A man is know to speak truth 3 out of 4 times. He throws a die and reports that it is 6 . Then the…
A man is know to speak truth 3 out of 4 times. He throws a die and reports that it is 6 . Then the probability that it is actually 6 is
$\frac{3}{4}$
$\frac{1}{4}$
$\frac{3}{8}$
$\frac{5}{6}$
Solution
Probability of man speaking truth $=\frac{3}{4} \Rightarrow$ Probability of telling
$\text { lies }=\frac{1}{4}$
Probability of a die actually showing $6=\frac{\left(\frac{1}{6} \times \frac{3}{4}\right)}{\left(\frac{1}{6} \times \frac{3}{4}\right)+\left(\frac{5}{6} \times \frac{1}{4}\right)}=\frac{3}{8}$