A and B are two students. Their chances of solving a problem correctly are $\frac{1}{3}$ and $\frac{1}{4}$, respectively. If the probability of their making a common error is, $\frac{1}{{20}}$ and they obtain the same answer, then the probability of their answer to be correct is

Let ${E_1} =$ Event that both A and B solve the problem $\therefore$ $P\left( {{E_1}} \right) = \frac{1}{3} \times \frac{1}{4} = \frac{1}{{12}}$.

Let ${E_2} =$ Event that both $A$ and $B$ got incorrect

Solution

of the problem
$\therefore$ $P\left( {{E_2}} \right) = \frac{2}{3} \times \frac{3}{4} = \frac{1}{2}$

Let $E =$ Event that they got same answer
Here, $P\left( {E/{E_1}} \right) = 1,P\left( {E/{E_2}} \right) = \frac{1}{{20}}$

$\therefore$ $P\left( {{E_1}/E} \right) = \frac{{P\left( {{E_1} \cap E} \right)}}{{P(E)}} = \frac{{P\left( {{E_1}} \right) \cdot P\left( {E/{E_1}} \right)}}{{P\left( {{E_1}} \right) \cdot P\left( {E/{E_1}} \right) + P\left( {{E_2}} \right)P\left( {E/{E_2}} \right)}}$

$= \frac{{\frac{1}{{12}} \times 1}}{{\frac{1}{{12}} \times 1 + \frac{1}{2} \times \frac{1}{{20}}}} = \frac{{1/12}}{{\frac{{10 + 3}}{{120}}}} = \frac{{120}}{{12 \times 13}} = \frac{{10}}{{13}}$