Posted by **Anonymous** on Sunday, July 22, 2012 at 10:48pm.

The J.O. Supplies Company buys calculators from a Korean supplier. The probability of a defective calculator is 10%. If 15 calculators are selected at random, what is the probability that 4 or more of the calculators will be defective

- probability -
**Reiny**, Sunday, July 22, 2012 at 11:51pm
prob of defect = .1

prob of NOT defect = .9

4 or more defective means

exclude cases of 0, 1, 2, or 3 defective

prob of none defective = C(15,0) (.1^0)( .9^15) = .20589

prob of one defective = C(15,1) (.1)^1 (.9)^14 =.34315

prob of two defective = C(15,2) (.1^2)(.9^13) = .266896

prob of three defective = C(15,3) (.1^3)(.9^12) = .128505

So prob of 4 or more defective

= 1 - sum of the above cases

= .944444

prob of three defective = C(15,3) (.1^12)(.9^3) =

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