Question

A and B are two students. Their chances of solving a problem correctly are $\frac{1}{5}$ and $\frac{1}{6}$
respectively. If the probability of their making a common error is, $\frac{1}{20}$ and they obtain the same answer, then the probability of their answer to be correct is

• A. $\frac{1}{2}$
• B. $\frac{5}{11}$
• C. $\frac{6}{11}$
• D. 1

Answer 1

The correct option is A $\frac{1}{2}$

$\mathrm{Let}E\mathrm{denotes}\mathrm{the}\mathrm{event}\mathrm{that}A\mathrm{solves}\mathrm{the}\mathrm{problem}\mathrm{correctly}.\mathrm{and}F\mathrm{denotes}\mathrm{the}\mathrm{event}\mathrm{that}B\mathrm{solves}\mathrm{the}\mathrm{problem}\mathrm{correctly}.\mathrm{Then}P\left(E\right)=\frac{1}{5}\mathrm{and}P\left(F\right)=\frac{1}{6}\mathrm{We}\mathrm{can}\mathrm{observe}\mathrm{that}E\mathrm{and}F\mathrm{are}\mathrm{independent}.\mathrm{Let}{E}_1\mathrm{be}\mathrm{the}\mathrm{event}\mathrm{that}\mathrm{both}\mathrm{solves}\mathrm{the}\mathrm{problem}\mathrm{correctly}.\mathrm{Then}P\left(E_1\right)=P\left(E\cap F\right)=P\left(E\right)P\left(F\right)=\frac{1}{5}\times \frac{1}{6}=\frac{1}{30}\mathrm{Let}{E}_2\mathrm{denotes}\mathrm{that}\mathrm{both}\mathrm{don}\text{'}t\mathrm{solve}\mathrm{the}\mathrm{problem}\mathrm{correctly}.\mathrm{Then}P\left(E_2\right)=P\left(E^C\cap F^C\right)=P\left(E^C\right)P\left(F^C\right)=\frac{4}{5}\times \frac{5}{6}=\frac{2}{3}$
$\mathrm{Let}S\mathrm{denote}\mathrm{the}\mathrm{event}\mathrm{of}\mathrm{getting}\mathrm{same}\mathrm{answer}.\mathrm{If}\mathrm{both}\mathrm{are}\mathrm{making}\mathrm{same}\mathrm{error},\mathrm{we}\mathrm{are}\mathrm{sure}\mathrm{that}\mathrm{the}\mathrm{answer}\mathrm{coming}\mathrm{out}\mathrm{is}\mathrm{wrong}.\mathrm{Then}P\left(S|E_2\right)=\frac{1}{20}\mathrm{And}\mathrm{if}\mathrm{their}\mathrm{answer}\mathrm{is}\mathrm{correct},\mathrm{both}\mathrm{will}\mathrm{get}\mathrm{the}\mathrm{same}\mathrm{answer}.\therefore P\left(S|E_1\right)=1\mathrm{We}\mathrm{need}\mathrm{to}\mathrm{find}\mathrm{the}\mathrm{probability}\mathrm{of}\mathrm{getting}a\mathrm{correct}\mathrm{answer}\mathrm{given}\mathrm{both}\mathrm{committed}a\mathrm{common}\mathrm{error}\mathrm{and}\mathrm{got}\mathrm{the}\mathrm{same}\mathrm{answer}.i.e.,\mathrm{we}\mathrm{need}\mathrm{to}\mathrm{find}P\left(E_1|S\right).\mathrm{By}\mathrm{Bayes}\text{'}\mathrm{theorem},P\left(E_1|S\right)=\frac{P\left(E_1\right)P\left(S|E_1\right)}{P\left(E_1\right)P\left(S|E_1\right)+P\left(E_2\right)P\left(S|E_2\right)}=\frac{\frac{1}{30}\times 1}{\frac{1}{30}\times 1+\frac{2}{3}\times \frac{1}{20}}=\frac{1}{2}\mathrm{Hence},\mathrm{required}\mathrm{probability}\mathrm{is}\frac{1}{2}.$