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 6 is 

$E_1:$ die shows a six

$E_2:$ die doesn't show a six 

$\mathrm{A}:$ man reports it is a six 

We have 

$\begin{array}{l}P\left(E_1\right)=\frac{1}{6},P\left(E_2\right)=\frac{5}{6}\\ P\left(A\mid E_1\right)=\frac{3}{4},P\left(A\mid E_2\right)=\frac{1}{4}\end{array}$

By the Bayes' rule 

$\begin{array}{l}P\left(E_1\mid A\right)=\frac{P\left(E_1\right)P\left(A\mid E_1\right)}{P\left(E_1\right)P\left(A\mid E_1\right)+P\left(E_2\right)P\left(A\mid E_2\right)}\\ =\frac{(1/6)(3/4)}{(1/6)(3/4)+(5/6)(1/4)}\\ =3/8\end{array}$