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

All possible outcomes are BBB, BBG, BGB, GBB, BGG, GBG, GGB and GGG.

Number of all possible outcomes = 8

Let E be the event of having at least one boy.

Then the outcomes are BBB, BBG, BGB, GBB, BGG, GBG and GGB.

Number of possible outcomes = 7

$\therefore P$ (Having at least one boy) $=P(E)=\frac{\text { Number of outcomes favourable to } E}{\text { Number of all possible outcomes }}$

$=\frac{7}{8}$

Thus, the probability of having at least one boy is $\frac{7}{8}$.