Explanation

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