prob(defect) = .16
prob(not defect) = .84
prob(at least 1 defective)
= 1 - prob(none defective)
= 1 - C(5,0) (.16)^0 (.84)^5
= 1 - .84^5
= appr.582
Suppose that 16% of the CD's that are distributed are defective. If you choose 5 CD's randomly, what is the probability that at least 1 of them is defective?
2 answers
90