Question

A man is known to speak the truth $3$ out of 4 times. He throws a die and reports that it is a six. The probability that it is actually a six is $\frac{k}{8}$. Find the value of $k$

Answer 1

Favourable outcome$=$probability that it's a six and he is telling the truth$=\left(\frac{1}{6}\right)\left(\frac{3}{4}\right)=\frac{3}{24}$
Total possible outcomes$=$probability that it is a six and he is telling the truth or it is not a six and he is telling a lie$=\left(\frac{1}{6}\right)\left(\frac{3}{4}\right)+\left(\frac{5}{6}\right)\left(\frac{1}{4}\right)=\frac{8}{24}$
Hence,$P\left(\frac{E}{S}\right)=\frac{\frac{3}{24}}{\frac{8}{24}}=\frac{3}{8}$
$\Rightarrow$ $k=3$