 4 \cdot (1-2\alpha)$

$E(X^2) = \alpha + 4 - 8\alpha$

$E(X^2) = 4 - 7\alpha$

Statement (B) claims $E(X^2) = 4 + 7\alpha$, which is incorrect.

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

$4 - 7\alpha = 2 - 3\alpha$

$4 - 2 = 7\alpha - 3\alpha$

$2 = 4\alpha$

$\alpha = \frac{1}{2}$

Statement (C) is correct.

Statement (D) claims $\alpha = \frac{1}{5}$, which is incorrect.

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Therefore, statements (A) and (C) are correct.