A piece of electronic equipment contains 6 computer chips, 2 of which are defective. 3 chips are selected at random and inspected.
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P(at least one defective) = 1 - P(none defective)
P(none defective) = (4 choose 3)(2 choose 0)/(6 choose 3)
= 4*1/20
= 1/5
So P(at least one defective) = 1 - (1/5) = 4/5
P(none defective) = (4 choose 3)(2 choose 0)/(6 choose 3)
= 4*1/20
= 1/5
So P(at least one defective) = 1 - (1/5) = 4/5