Question

A man speaks 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.

• A. $\frac{3}{8}$
• B. $\frac{5}{8}$
• C. $\frac{7}{8}$
• D. $\frac{1}{12}$

Answer 1

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

Let $E_1,E_2$ and A be the events defined as follows:
$E_1$ = six occurs, $E_2$ = six does not occur and A = the man reports that it is a six.
We have, $P\left(E_1\right)=\frac{1}{6},P\left(E_2\right)=\frac{5}{6}$
Now
$P\left(\frac{A}{E_1}\right)$=probability that the man reports that there is a six on the die given that six has occured on the die.
= probability that the man speak truth = $\frac{3}{4}$ and $P\left(\frac{A}{E_2}\right)$ = probability that the man reports that there is a six on the die given that six has not occured on the die.
= probability that the man does not speak truth
$=1-\frac{3}{4}=\frac{1}{4}$
We have to find $P\left(\frac{E_1}{A}\right)$
By Baye's rule, we have
$P\left(\frac{E_1}{A}\right)=\frac{P\left(E_1\right)P\left(\frac{A}{E_1}\right)}{P\left(E_1\right)P\left(\frac{A}{E_1}\right)+P\left(E_2\right)P\left(\frac{A}{E_2}\right)}=\frac{\frac{1}{6}\times \frac{3}{4}}{\frac{1}{6}\times \frac{3}{4}+\frac{5}{6}\times \frac{1}{4}}=\frac{3}{8}$