Probability that a man speaks truth is $\frac{3}{4}$. He throws a die and reports that it is a six. The probability that it is actually a six is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 3/8
Let:
$E_1$ = event that a six actually appears.
$E_2$ = event that man reports a six.
Given:
$P(E_1)=\frac{1}{6}$, $P(E_1')=\frac{5}{6}$, $P(\text{truth})=\frac{3}{4}$, $P(\text{lie})=\frac{1}{4}$.
$P(E_2|E_1)=\frac{3}{4}$ (he speaks truth when a six appears)
$P(E_2|E_1')=\frac{1}{4}$ (he lies and reports six when it’s not six)
$P(E_1|E_2)=\frac{P(E_1)P(E_2|E_1)}{P(E_1)P(E_2|E_1)+P(E_1')P(E_2|E_1')}$
$=\frac{\frac{1}{6}\cdot\frac{3}{4}}{\frac{1}{6}\cdot\frac{3}{4}+\frac{5}{6}\cdot\frac{1}{4}}$
$=\frac{\frac{3}{24}}{\frac{3}{24}+\frac{5}{24}}=\frac{3}{8}$