Let a random variable X take values {0, 1, 2, | vedclass

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(B) The sum of probabilities must be $1$: $P(X=0) + P(X=1) + P(X=2) + P(X=3) = 1$ $p + p + q + q = 1 \implies 2p + 2q = 1 \implies p + q = \frac{1}{2}

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$2$

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(B) The sum of probabilities must be $1$: $P(X=0) + P(X=1) + P(X=2) + P(X=3) = 1$ $p + p + q + q = 1 \implies 2p + 2q = 1 \implies p + q = \frac{1}{2}$ Now,calculate the expected value $E(X)$: $E(X) = \sum x_i P(X=x_i) = 0(p) + 1(p) + 2(q) + 3(q) = p + 5q$ Calculate the expected value $E(X^2)$: $E(X^2) = \sum x_i^2 P(X=x_i) = 0^2(p) + 1^2(p) + 2^2(q) + 3^2(q) = p + 13q$ Given $E(X^2) = 2E(X)$: $p + 13q = 2(p + 5q)$ $p + 13q = 2p + 10q$ $p = 3q$ Substitute $p = 3q$ into $p + q = \frac{1}{2}$: $3q + q = \frac{1}{2} \implies 4q = \frac{1}{2} \implies q = \frac{1}{8}$ Then $p = 3(\frac{1}{8}) = \frac{3}{8}$ Finally,calculate $8p - 1$: $8(\frac{3}{8}) - 1 = 3 - 1 = 2$

Let a random variable $X$ take values $\{0, 1, 2, 3\}$ with $P(X=0) = P(X=1) = p$,$P(X=2) = P(X=3) = q$,and $E(X^2) = 2E(X)$. Then the value of $8p - 1$ is:

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