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Question
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 common error is `1/20` and they obtain the same answer, then the probability of their answer to be correct is
Options
`10/13`
`13/120`
`1/40`
`1/12`
Solution
E1 = they solve correctly.
E2 = they solve incorrectly.
A = they obtain the same result.
\[P\left( \frac{E_1}{A} \right) = \frac{P\left( E_1 \right) P\left( \frac{A}{E_1} \right)}{P\left( E_1 \right) P\left( \frac{A}{E_1} \right) + P\left( E_2 \right) P\left( \frac{A}{E_2} \right)}\]
\[ = \frac{\frac{1}{12} \times 1}{\frac{1}{12} \times 1 + \frac{6}{12} \times \frac{1}{20}}\]
\[ = \frac{\frac{1}{12} \times 1}{\frac{1}{12} \times 1 + \frac{6}{12} \times \frac{1}{20}}\]
\[ = \frac{20}{26} = \frac{10}{13}\]
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