Question

Mr. $A$ is known to speak truth $3$ out of $4$ times and Mr. $B$ speaks truth $4$ out of $5$ times. A pair of dice is thrown. Both reported that the result (sum of the numbers) is $9$. Then the probability that the result is actually $9$ is:

• A. $\frac{3}{5}$
• B. $\frac{2}{5}$
• C. $\frac{4}{5}$
• D. $\frac{1}{5}$

Answer 1

The correct option is A $\frac{3}{5}$

Let $E_1$ be the event of sum is actually $9$ on pair of dice and $E_2$ be the event sum is actually not $9$. Then $P\left(E_1\right)=\frac{4}{36}=\frac{1}{9}$ and $P\left(E_2\right)=\frac{8}{9}$.
Let $A$ be the event that both $A$ and $B$ reports that (sum is $9$)
$P\left(\frac{A}{E_1}\right)=\frac{3}{4}\times \frac{4}{5}$and$P\left(\frac{A}{E_2}\right)=\frac{1}{4}\times \frac{1}{5}$
$P\left(E_1/A\right)=\left[\frac{P\left(E_1\right).P\left(A/E_1\right)}{P\left(E_1\right).P\left(A/E_1\right)+P\left(E_2\right).P\left(A/E_2\right)}\right]=\frac{(\frac{1}{9}.\frac{3}{4}.\frac{4}{5})}{(\frac{1}{9}.\frac{3}{4}.\frac{4}{5})+(\frac{8}{9}.\frac{1}{4}.\frac{1}{5})}=\frac{12}{12+8}=\frac{12}{20}=\frac{3}{5}$