Question

A man is know to speak the truth $3$ out of $4$ times. He throws a die and reports that it is a 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 $E$ denotes the event that a six occurs and

$A$ the event that the man reports that it is a six.

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 $P\left(\frac{A}{E^{\prime }}\right)=\frac{1}{4}$

$\therefore$By baye's therefore $P\left(\frac{E}{A}\right)=\frac{P\left(E\right)P\left(\frac{A}{E}\right)}{P\left(E\right)P\left(\frac{A}{E}\right)+P\left(E^{\prime }\right)P\left(\frac{A}{E^{\prime }}\right)}$

$\Rightarrow \frac{\frac{1}{6}\times \frac{3}{4}}{\frac{1}{6}\times \frac{3}{4}+\frac{5}{6}\times \frac{1}{4}}=\frac{3}{8}$