1. **State the problem:** We want to find the probability of having 7 girls in a family with 7 children, assuming each child is equally likely to be a girl or a boy. 2. **Formula used:** The probability of having exactly $k$ girls in $n$ children, with each child having a probability $p$ of being a girl, follows the binomial distribution: $$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$$ 3. **Important rules:** Here, $n=7$, $k=7$, and $p=\frac{1}{2}$ since a girl is as likely as a boy. 4. **Calculate the probability:** $$P(X=7) = \binom{7}{7} \left(\frac{1}{2}\right)^7 \left(1-\frac{1}{2}\right)^{0}$$ 5. Simplify the binomial coefficient: $$\binom{7}{7} = 1$$ 6. Simplify the powers: $$\left(\frac{1}{2}\right)^7 = \frac{1}{2^7} = \frac{1}{128}$$ $$\left(1-\frac{1}{2}\right)^0 = 1$$ 7. Therefore, $$P(X=7) = 1 \times \frac{1}{128} \times 1 = \frac{1}{128}$$ **Final answer:** The probability of having 7 girls is $\frac{1}{128}$.