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(B) Step $1$: The mean $E(X) = \sum x_i P(x_i) = (0 \cdot q^3) + (1 \cdot 3pq^2) + (2 \cdot 3p^2q) + (3 \cdot p^3) = 3pq^2 + 6p^2q + 3p^3 = 3p(q^2 + 2

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$3pq$

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(B) Step $1$: The mean $E(X) = \sum x_i P(x_i) = (0 \cdot q^3) + (1 \cdot 3pq^2) + (2 \cdot 3p^2q) + (3 \cdot p^3) = 3pq^2 + 6p^2q + 3p^3 = 3p(q^2 + 2pq + p^2) = 3p(q+p)^2$. Since $q+p=1$, $E(X) = 3p$.Step $2$: Calculate $E(X^2) = \sum x_i^2 P(x_i) = (0^2 \cdot q^3) + (1^2 \cdot 3pq^2) + (2^2 \cdot 3p^2q) + (3^2 \cdot p^3) = 3pq^2 + 12p^2q + 9p^3$.Step $3$: Variance $Var(X) = E(X^2) - [E(X)]^2 = (3pq^2 + 12p^2q + 9p^3) - (3p)^2 = 3pq^2 + 12p^2q + 9p^3 - 9p^2$.Step $4$: Substitute $q = 1-p$: $Var(X) = 3p(1-p)^2 + 12p^2(1-p) + 9p^3 - 9p^2 = 3p(1 - 2p + p^2) + 12p^2 - 12p^3 + 9p^3 - 9p^2 = 3p - 6p^2 + 3p^3 + 3p^2 - 3p^3 = 3p - 3p^2 = 3p(1-p) = 3pq$.

The variance of the following probability distribution of $X$ is given by:
$X: 0, 1, 2, 3$
$P(X): q^3, 3pq^2, 3p^2q, p^3$
where $0 < p < 1$ and $q = 1 - p$.

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The variance of the following probability distribution of $X$ is given by: $X: 0, 1, 2, 3$$P(X): q^3, 3pq^2, 3p^2q, p^3$where $0 < p < 1$ and $q = 1 - p$.