Suppose that $5 \%$ of men and $0.25 \%$ of women have gray hair. A gray hair person is selected at random.…
Suppose that $5 \%$ of men and $0.25 \%$ of women have gray hair. A gray hair person is selected at random. If there are equal number of males and females, then the probability that the person selected being men is
$\frac{20}{21}$
$\frac{10}{21}$
$\frac{1}{21}$
$\frac{11}{21}$
Solution
A grey haired person is selected at eandom, the probability that this person is male $=\mathrm{P}(\mathrm{M} / \mathrm{G})$ Here $M$ is men, $W$ is women, $G$ is grey hair.
$\begin{aligned}
\therefore \mathrm{P}(\mathrm{M} / \mathrm{G}) &=\frac{\mathrm{P}(\mathrm{M}) \times \mathrm{P}(\mathrm{G} / \mathrm{M})}{\mathrm{P}(\mathrm{M}) \times \mathrm{P}(\mathrm{G} / \mathrm{M})+\mathrm{P}(\mathrm{W}) \times \mathrm{P}(\mathrm{G} / \mathrm{W})} \\
&=\frac{\frac{1}{2} \times \frac{5}{100}}{\left(\frac{1}{2} \times \frac{5}{100}\right)+\left(\frac{1}{2} \times \frac{0.25}{100}\right)}=\frac{20}{21}
\end{aligned}$