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 same answer, then the probability that their answer is correct is :

$\frac{10}{13}$

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

$P(A)=\frac{1}{3}$

$P(B)=\frac{1}{4}$

P(Both solve the question) = $\frac{1}{3}×\frac{1}{4}=\frac{1}{12}=P(P)$

P(Both solve incorrectly) = $(1-\frac{1}{3})(1-\frac{1}{4})=\frac{1}{2}=P(Q)$

E → Both get same answer

so $P(E|P)=1$

$P(E|Q)=\frac{1}{20}$

$P(E|P)=\frac{P(P)P(E|P)}{P(P)P(E|P)+P(Q)P(E|Q}$

$=\frac{10}{13}$