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`

Answer 1

E₁ = they solve correctly.
E₂ = they solve incorrectly.
A = they obtain the same result.

\[P\left( \frac{E_1}{A} \right) = \frac{P\left( E_1 \right) P\left( \frac{A}{E_1} \right)}{P\left( E_1 \right) P\left( \frac{A}{E_1} \right) + P\left( E_2 \right) P\left( \frac{A}{E_2} \right)}\]

\[ = \frac{\frac{1}{12} \times 1}{\frac{1}{12} \times 1 + \frac{6}{12} \times \frac{1}{20}}\]

\[ = \frac{\frac{1}{12} \times 1}{\frac{1}{12} \times 1 + \frac{6}{12} \times \frac{1}{20}}\]

\[ = \frac{20}{26} = \frac{10}{13}\]