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 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?

Answer 1

0.2700

Explanation:

-Notice that this is a binomial distribution with p=0.52, n=4 and x=3.

-The formula for the binomial probabilities is given as:

`P(X=x)={n\choose x}p^x(1-p)^(n-x)`

The probability than in 4, exactly 3 order their food on the go is:

`P(X=x)={n\choose x}p^x(1-p)^(n-x)\\\\P(X=3)={4\choose3}0.52^3(1-0.52)^1\\\\=0.2700`

Hence, the probability that exactly 3 order their food on the go is 0.2700