Question

In Problems 37–40, assume equally likely outcomes.

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

Short answer

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

Step-by-step solution

Step 1. Given information.

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

Step 2. Find the probability.

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

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

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

localid="1647410635798" $S=\left\{\mathrm{BBB},\mathrm{BBC},\mathrm{BGB},\mathrm{BGG},\mathrm{GBB},\mathrm{GBG},\mathrm{GGB},\mathrm{GGG}\right\}$

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

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

Step 3. Substitute the values.

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

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

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

Step 4. Conclusion.

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