Posted by **Jason L** on Monday, February 26, 2007 at 4:19pm.

A company claims that no more than 5% of a particular product has a defect. In a shipment of 15 units what is the probability of receiving 3 or 4 defective units?

You can probably use a binomial probability function for this problem.

P(x) = (nCx)(p^x)[q^(n-x)]

n = 15

x = 3, 4

p = .05

q = 1 - p = .95

Find P(3), P(4).

P(3) = (15C3)(.05^3)(.95^12)

P(4) = (15C4)(.05^4)(.95^11)

Finish the calculations; then add together for your probability.

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