Anita's a fast food chain specializing in hotdogs and garlic fries, keeps track of the proportion of its customers who decide to eat in the restaurant so it can make decisions reguarding possible construction of the in-store play areas.the attendance of its mascot Stacy at the franchise locations, and so on. Anitas reports that 52% of its customers order their food to go. If the proportion is correct, what is the probability that, in a random sample of 5 customers at anitas, exactly 4 order their food to go?

Round your response to at least 3 decimal.

If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events.

(.52)^4 * (1-.52) = ?

To solve this problem, we can use the binomial probability formula:

P(x) = C(n, x) * p^x * (1 - p)^(n - x)

Where:
P(x) is the probability of getting exactly x successes,
n is the total number of trials,
p is the probability of a success in one trial,
C(n, x) is the number of combinations of n items taken x at a time.

In this case, we want to find the probability that exactly 4 out of 5 customers order their food to go. The proportion of customers who order to go is 52%, so the probability of a customer ordering to go (p) is 0.52.

Substituting the values into the formula, we have:

P(4) = C(5, 4) * (0.52)^4 * (1 - 0.52)^(5 - 4)

C(5, 4) = 5! / (4! * (5 - 4)! ) = 5

P(4) = 5 * (0.52)^4 * (1 - 0.52)^(5 - 4)

P(4) = 5 * 0.52^4 * 0.48

P(4) = 0.41472

Rounding the answer to three decimal places, the probability that exactly 4 out of 5 customers order their food to go is approximately 0.415.