In Problems 37–40, assume equally likely outcomes.

Determine the probability of having 3 girls in a 3-child family.

Determine the probability of having 3 girls in a 3-child family.

Assuming that having a boy, and having a girl are equally likely outcomes. To get the probability of having 3 girls, we have to use the equation

$P\left(3\mathrm{Girls}\right)=\frac{n\left(3\mathrm{Girls}\right)}{n\left(\mathrm{S}\right)}$

Looking for $n\left(\mathrm{S}\right)$, given that a family has only 3 children, and two choices of gender, we have

$n\left(\mathrm{S}\right)=2^3=2\times 2\times 2=8$

The number of possible outcomes in a sample space, $n\left(\mathrm{S}\right)=8$.

Listing the possible outcomes where B stands for Boy and G stands for Girl, we have

$S=\left\{\mathrm{BBB},\mathrm{BBC},\mathrm{BGB},\mathrm{BGG},\mathrm{GBB},\mathrm{GBG},\mathrm{GGB},\mathrm{GGG}\right\}$

From the list, we can see 1 possible outcome where there is 3 Girls which is $\left\{\mathrm{GGG}\right\}$.

Thus,$n\left(3\mathrm{Girls}\right)=1$

Substitute the values of $n\left(3\mathrm{Girls}\right)\mathrm{and}n\left(\mathrm{S}\right)$into the equation.

$P\left(3\mathrm{Girls}\right)=\frac{n\left(3\mathrm{Girls}\right)}{n\left(\mathrm{S}\right)}=\frac{1}{8}$

Therefore, the probability of having 3 girls in a 3-child family is$\frac{1}{8}$.

The probability of having 3 girls in a 3-child family is$\frac{1}{8}$.