Question

In a group of $10$ children, there are $6$ boys and $4$ girls, $3$ children are selected at random. Find the probability that the selected group have only one special girl.

Answer 1

Total number of boys and girls$=6+4=10$
No. of ways of selecting $3$ out of $10$ boys and girls
$={{}^{10}C}_3=\frac{10.9\times 4}{\mathrm{3.2.1}}=120$
No. of ways of selecting $1$ girl and $2$ boys
$={{}^{6}C}_2{{}^{4}C}_1=\frac{6\times 4}{2\times 1}\times \frac{4}{1}=48$
$\therefore$ the required probability
$=\frac{48}{120}=\frac{6}{15}=\frac{2}{5}$