Explanation

(c)
Given, $n = 6$ and
$P(X = 2) = 9P(X = 4)$
$\Rightarrow {\ ^{6}C}_{2}p^{2}q^{4} = {9.}^{6}C_{4}p^{4}q^{2}$
$\Rightarrow 9p^{2} = q^{2}$
$\Rightarrow P = \frac{1}{3}q$
$\therefore$ We know that $p + q = 1$
$\Rightarrow \ \frac{q}{3} + q = 1$
$\Rightarrow \ q = \frac{3}{4}\ and\ p = \frac{1}{4}$
$\therefore$ Variance $= npq$
$= 6.\frac{1}{4}.\frac{3}{4} = \frac{9}{8}$