$A$ 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

$\frac{10}{13}$

The correct answer is Option (4) → $\frac{10}{13}$ ##

Let $E_1 =$ Event that both $A$ and $B$ solve the problem

$∴P(E_1) = \frac{1}{3} \times \frac{1}{4} = \frac{1}{12}$

Let $E_2 =$ Event that both $A$ and $B$ got incorrect solution of the problem

$∴P(E_2) = [1 - P(A)][1 - P(B)] = \frac{2}{3} \times \frac{3}{4} = \frac{1}{2}$

Let $E =$ Event that they got same answer

Here, $P(E|E_1) = 1, P(E|E_2) = \frac{1}{20}$

$∴P(E_1|E) = \frac{P(E_1 \cap E)}{P(E)} = \frac{P(E_1) \cdot P(E|E_1)}{P(E_1) \cdot P(E|E_1) + P(E_2) \cdot P(E|E_2)}$

$= \frac{\frac{1}{12} \times 1}{\frac{1}{12} \times 1 + \frac{1}{2} \times \frac{1}{20}}$

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

$= \frac{10}{13}$