Let E1, E2 and A be the events defined as follows:
E1 = Six occurs,
E2 = Six does not occur
A = man reports it is a six
Then, P(E1) = \(\frac{1}{6}\) , P(E2) = 1- \(\frac{1}{6}\) = \(\frac{5}{6}\)
P(A/E1) = Probability of man reporting it a six when six occurs = Probability of speaking truth = \(\frac{3}{4}\)
P(A/E2) = Probability of man reporting a six when six does not occur
= Probability of not speaking truth = 1-\(\frac{3}{4}\) = \(\frac{1}{4}\)
∴ P(Throw is actually a six) =\(\frac{P(E_1)\times P(A/E_1)}{P(E_1)\times P(A/E_1)+P(E_2)\times P(A/E_2)}\) ... (Baye’s Theorem)
= \(\frac{\frac{1}{6}\times \frac{3}{4}}{\frac{1}{6}\times \frac{3}{4}+\frac{5}{6}\times \frac{1}{4}}\) = \(\frac{\frac{3}{24}}{\frac{8}{24}}\) = \(\frac{3}{8}\)
