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

The correct answer is $\dfrac{3}{8}$

Let $E_1$ be the event that six occurs on the die and $E_2$ be the event that six does not occur.

Probability of $E_1$ = $P(E_1) = \dfrac{1}{6}$


Probability of $E_2$ = $P(E_2) = \dfrac{5}{6}$


Let $A$ be the condition that the man reports that it is a six.

∴ $P(A|E_1)$ = $\dfrac{3}{4}$ and $P(A|E_2)$ = $\dfrac{1}{4}$


Probability that it is actually a six $= P(E_1|A) =$ $\dfrac{P(E_1) P(A|E_1)}{P(E_1) P(A|E_1) + P(E_2) P(A|E_2)}$


= $\dfrac{\dfrac{1}{6} \dfrac{3}{4}}{\dfrac{1}{6} \dfrac{3}{4} + \dfrac{5}{6} \dfrac{1}{4}}$ $= \dfrac{3}{8}$