Question

A, B are two inaccurate arithmeticians whose chances of solving a given question correctly are (1/8) and (1/12) 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

• A. $\frac{13}{14}$
• B. $\frac{2}{7}$
• C. $\frac{1}{8}$
• D. $\frac{11}{12}$

Answer 1

The correct option is A $\frac{13}{14}$

$A$ : $A$ solves correctly $B:B$ solves correctly
$E$ : Commit same mistake
$F$ : same result
$P\left(AB/F\right)=\frac{P\left(AB\right)}{P\left(AB\right)+P\left(\overline{A}\overline{B}E\right)}=\frac{\frac{1}{8}.\frac{1}{12}}{\frac{1}{8}.\frac{1}{12}+\frac{7}{8}.\frac{11}{12}.\frac{1}{1001}}$
$=\frac{1001}{1078}=\frac{13}{14}$