Explanation

(a)
Let the probability of success and failure are $p$ and $q$ respectively
$\therefore\ p = 2q$ and $p + q = 1$
$\Rightarrow \ 3q = 1\ \Rightarrow \ q = \frac{1}{3}\ and\ p = \frac{2}{3}$
$\therefore$ Required probability
$= {\ ^{6}C}_{4}\left( \frac{2}{3} \right)^{4}\left( \frac{1}{3} \right)^{2} +^{6}C_{5}\left( \frac{2}{3} \right)^{5}\left( \frac{1}{3} \right) +^{6}C_{6}\left( \frac{2}{3} \right)^{6}$
$= \frac{240}{729} + \frac{192}{729} + \frac{64}{729} = \frac{496}{729}$