Explanation

(a)
Probability of getting $4 = \frac{1}{6}$
Probability of not getting $4 = 1 - \frac{1}{6} = \frac{5}{6}$
$\therefore$ Probability of getting 4 at least once in two throw of dice
$= {\ ^{2}C}_{1}\left( \frac{1}{6} \right)\left( \frac{5}{6} \right) +^{2}C_{2}\left( \frac{1}{6} \right)^{2}$
$= 2.\frac{5}{36} + 1.\frac{1}{36} = \frac{11}{36}$