1. **State the problem:** We want to find the probability of having exactly 4 boys and 1 girl in a family of 5 children, assuming each child is equally likely to be a boy or a girl. 2. **Formula used:** This is a binomial probability problem. The probability of exactly $k$ successes (boys) in $n$ trials (children) is given by: $$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$$ where $p$ is the probability of a boy (success), $n=5$, and $k=4$. 3. **Important rules:** Since boys and girls are equally likely, $p = \frac{1}{2}$. The binomial coefficient $\binom{n}{k}$ counts the number of ways to choose which $k$ children are boys. 4. **Calculate the binomial coefficient:** $$\binom{5}{4} = \frac{5!}{4!1!} = 5$$ 5. **Calculate the probability:** $$P = 5 \times \left(\frac{1}{2}\right)^4 \times \left(\frac{1}{2}\right)^1 = 5 \times \frac{1}{16} \times \frac{1}{2}$$ 6. **Simplify the expression:** $$5 \times \frac{1}{16} \times \frac{1}{2} = 5 \times \frac{1}{\cancel{16}} \times \frac{1}{\cancel{2}} = 5 \times \frac{1}{32} = \frac{5}{32}$$ 7. **Final answer:** The probability of having exactly 4 boys and 1 girl is $$\boxed{\frac{5}{32}}$$