A study determined that 7% of children under 18 years old lived with their father only. Find the probability that none of the 15 children, selected at random from all children under 18 years old, lived with their father only. The probability that none of the children lived with their father only is ? (Do not round until the final answer. Then round to the nearest thousandth as needed.)

Step 1 ：This is a binomial distribution problem where we are trying to find the probability of 0 successes (children living with their father only) in 15 trials (children selected at random). The probability of success on each trial is 0.07 (7%).

Step 2 ：The formula for the probability mass function of a binomial distribution is: \(P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k))\), where: \(P(X=k)\) is the probability of k successes in n trials, \(C(n, k)\) is the number of combinations of n items taken k at a time, \(p\) is the probability of success on each trial, \(n\) is the number of trials.

Step 3 ：In this case, we want to find \(P(X=0)\), so \(k=0\). The number of trials, \(n\), is 15, and the probability of success, \(p\), is 0.07. The number of combinations of 15 items taken 0 at a time, \(C(15, 0)\), is 1. So we can substitute these values into the formula to find the probability.

Step 4 ：Substituting the values into the formula, we get: \(P = C(n, k) * (p^k) * ((1-p)^(n-k)) = 1 * (0.07^0) * ((1-0.07)^(15-0)) = 0.3367008620516138\)

Step 5 ：The probability that none of the 15 children, selected at random from all children under 18 years old, lived with their father only is approximately 0.337.

Step 6 ：Final Answer: The probability that none of the 15 children, selected at random from all children under 18 years old, lived with their father only is approximately \(\boxed{0.337}\).