91Ó°ÊÓ

Determine the probability of having 2 girls and 2 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 2 girls and 2 boys, we have to use the equation

$P\left(2\mathrm{girl}s\mathrm{and}2\mathrm{boys}\right)=\frac{n\left(2\mathrm{girl}s\mathrm{and}2\mathrm{boys}\right)}{n\left(\mathrm{S}\right)}$

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

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

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 6 possible outcomes where there is 2 Girls and 2 Boys. Those are $\left\{\mathrm{BBGG},\mathrm{BGBG},\mathrm{BGGB},\mathrm{GBBG},\mathrm{GBGB},\mathrm{GGBB}\right\}$.

Thus,$n\left(2\mathrm{girl}s\mathrm{and}2\mathrm{boys}\right)=6$.

$P\left(2\mathrm{girl}s\mathrm{and}2\mathrm{boys}\right)=\frac{n\left(2\mathrm{girl}s\mathrm{and}2\mathrm{boys}\right)}{n\left(\mathrm{S}\right)}=\frac{6}{16}=\frac{3}{8}$Substitute the values of $n\left(2\mathrm{girl}s\mathrm{and}2\mathrm{boys}\right)\mathrm{and}n\left(\mathrm{S}\right)$into the equation.

Therefore, the probability of having 2 girls and 2 boys in a 4-child family is$\frac{3}{8}$.

The probability of having 2 girls and 2 boys in a 4-child family is$\frac{3}{8}$.