Explanation

(a)
Let $E$ denote the event that a six occurs and $A$ the event that the man reports that it is a ‘6’. We have
$P(E) = \frac{1}{6},P^{'}(E) = \frac{5}{6}$
$P\left( \frac{A}{E} \right) = \frac{3}{4}and\ P\left( \frac{A}{E'} \right) = \frac{1}{4}$
From Baye’s theorem,
$P\left( \frac{E}{A} \right) = \frac{P(E).P\left( \frac{A}{E} \right)}{P(E).P\left( \frac{A}{E} \right) + P\left( E^{'} \right).P\left( \frac{A}{E^{'}} \right)}$
$= \frac{\frac{1}{6} \times \frac{3}{4}}{\frac{1}{6} \times \frac{3}{4} + \frac{5}{6} \times \frac{1}{4}} = \frac{3}{8}$