Question

A total of 60% of the customers of a fast food chain order a hamburger, French fries, and a drink. If a random sample of 15 cash register receipts is selected, what is the probability that 10 or more will show that the above three food items were ordered? (What are the steps used to solve this using Binomial probability distribution? It says the correct answer is .403, but I am getting .186) A. 1.000 B. 0.186 C. 0.403 D. 0.000

Answer 1

Answer:

C. 0.403

Step-by-step explanation:

Binomial distribution has two parameter n and p. Here, n=15 and p=0.60.

We have to find the probability of 10 or more will show that the hamburger, French fries and a drink were ordered. P(X≥10)=?

The binomial probability distribution function for random variable X  is

$P(X=x)=nCxp^{x} q^{n-x}$

where q=1-p=1-0.6=0.4,n=15 and p=0.6.

P(X≥10)=P(X=10)+P(X=11)+P(X=12)+P(X=13)+P(X=14)+P(X=15)

$P(X=10)=15C10(0.6^{10})( 0.4^{5})$ =0.185938

$P(X=11)=15C11(0.6^{11}) (0.4^{4})$ =0.126776

$P(X=12)=15C12(0.6^{12}) (0.4^{3})$ =0.063388

$P(X=13)=15C13(0.6^{13}) (0.4^{2})$ =0.021942

$P(X=14)=15C14(0.6^{14}) (0.4^{1})$ =0.004702

$P(X=15)=15C15(0.6^{15}) (0.4^{0})$ =0.000470

P(X≥10)= 0.185938 +0.126776 +0.063388 +0.021942 +0.004702 +0.000470

P(X≥10)=0.403216

So, the probability of 10 or more will show that the hamburger, French fries and a drink were ordered is 0.403.