SOLUTION: Just over half (0.6) of the homes built in the new upper-bracket snootyville subdivision included a room designed specifically to be a home theater. Four snootyville homes are sel

Question 619237: Just over half (0.6) of the homes built in the new upper-bracket snootyville subdivision included a room designed specifically to be a home theater. Four snootyville homes are selected at random.
what is the chance that all four homes will have the home theater room?
What is the chance that non of the homes in the sample of four will have the home theater room?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
p = .6
q = .4
p = probability they have a theater room.
q = probability they don't.
4 selected at random.
probablity all 4 have a theater room = .6^4 = .1296
all of the probabilities are:
p(0) = .6^0 * .4^4 * 4C0 = .0256
p(1) = .6^1 * .4^3 * 4C1 = .1536
p(2) = .6^2 * .4^2 * 4C2 = .3456
p(3) = .6^3 * .4^1 * 4C3 = .3456
p(4) = .6^4 * .4^0 * 4C4 = .1296
sum of all probabilities is equal to 1
nCx is equal to n! / (x! * (n-x)!)

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