In a certain city 40% of the people have brown hair, 25% have brown eyes and 15% have brown eyes as well as brown hair. If a person selected at random has brown hair, the probability that the also has brown eyes, is equal to

$\frac{3}{8}$

$E_1$ : Person has brown hair.

$E_2$ : Person has brown hair.

$P\left(E_1\right)=\frac{40}{100}=\frac{2}{5}$

$P\left(E_2\right)=\frac{25}{100}=\frac{1}{4}$

$P\left(E_1 \cap E_2\right)=\frac{15}{100}=\frac{3}{20}$

$P\left(E_2 / E_1\right)=\frac{P\left(E_1 \cap E_2\right)}{P\left(E_1\right)}$

$=\frac{3 / 20}{2 / 5}=\frac{3}{8}$