A man known to speak truth $3$ out of $4$ times throws a die and reports that it is a six. The probability that it is actually a six is:

Let: $T$ = speaks truth, $P(T) = \dfrac{3}{4}$ $F$ = lies, $P(F) = \dfrac{1}{4}$ $S$ = shows six on die, $P(S) = \dfrac{1}{6}$. We want $P(S|R)$ where $R$ = reports six. By Bayes’ theorem: $P(S|R) = \dfrac{P(R|S)P(S)}{P(R|S)P(S) + P(R|\bar S)P(\bar S)}$ $P(R|S)=\dfrac{3}{4},\ P(R|\bar S)=\dfrac{1}{4}$ $\Rightarrow P(S|R) = \dfrac{(\tfrac{3}{4})(\tfrac{1}{6})} {(\tfrac{3}{4})(\tfrac{1}{6}) + (\tfrac{1}{4})(\tfrac{5}{6})} = \dfrac{3}{8}.$