Question

$A$ speaks the truth '$3$ times out of $4$' and $B$ speaks the truth '$2$ times out of $3$', $A$ die is thrown. Both assert that the number turned up is $2.$ Find the probability of the truth of their assertion.

Answer 1

Let $A$ and $B$ be the events '$A$ speaks the truth' and '$B$ speaks the truth' repectively. Let $C$ be the event 'the number turned up is not $2$ but both agree to the same conclustion that the die has turned up $2.$ Then $P\left(A\right)=\frac{3}{4},P\left(B\right)=\frac{2}{3}$ and $P\left(C\right)=\frac{1}{5}\times \frac{1}{5}$
There are two hypotheses
(i) the die turns up $2$
(ii) the die does not turns up $2$
Let these be the events $E_1$ and $E_2$ respectively, then
$=\left(E_1\right)=\frac{1}{6},P\left(E_2\right)=\frac{5}{6}$ (a priori probabilities)
Now let $E$ be the event the statement made by $A$ and $B$ agree to the same conclusion.
then $P\left(E/E_1\right)=P\left(A\right).P\left(B\right)=\frac{3}{4}.\frac{2}{3}=\frac{1}{2}$
$P\left(E/E_2\right)=P\left(\overline{A}\right).P\left(\overline{B}\right).P\left(C\right)=\frac{1}{4}.\frac{1}{3}.\frac{1}{25}=\frac{1}{300}$
Thus $P\left(E\right)=P\left(E_1\right)P\left(E/E_1\right)+P\left(E_2\right)P\left(E/E_2\right)=\frac{1}{6}\times \frac{1}{2}+\frac{5}{6}\times \frac{1}{300}=\frac{31}{360}$
$\therefore$ $P\left(E_1/E\right)=\frac{P\left(E_1\right)P\left(E/E_1\right)}{P\left(E\right)}=\frac{30}{31}$