1. **State the problem:** We want to find the probability of having exactly 6 boys and 1 girl in a family of 7 children, where 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 (here $p=\frac{1}{2}$), $n=7$, and $k=6$. 3. **Calculate the binomial coefficient:** $$\binom{7}{6} = \frac{7!}{6!1!} = 7$$ 4. **Calculate the probability:** $$P = 7 \times \left(\frac{1}{2}\right)^6 \times \left(\frac{1}{2}\right)^1 = 7 \times \left(\frac{1}{2}\right)^7$$ 5. **Simplify:** $$P = 7 \times \frac{1}{128} = \frac{7}{128}$$ **Final answer:** The probability of having exactly 6 boys and 1 girl in any order is $\boxed{\frac{7}{128}}$.