Anita's, a fast-food chain specializing in hot dogs and garlic fries, keeps track of the proportion of its customers who decide to eat in the restaurant (as opposed to ordering the food "to go"), so it can make decisions regarding the possible construction of in-store play areas, the attendance of its mascot Sammy at the franchise locations, and so on. Anita's reports that $52 \%$ of its customers order their food to go. If this proportion is correct, what is the probability that, in a random sample of 4 customers at Anita's, exactly 3 order their food to go? Round your response to at least three decimal places. (If necessary, consult a list of formulas.)

Step 1 ：This problem is a binomial probability problem. The binomial distribution model is appropriate for a statistical experiment if the following conditions are met: 1. The experiment consists of n repeated trials. 2. Each trial can result in just two possible outcomes. We call one of these outcomes a success and the other, a failure. 3. The probability of success, denoted by P, is the same on every trial. 4. The trials are independent; that is, the outcome on one trial does not affect the outcome on other trials.

Step 2 ：In this case, we have n=4 (the number of customers), k=3 (the number of customers who order their food to go), and P=0.52 (the probability of a customer ordering their food to go).

Step 3 ：We can use the formula for binomial probability: \(P(X=k) = C(n, k) * (P^k) * ((1-P)^(n-k))\), where C(n, k) is the number of combinations of n items taken k at a time, P is the probability of success, and k is the number of successes.

Step 4 ：Substituting the given values into the formula, we get \(P(X=3) = C(4, 3) * (0.52^3) * ((1-0.52)^(4-3))\).

Step 5 ：Calculating the above expression, we get a probability of approximately 0.270.

Step 6 ：Thus, the probability that, in a random sample of 4 customers at Anita's, exactly 3 order their food to go is approximately \(\boxed{0.270}\).