1. **Problem statement:** A family has three children, and it is known that at least one of the children is a girl. We want to find the probability that all three children are girls given this information. 2. **Formula and rules:** We use conditional probability. The probability that all three children are girls given that at least one is a girl is $$P(\text{3 girls} \mid \text{at least 1 girl}) = \frac{P(\text{3 girls and at least 1 girl})}{P(\text{at least 1 girl})}.$$ Since having 3 girls implies at least 1 girl, this simplifies to $$P(\text{3 girls} \mid \text{at least 1 girl}) = \frac{P(\text{3 girls})}{P(\text{at least 1 girl})}.$$ Important rules: - Each child is equally likely to be a boy or a girl. - The gender of each child is independent. 3. **Calculate probabilities:** - Total possible outcomes for 3 children: $2^3 = 8$. - Probability of 3 girls: $P(\text{3 girls}) = \frac{1}{8}$ (only one outcome: GGG). - Probability of at least 1 girl: $P(\text{at least 1 girl}) = 1 - P(\text{no girls}) = 1 - P(\text{3 boys}) = 1 - \frac{1}{8} = \frac{7}{8}$. 4. **Calculate conditional probability:** $$P(\text{3 girls} \mid \text{at least 1 girl}) = \frac{\frac{1}{8}}{\frac{7}{8}} = \frac{1}{7}.$$ 5. **Answer:** The probability that all three children are girls given that at least one is a girl is $\boxed{\frac{1}{7}}$.