Let E_{1}, E_{2} and A be the events defined as follows:

**E**_{1} = Six occurs,

**E**_{2} = Six does not occur

**A = man reports it is a six **

Then, P(E_{1}) = \(\frac{1}{6}\) , P(E_{2}) = 1- \(\frac{1}{6}\) = \(\frac{5}{6}\)

P(A/E_{1}) = Probability of man reporting it a six when six occurs = Probability of speaking truth = \(\frac{3}{4}\)

P(A/E_{2}) = 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**)

### = \(\frac{\frac{1}{6}\times \frac{3}{4}}{\frac{1}{6}\times \frac{3}{4}+\frac{5}{6}\times \frac{1}{4}}\) = \(\frac{\frac{3}{24}}{\frac{8}{24}}\) = **\(\frac{3}{8}\)**