A husband and wife discover that there is a \( 5 \% \) 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 of the three children will inherit the disease, we first calculate the probability that none of the children inherit the disease. The probability of not passing on the disease to a child is \( 95\% \) or \( 0.95 \). If they have three children, the probability that none inherit the disease is: \[ P(\text{none}) = (0.95)^3 = 0.857375 \] Now, the probability that at least one child inherits the disease is the complement of the probability that none inherit it: \[ P(\text{at least one}) = 1 - P(\text{none}) = 1 - 0.857375 = 0.142625 \] To convert this into a percentage, we multiply by 100: \[ P(\text{at least one}) = 0.142625 \times 100 \approx 14.3\% \] Thus, the probability that at least one child will inherit the disease is \( 14.3\% \).