$P = \left( \dfrac{1}{2} \right)^3 \left( \dfrac{1}{2} \right)^2 ({^5}C_3)$
$P = \dfrac{5}{16}$ ← answer
Note:
Among the 5 children, 3 are boys = ${^5}C_3$
among the 5 children, 2 are girls = ${^5}C_2$
in 5 children, 3 are boys and 2 are girls = $\dfrac{5!}{3! \, 2!}$
and that ${^5}C_3 = {^5}C_2 = \dfrac{5!}{3! \, 2!}$