A man is known to speak the truth 2 out of 3 times. If he throws a die and reports that it is six, then the…
A man is known to speak the truth 2 out of 3 times. If he throws a die and reports that it is six, then the probability that it is actually five, is
$\frac{3}{8}$
$\frac{1}{7}$
$\frac{2}{7}$
$\frac{4}{5}$
Solution
Let the events
$E_1=$ six occurs
$E_2=$ six does not occurs
$A=$ the mean reports that it is six.
We have,
$
\begin{aligned}
& P\left(E_1\right)=\frac{1}{6} \\
& P\left(E_2\right)=\frac{5}{6}
\end{aligned}
$
Now, $P\left(\frac{A}{E_1}\right)=\frac{2}{3}, P\left(\frac{A}{E_2}\right)=\frac{1}{3}$
$
\begin{aligned}
\therefore \quad P\left(\frac{E_1}{A}\right) & =\frac{P\left(E_1\right) \times P\left(\frac{A}{E_1}\right)}{P\left(E_1\right) \times P\left(\frac{A}{E_1}\right)+P\left(E_2\right) \times P\left(\frac{A}{E_2}\right)} \\
& =\frac{\frac{1}{6} \times \frac{2}{3}}{\frac{1}{6} \times \frac{2}{3}+\frac{5}{6} \times \frac{1}{3}}=\frac{2}{7}
\end{aligned}
$
The probability that it is actually six $=\frac{2}{7}$
The probability that it is not actually six $=1-\frac{2}{7}=\frac{5}{7}$
The probability that is actually five $=\frac{1}{5} \times \frac{5}{7}=\frac{1}{7}$