Question

A crime is committed by one of two suspects, $A$ and $B$. Initially, there is equal evidence against both of them. In further investigation at the crime scene, it is found that the guilty party had a blood type found in $20\%$ of the population. If the suspect $A$ does match this blood type, whereas the blood type of suspect $B$ is unknown, then the probability that $A$ is the guilty party, is

• A. $ \frac{3}{5} $
• B. $ \frac{5}{6} $
• C. $ \frac{1}{3} $
• D. $ \frac{2}{3} $

Answer 1

Answer: (B)

Consider, the events $ {{E}_{1}},\,{{E}_{2}} $ and A $ P({{E}_{1}})=\frac{1}{2}.\,P({{E}_{2}})=\frac{1}{2} $
$ P({{E}_{1}}/A)=\frac{20\times 100}{100\times 100},\,\,P({{E}_{2}}/A)=\frac{20\times 20}{100\times 100} $
$ \therefore $ Required probability $ P(A/{{E}_{1}})=\frac{P({{E}_{1}})\times P({{E}_{1}}/A)}{P({{E}_{1}})\times P({{E}_{1}}/A)+P({{E}_{2}})\times P({{E}_{2}}/A)} $
$=\frac{\frac{1}{2}\times \frac{20\times 100}{100\times 100}}{\frac{1}{2}\times \frac{20\times 100}{100\times 100}+\frac{1}{2}\times \frac{20\times 20}{100\times 100}} $
$=\frac{20\times 100}{20\times (100+20)}=\frac{100}{120}=\frac{5}{6} $