assume that boys and girls are equally likely. find the probability that when a couple has three children…

Explanation:

Step1: Determine probability of one - child case

The probability of having a girl in a single - child birth is $p=\frac{1}{2}$ since boys and girls are equally likely.

Step2: Use the binomial probability formula

The binomial probability formula is $P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}$, where $n$ is the number of trials, $k$ is the number of successful trials, $p$ is the probability of success in a single trial, and $C(n,k)=\frac{n!}{k!(n - k)!}$. Here, $n = 3$ (number of children), $k = 3$ (number of girls), and $p=\frac{1}{2}$. First, calculate $C(3,3)=\frac{3!}{3!(3 - 3)!}=\frac{3!}{3!0!}=1$. Then, $p^{k}=(\frac{1}{2})^{3}$ and $(1 - p)^{n - k}=(1-\frac{1}{2})^{3 - 3}=1$. So $P(X = 3)=1\times(\frac{1}{2})^{3}\times1$.

Step3: Calculate the final probability

$P(X = 3)=(\frac{1}{2})^{3}=\frac{1}{8}$.

Answer:

$\frac{1}{8}$