Question

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 2 order their food to go?

Answer 1

Explanation:

To calculate the probability of exactly 2 out of 4 customers ordering their food to go, we can use the binomial probability formula. The binomial probability formula calculates the probability of getting exactly k successes in n independent Bernoulli trials.

The formula for the binomial probability is:

P(X = k) = (n C k) * p^k * (1 - p)^(n - k)

Where:

P(X = k) is the probability of getting exactly k successes,

n is the number of trials,

k is the number of successes,

p is the probability of success on a single trial,

(1 - p) is the probability of failure on a single trial,

and (n C k) is the binomial coefficient, calculated as n! / (k! * (n - k)!)

In this case:

n = 4 (number of customers in the sample),

k = 2 (number of customers ordering their food to go),

p = 0.52 (proportion of customers ordering their food to go).

Let's calculate the probability:

P(X = 2) = (4 C 2) * 0.52^2 * (1 - 0.52)^(4 - 2)

Using the binomial coefficient:

(4 C 2) = 4! / (2! * (4 - 2)!) = 6

Calculating the probability:

P(X = 2) = 6 * 0.52^2 * (1 - 0.52)^(4 - 2)

= 6 * 0.2704 * 0.2704

= 0.4374 (rounded to four decimal places)

Therefore, the probability that exactly 2 out of 4 customers at Anita's order their food to go is approximately 0.4374, or 43.74%.