Question

A man is known to speak the truths $3$ out of $4$ times. He throw a die and report that it is six. The probability that it is actually a six, is

• A. $\frac{3}{8}$
• B. $\frac{1}{5}$
• C. $\frac{3}{4}$
• D. None of these

Answer 1

The correct option is A $\frac{3}{8}$

Let $E$ denote the event that a six occurs and $A$ be the event that the man reports that it is a $6$.
We have,
$P\left(E\right)=\frac{1}{6},P\left(E^{\prime }\right)=\frac{5}{6}$
$P\left(\frac{A}{E}\right)=\frac{3}{4}$ and $P\left(\frac{A}{E^{\prime }}\right)=\frac{1}{4}$
Using Baye's theorem, we get
$P\left(\frac{E}{A}\right)=\frac{P\left(E\right)\cdot P\left(\frac{A}{E}\right)}{P\left(E\right)\cdot P\left(\frac{A}{E}\right)+P\left(E^{\prime }\right)\cdot P\left(\frac{A}{E}\right)}$
$=\frac{\frac{1}{6}\times \frac{3}{4}}{\frac{1}{6}\times \frac{3}{4}+\frac{5}{6}\times \frac{1}{4}}=\frac{3}{8}$.