Question

A man is known to speak truth 3 out of 4 times. He throws a dice and reports that it is a six. Find the probability that it is actually a six.

Answer 1

Let $E_1,E_2$ and A be the events that six occurs, six does not occur and the man reports that it is six

$\Rightarrow P\left(E_1\right)=\frac{1}{6},P\left(E_2\right)=\frac{5}{6}P\left(A/E_1\right)=\frac{3}{4},P\left(A/E_2\right)=\frac{1}{4}P\left(\frac{E_1}{A}\right)=\frac{P\left(E_1\right)P\left(A/E_1\right)}{P\left(E_1\right)P\left(A/E_1\right)+P\left(E_2\right)\left(A/E_2\right)}=\frac{3}{8}$