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

$P\left(truth\right)=P\left(T\right)=\frac{3}{4}\text{\hspace{0.17em}⇒}P\left(False\right)=P\left(F\right)=\frac{1}{4}$    $P\left(gets6\right)=\frac{1}{6}$ 

  $P\left(E\right)=P\left(getting6\right)=\frac{1}{6}\Rightarrow P\left(\overline{E}\right)=P\left(notgetting6\right)=\frac{5}{6}$

A= Reporting  that the result is  6

           =$$\begin{gathered} P(E/A)=\frac{P\left(E\right)P(A/E)}{P\left(E\right)P(A/E)+P\left(\overline{E}\right)P(A/\overline{E})} \\ P(E/A)=\frac{\frac{3}{4}.\frac{1}{6}}{\frac{3}{4}.\frac{1}{6}+\frac{1}{4}.\frac{5}{6}} \end{gathered}$$

           $=\frac{3}{8}$

(by Bayes' theorem)