A husband and wife discover that there is a \( 4 \% \) probability of their passing on a hereditary disease to any of their children. If they plan to have three children, what is the probability of the event that at least one child will inherit the disease? The probability is \( \square \% \). (Round to one decimal place as needed.)

To find the probability that at least one child will inherit the hereditary disease, we can first find the probability that none of the children inherit the disease and then subtract that value from 1. The probability that a child does not inherit the disease is \( 1 - 0.04 = 0.96 \). If they have three children, the probability that none of them inherit the disease is: \[ P(\text{none inherit}) = (0.96)^3 = 0.96 \times 0.96 \times 0.96 \approx 0.884736. \] Now, the probability that at least one child inherits the disease is: \[ P(\text{at least one inherits}) = 1 - P(\text{none inherit}) \approx 1 - 0.884736 \approx 0.115264. \] To express this as a percentage, we multiply by 100: \[ P(\text{at least one inherits}) \approx 0.115264 \times 100 \approx 11.5\%. \] So, the probability is \( 11.5\% \).