A man is known to speak the truth on an average 4 out of 5 times. He throws a die and reports that it is a five. The probability that it is actually a five is

Let A be the event that the die shows a five. Let B be the event that the die does not show a five. And let E be the event that the person reports that the die shows a five.

Probability of getting a five $$P\left(A\right)=\dfrac{1}{6}$$

Probability of not getting a five $$P\left(B\right)=\dfrac{5}{6}$$

Now, we will find the conditional probabilities.

The probability of the number being reported as five, and it is five (he is telling the truth) $$P(E|A)=\dfrac{4}{5}$$

The probability of the number being reported as five, but it is not five (he is telling the lie) $$P(E|B)=\dfrac{1}{5}$$

Now, we will apply Bayes' rule. Since events A and B are mutually exclusive -

=> $$P(A|E)=\dfrac{P(E|A)\cdot P\left(A\right)}{P(E|B)\cdot P\left(B\right)+P(E|A)\cdot P\left(A\right)}$$

=> $$P(A|E)=\dfrac{\frac{4}{5}\times\frac{1}{6}}{\frac{1}{5}\times\frac{5}{6}+\frac{4}{5}\times\frac{1}{6}}$$

=> $$P(A|E)=\dfrac{\frac{4}{30}}{\frac{9}{30}}$$

=> $$P(A|E)=\dfrac{4}{9}$$