The Variance of the following probability distribution of X is

X	$0$	$1$	$2$	$3$
P(X)	$q^3$	$3q^2 p$	$3qp^2$	$p^3$

where $0 < p < 1, q = 1 - p$

The given probability distribution represents a binomial distribution with parameters $n = 3$ and probability of success $p$, since $P(X = x) = ^{3}C_{x} p^x q^{3-x}$ for $x = 0, 1, 2, 3$.

For a binomial distribution, the variance is given by $npq$.

Substituting $n = 3$, the variance is $3pq$.

Alternatively, calculating the variance using expectations:

$E(X) = \sum x P(x) = 0(q^3) + 1(3q^2 p) + 2(3qp^2) + 3(p^3)$

$E(X) = 3p(q^2 + 2pq + p^2) = 3p(q+p)^2 = 3p$

$E(X^2) = \sum x^2 P(x) = 0(q^3) + 1(3q^2 p) + 4(3qp^2) + 9(p^3)$

$E(X^2) = 3p(q^2 + 4pq + 3p^2) = 3p((q+p)^2 + 2p(q+p)) = 3p(1 + 2p) = 3p + 6p^2$

$\text{Variance} = E(X^2) - (E(X))^2$

$\text{Variance} = 3p + 6p^2 - (3p)^2 = 3p - 3p^2 = 3p(1-p) = 3pq$

Answer: $3pq$