Question 886220
if the probability of a defective manufacturing product is 0.05, and if there 18 of this product, what is the probability that
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Binomial Problem with n = 18 and p(defective) = 0.05 ; p(not) = 0.95
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 a. exactly 4 are defective?
P(x = 4) = 18C4(0.05)^4*(0.95)^14 = binompdf(18,0.05,4) = 0.0093
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 b. at least 15 are not defective?
Note: P(<= 3 defective)+P(>= 15 not defective) = 1 
So, P
P(not defective >= 15) = 1 - p(defect <= 3) = 1-binomcdf(18,0.05,3) = 0.0109
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Cheers,
Stan H.
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