Question

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 common error is $\frac{1}{20}$ and they obtain the same answer, then the probability of their answer to be correct is
(a) $\frac{10}{13}$ (b) $\frac{13}{120}$ (c) $\frac{1}{40}$ (d) $\frac{1}{12}$

Answer 1

E₁ = they solve correctly.
E₂ = they solve incorrectly.
A = they obtain the same result.
$$\begin{gathered} P\left(\raisebox{1ex}{$E_1$}\!\left/ \!\raisebox{-1ex}{$A$}\right.\right)=\frac{P\left(E_1\right)P\left({\displaystyle \raisebox{1ex}{$A$}\!\left/ \!\raisebox{-1ex}{$E_1$}\right.}\right)}{P\left(E_1\right)P\left(\raisebox{1ex}{$A$}\!\left/ \!\raisebox{-1ex}{$E_1$}\right.\right)+P\left(E_2\right)P\left(\raisebox{1ex}{$A$}\!\left/ \!\raisebox{-1ex}{$E_2$}\right.\right)} \\ =\frac{{\displaystyle \frac{1}{12}}\times 1}{{\displaystyle \frac{1}{12}}\times 1+{\displaystyle \frac{6}{12}}\times {\displaystyle \frac{1}{20}}} \\ =\frac{{\displaystyle \frac{1}{12}\times 1}}{{\displaystyle \frac{1}{12}\times 1+\frac{6}{12}\times \frac{1}{20}}} \\ =\frac{{\displaystyle 20}}{{\displaystyle 26}}=\frac{10}{13} \end{gathered}$$
Hence, the correct answer is option (a).