Question 795394: A lot of 100 semiconductor chips contain 20 that are defective.
a. two are selected , at random, without replacement, from the lot. Determine the probability that the second chip is defective.
b. three are selected at random, without replacement, from the lot. Determine the probability that all are defective.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A lot of 100 semiconductor chips contain 20 that are defective.
a. two are selected , at random, without replacement, from the lot. Determine the probability that the second chip is defective.
P(2nd not defective) = [(20/100)(80/99)+(80/100)(79/99)] = 0.8
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b. three are selected at random, without replacement, from the lot. Determine the probability that all are defective.
P(3 defective) = 20C3/100C3 = 0.0071
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Cheers,
Stan H.
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