Question 579359
A batch of 20 used automobile alternators contains 4 defectives and 16 good ones.
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If 3 alternators are selected at random, find the probability of the events (Using the Rule of combinations): 
A= None of the defective appears
P(no defectives) = P(3 good ones) = 16C3/20C3 = 0.4912
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B= Exactly two defective appears
P(x = 2) = 3C2(4/20)^2(16/20) = 3*(1/25)(4/5) = 12/125 = 0.0960
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Note: The answer to "B" is not 0.084
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Cheers,
Stan H.
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