Suppose 10% of men and 0.5% of women have grey hair. A grey haired person is selected at random. If there are equal number of males and females, then the probability of the selected person being a female is

$\frac{1}{21}$

The correct answer is Option (2) → $\frac{1}{21}$

$P(G|M)=10\%=0.10$

$P(G|F)=0.5\%=0.005$

$P(M)=P(F)=\frac{1}{2}$

$P(F|G)=\frac{P(G|F)P(F)}{P(G|M)P(M)+P(G|F)P(F)}$

$=\frac{0.005×0.5}{0.0525}=0.0476=\frac{1}{21}$