Explanation

(a)
Probability of occurrence of $4 = \frac{1}{6}$
Probability of non-occurrence of $4 = \frac{5}{6}$
Required probability
$= {\ ^{2}C}_{1}\left( \frac{1}{6} \right)\ \left( \frac{5}{6} \right) + {\ ^{2}C}_{2}\left( \frac{1}{6} \right)^{2}\left( \frac{5}{6} \right)^{0}$
$= 2 \bullet \frac{5}{36} + 1 \bullet \frac{1}{36} = \frac{10}{36} + \frac{1}{36} = \frac{11}{36}$