Question

A man is known 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 really a six is

• A. $\frac{3}{4}$
• B. $\frac{1}{2}$
• C. $\frac{3}{5}$
• D. $\frac{3}{8}$

Answer 1

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

Let $E_1,E_2$ and $A$ be the events defined as follows:

$E_1=$ die shows six, i.e., six has occured.

$E_2=$ dei does not show six i.e., six has not occured.

$A=$ the man reports that six has occured.

We have to calculate that six has actually occured

given that the man reports that six occurs i.e., $P\left(\frac{E_1}{A}\right)$

Now, $P\left(E_1\right)=\frac{1}{6},P\left(E_2\right)=\frac{5}{6}$

$P\left(\frac{A}{E_1}\right)=$ probability that the man reports that six occured given that six has occured.

$=$ probability that the man is reporting the truth.$=\frac{3}{4}$

And $P\left(\frac{A}{E_2}\right)=$probability that the man reports that six occured given that six has not occured.

$=$ probability that the man does not speak truth $=\frac{1}{4}$

By Bayes' Theorem,

$P\left(\frac{E_1}{A}\right)=\frac{P\left(E_1\right)P\left(\frac{A}{E_1}\right)}{P\left(E_1\right)P\left(\frac{A}{E_1}\right)+P\left(E_2\right)P\left(\frac{A}{E_2}\right)}$

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