Assume there are 19 homes in the Quail Creek area and 9 of them have a security
ID: 3152744 • Letter: A
Question
Assume there are 19 homes in the Quail Creek area and 9 of them have a security system. Five homes are selected at random: What is the probability all five of the selected homes have a security system? (Round your answer to 4 decimal places.) What is the probability none of the five selected homes has a security system? (Round your answer to 4 decimal places.) What is the probability at least one of the selected homes has a security system? (Round your answer to 4 decimal places.) Are the events dependent or independent?Explanation / Answer
Given that,
n = 5
p = 9/19 = 0.47
Let X be the random variable that number of homes have a security system.
X ~ Binomial(n=5, p=0.47)
The pmf of X is,
P(X=x) = (5 C x) * 0.47x * (1-0.47)5-x
a) We have to find P(all five of the selected homes have a security system).
P(X = 5) = (5 C 5)* 0.475 * 0.535-5 = 0.0229
b) We have to find P(none of the five selected homes has a security system).
P(X=0) = (5 C 0) * 0.470 * 0.535-0 = 0.0418
c) We have to find P(atleast one of the selected homes have a security system).
That is we have to find P(X 1).
P(X 1) = 1 - P(X<1) = 1- P(X=0) = 1 - 0.0418 = 0.9582