Question

A man is known to speak the truth $3$ out of $5$ times. He throws a die and reports that it is ${}^{\prime }{6}^{\prime }$. The probability that it is actually '$6$' is $\frac{3}{k}$. Find the value of $k$.

Answer 1

Let $E$ be the event that the man reports that one occurs in throwing of the die and let $S_1$ be the event that one occurs and $S_2$ be the event that one does not occur.
Then, $P\left(S_1\right)=\frac{1}{6}$
$P\left(S_2\right)=\frac{5}{6}$
$P\left(E|S_1\right)$= Probability that the man reports that $6$ occurs when six has actually occured on die = $\frac{3}{5}$= probability that man speaks truth.
$P\left(E|S_2\right)$= Probability that the man reports that $6$ occurs when $6$ has not actually occured on the die = $1-\frac{3}{5}=\frac{2}{5}$= probability that man does not speak truth.
By Baye's theorem,
$P\left(S_1|E\right)=\frac{P\left(S_1\right)P\left(E|S_1\right)}{P\left(S_1\right)P\left(E|S_1\right)+P\left(S_2\right)P\left(E|S_2\right)}=\frac{\frac{1}{6}\times \frac{3}{5}}{\frac{1}{6}\times \frac{3}{5}+\frac{5}{6}\times \frac{2}{5}}=\frac{3}{13}$
Hence, required probability is $\frac{3}{13}$.