In a family of 3 children , the probability of having at least one boy is

$n\left(s\right)=2^3=8$

Let E be the event of getting at least one boy

$p\left(E\right)=1-p\left(\overline{E}\right)$

$=1-$ probability of getting no boy

$=1-\frac{1}{8}=\frac{7}{8}$