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**apcalculus** Let X be the number of defective units in the sample of 10.

P(X= or > 1)= P(X=1) + P(X=2) +...+ P(X=9) since 9 is the maximum number of defective units in the shipment.

P(X=1) is the probability of getting EXACTLY one defective unit, then all the other nine work.

= (9/60) (51/59)(50/58) (49/57) continue until you have 10 terms.

Doing this for all nine cases is a lot of work so we write:

P(X>=1) = 1 - P(X <1)

P(X < 1) = P(X=0) which is the probability of getting 0 defective units. This means all 10 units chosen at random will work:

51/60 is the probability of the first one working

50/59 is the probability of the second one working

... write them all out, and multiply them since we are assuming independence.

1 - the product above is your answer.

Good luck!