Find the probability that when a couple has six children, at least one of them is a boy. (Assume that boys and girls are equally likely.) The probability is \( \square \) that at least one of the six children is a boy. (Simplify your answer. Do not round.)

To find the probability that at least one of the six children is a boy, we can use the complement rule. First, we determine the probability that all six children are girls. The probability of having a girl for each child is \( \frac{1}{2} \). Therefore, the probability of having six girls is: \[ P(\text{all girls}) = \left( \frac{1}{2} \right)^6 = \frac{1}{64}. \] Now, to find the probability that at least one child is a boy, we subtract the probability of having all girls from 1: \[ P(\text{at least one boy}) = 1 - P(\text{all girls}) = 1 - \frac{1}{64} = \frac{64}{64} - \frac{1}{64} = \frac{63}{64}. \] Thus, the probability that at least one of the six children is a boy is \( \frac{63}{64} \).