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 $P(A/E_1)=\frac{P(E_1)\times P(E_1/A)}{P(E_1)\times P(E_1/A)+P(E_2)\times P(E_2/A)}$
$=\frac{\frac{1}{2}\times \frac{20\times 100}{100\times 100}}{\frac{1}{2}\times \frac{20\times 100}{100\times 100}+\frac{1}{2}\times \frac{20\times 20}{100\times 100}}$
$=\frac{20\times 100}{20\times (100+20)}=\frac{100}{120}=\frac{5}{6}$