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

$\frac{3}{8}$

Let E denote the event that a six occurs and A the event that the man reports that it is a ‘6’, we have

$P(E)=\frac{1}{6},P(E')=\frac{5}{6}, P(\frac{A}{E})=\frac{3}{4}$ and $P(\frac{A}{E'})=\frac{1}{4}.$

By Baye's theorem , $P(\frac{E}{A})=\frac{P(E).P(A/E)}{P(E).P(A/E)+P(E')P(A/E')}=\frac{\frac{1}{6}×\frac{3}{4}}{\frac{1}{6}×\frac{3}{4}+\frac{5}{6}×\frac{1}{4}}=\frac{3}{8}.$