Question

set up 0: a total of 75% of the customers of a fast-food chain order a hamburger, french fries, and a drink. if a random sample of 20 cash register receipts is selected. what is the probability that 10 will show that the three food items were ordered?

Answer 1

The probability that 10 out of 20 cash register receipts will show that the three food items were ordered is approximately 0.00992.

Let's find the probability that 10 out of 20 cash register receipts will show that the three food items were ordered.

We can use the binomial probability formula to calculate this probability:

P(X=k)=( n/k​ )⋅p^k ⋅(1−p)^n−k

where:

X is the number of successes (receipts showing that the three food items were ordered)

n is the number of trials (cash register receipts)

p is the probability of success (customer ordering the three food items)

In this case, we have:

n=20

p=0.75

k=10

Plugging these values into the formula, we get:

P(X=10)=( 20/10 )⋅0.75^10 ⋅0.25^10

Evaluating this expression, we get:

P(X=10)≈0.00992

Therefore, the probability that 10 out of 20 cash register receipts will show that the three food items were ordered is approximately 0.00992.