(a)
Here, $n = 6$
According to the question
$\ ^{6}C_{2}p^{2}q^{4} = 4.{\ ^{6}C}_{4}p^{4}q^{2}$
$\Rightarrow \ q^{2} = 4p^{2}$
$\Rightarrow \ (1 - p)^{2} = 4p^{2}$
$\Rightarrow \ 3p^{2} + 2P - 1 = 0$
$\Rightarrow \ (p + 1)(3p - 1) = 0$
$\Rightarrow p = \frac{1}{3}(\because p\ cannot\ be\ negative)$