Question

$A,B$ are two inaccurate arithmeticians whose chance of solving a given question correctly are $\left(1/8\right)$ and $\left(1/12\right)$ respectively. They solve a problem and obtained the same result. If it is $1000$ to $1$ against their making the same mistake, find the chance that the result is correct.

Answer 1

The prohability that both of them getting the same correct answer is

$\frac{1}{12}\times \frac{1}{8}\times \frac{999}{1000}$

$\frac{1}{12}\times \frac{1}{8}\times \frac{999}{1000}+\frac{11}{12}\times \frac{7}{8}\times \frac{1}{1000}$

$=\frac{\frac{999}{96000}}{\frac{999+77}{96000}}$

$=\frac{999}{1076}$