Question

A man is know to speak truth $3$ out of $4$ times. He throws a dice and reports that it is six. The probability that it is actually six is

• A. $\frac{3}{8}$
• B. $\frac{1}{5}$
• C. $\frac{3}{5}$
• D. None of these

Answer 1

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

Let $A$ denotes the event that a six occurs and $B$ the events that the man reports that it is a six.
Then the probability that it is actually a six is given by
$P\left(\frac{A}{B}\right)=\frac{P(A\cap B)}{P\left(B\right)}$
Now $P(A\cap B)=\frac{1}{6}\times \frac{3}{4}=\frac{3}{24}$
$P\left(B\right)=P(A\cap B)+P(\overline{A}\cap B)=\frac{1}{6}\times \frac{3}{4}+\frac{5}{6}\times \frac{1}{4}=\frac{8}{24}$
Hence $P\left(\frac{A}{B}\right)=\frac{\frac{3}{24}}{\frac{8}{24}}=\frac{3}{8}$