In Problems 37–40, assume equally likely outcomes.

Determine the probability of having 1 girl and 3 boys in a 4-child family.

Determine the probability of having 1 girl and 3 boys in a 4-child family.

Assuming that having a boy, and having a girl are equally likely outcomes. To get the probability of having 1 girl and 3 boys, we have to use the equation

$P\left(1\mathrm{girl}\mathrm{and}3\mathrm{boys}\right)=\frac{n\left(1\mathrm{girl}\mathrm{and}3\mathrm{boys}\right)}{n\left(\mathrm{S}\right)}$

Looking for $n\left(\mathrm{S}\right)$, given that a family has only 4 children, and two choices of gender, we have

$n\left(\mathrm{S}\right)=2^4=2\times 2\times 2\times 4=16$

The number of possible outcomes in a sample space, $n\left(\mathrm{S}\right)=16$

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

$S=\{\mathrm{BBBB},\mathrm{BBBG},\mathrm{BBGB},\mathrm{BBGG},\mathrm{BGBB},\mathrm{BGBG},\mathrm{BGGB},\mathrm{BGGG},\mathrm{GBBB},\mathrm{GBBG},\mathrm{GBGB},\mathrm{GBGG},\mathrm{GGBB},\mathrm{GGBG},\mathrm{GGGB},\mathrm{GGGG}\}$

From the list , we can see 4 possible outcomes where there is 1 Girl and 3 Boys. Those are $\{\mathrm{BBBG},\mathrm{BBGB},\mathrm{BGBB},\mathrm{GBBB}\}$.

Thus,role="math" localid="1647412637993" $n\left(1\mathrm{girl}\mathrm{and}3\mathrm{boys}\right)=4$

Substitute the values of $n\left(1\mathrm{girl}\mathrm{and}3\mathrm{boys}\right)\mathrm{and}n\left(\mathrm{S}\right)$into the equation.

role="math" localid="1647413246131" $P\left(1\mathrm{girl}\mathrm{and}3\mathrm{boys}\right)=\frac{n\left(1\mathrm{girl}\mathrm{and}3\mathrm{boys}\right)}{n\left(\mathrm{S}\right)}=\frac{4}{16}=\frac{1}{4}$

Therefore, the probability of having 1 girl and 3 boys in a 4-child family is$\frac{1}{4}$.

The probability of having 1 girl and 3 boys in a 4-child family is$\frac{1}{4}$.