Question 940211:
Suppose 40% of the restaurants in a certain part of a town are in violation of the health code. A health inspector randomly selects eight of the restaurants for inspection. (Round your answers to four decimal places.)
(a) What is the probability that none of the restaurants are in violation of the health code?
(b) What is the probability that one of the restaurants is in violation of the health code?
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(c) What is the probability that at least two of the restaurants are in violation of the health code?
Answer by ewatrrr(24785)   (Show Source):
You can put this solution on YOUR website! p(in violation) = .40, n = 8
....
x the number in violation
P(x = 0) = (.60)^8
p(x = 1) = binompdf(8,.40,1)
P(x ≥ 2) = 1 - binomcdf(8,.40,1)
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