Question

A is known to speak truth 3 times out of 5 times. He throws a die and reports that it is one. Find the probability that it is actually one.

Answer 1

Let A, E₁ and E₂ denote the events that the man reports the appearance of 1 on throwing a die, 1 occurs and 1 does not occur, respectively.

$$\begin{gathered} \therefore P\left(E_1\right)=\frac{1}{6} \\ P\left(E_2\right)=\frac{5}{6} \\ \mathrm{Now}, \\ P\left(A/E_1\right)=\frac{3}{5} \\ P\left(A/E_2\right)=\frac{2}{5} \\ \mathrm{Using}\mathrm{Bayes}\text{'}\mathrm{theorem},\mathrm{we}\mathrm{get} \\ \mathrm{Required}\mathrm{probability}=P\left(E_1/A\right)=\frac{P\left(E_1\right)P\left(A/E_1\right)}{P\left(E_1\right)P\left(A/E_1\right)+P\left(E_2\right)P\left(A/E_2\right)} \\ =\frac{{\displaystyle \frac{1}{6}}\times {\displaystyle \frac{3}{5}}}{{\displaystyle \frac{1}{6}}\times {\displaystyle \frac{3}{5}}+{\displaystyle \frac{5}{6}}\times {\displaystyle \frac{2}{5}}} \\ =\frac{3}{3+10}=\frac{3}{13} \end{gathered}$$