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

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13/120

10/13

1/12

1/40

answer is D.

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Detailed Solution

Explanation: Let E₁= Event that both A and B solve the problem
$\begin{array}{l} P (E_{1}) = \frac{1}{3} \times \frac{1}{4} \\ = \frac{1}{12} \end{array}$
Let E₂= Event that both A and B got incorrect solution of the problem
$\begin{array}{l} P (E_{2}) = \frac{2}{3} \times \frac{3}{4} \\ = \frac{1}{2} \end{array}$
Let E= Event that they got same answer
Here,
$\begin{array}{l} P (E / E_{1}) = 1, \\ P (E / E_{2}) = \frac{1}{20} \\ P (E_{1} / E) = \frac{P (E_{1} \cap E)}{P (E)} \\ = \frac{P (E_{1}) \cdot P (E / E_{1})}{P (E_{1}) \cdot P (E / E_{1}) + P (E_{2}) P (E / E_{1})} \\ = \frac{\frac{1}{12} \times 1}{\frac{1}{12} \times 1 + \frac{1}{2} \times \frac{1}{20}} \end{array}$
$\begin{array}{l} = \frac{\frac{1}{12}}{\frac{10 + 3}{120}} \\ = \frac{120}{12 \times 13} \\ = \frac{10}{13} \end{array}$