A carton contains 12 bulbs of which 3 are defective. Three bulbs are drawn at random from the carton. What is the probability that at least one of them is defective?

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Accepted Answer

34/55. Total selections = C(12,3) = 220. Selections with no defective bulb = C(9,3) = 84, a probability of 84/220 = 21/55. By the complement rule P(at least one defective) = 1 − 21/55 = 34/55.

A carton contains 12 bulbs of which 3 are defective. Three bulbs are drawn at random from the carton. What is the probability that at least one of them is defective?

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