Posted by Anonymous on .
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,
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) =