A man is known to speak truth 3 out of 4 times. He throws a die and r…

The correct answer is Option (3) → $\frac{3}{8}$

Let $T$ be the event that the man speaks truth

$P(T)=\frac{3}{4},\;P(T')=\frac{1}{4}$

Let $A$ be the event that the die actually shows $4$

$P(A)=\frac{1}{6},\;P(A')=\frac{5}{6}$

Let $B$ be the event that the man reports $4$

If the man speaks truth, he reports $4$ only when the die is actually $4$

$P(B|A)=\frac{3}{4}$

If the man lies, he reports $4$ even when the die is not $4$

$P(B|A')=\frac{1}{4}$

Using Bayes’ theorem

$P(A|B)=\frac{P(B|A)P(A)}{P(B|A)P(A)+P(B|A')P(A')}$

$=\frac{\frac{3}{4}\cdot\frac{1}{6}}{\frac{3}{4}\cdot\frac{1}{6}+\frac{1}{4}\cdot\frac{5}{6}}$

$=\frac{\frac{3}{24}}{\frac{3}{24}+\frac{5}{24}}$

$=\frac{3}{8}$

The probability that the number is actually $4$ is $\frac{3}{8}$.