Question

A man is known to speak truth $3$ out of $4$ times. He throws a die and reports that it is a six. Find the probability that it is actually a six.

Answer 1

Let $E$ be the event that the man reports that six occurs in throwing of the die and let $S_1$ be the event that six occurs and $S_2$ be the event that six does not occur.

Then,
$P\left(S_1\right)=$ Probability that six occurs $=\frac{1}{6}$

$P\left(S_2\right)=$ Probability that six doesn't occur $=\frac{5}{6}$

$P\left(E|S_1\right)=$ Probability that the man reports six when six has actually occurred
$=$ ​​Probability that the man speaks the truth $=\frac{3}{4}$

$P\left(E|S_2\right)=$ Probability that the man reports six when six has not actually occurred
$=$ ​​Probability that the man does not speaks the truth
$=1-\frac{3}{4}=\frac{1}{4}$

Therefore, by Bayes' Theorem,
Probability that the report of the man that six has occurred is actually a six is given by,

$P\left(S_1|E\right)=\frac{P\left(S_1\right)P\left(E|S_1\right)}{P\left(S_1\right)P\left(E|S_1\right)+P\left(S_2\right)P\left(E|S_2\right)}$
$=\frac{\frac{1}{6}\times \frac{3}{4}}{\frac{1}{6}\times \frac{3}{4}+\frac{5}{6}\times \frac{1}{4}}$
$=\frac{3}{8}$