SOLUTION: A batch of 20 used automobile alternators contains 4 defectives. If 3 alternators are selected at random, find the probability of the events (Using the Rule of combinations):
A=
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Question 579359: A batch of 20 used automobile alternators contains 4 defectives. If 3 alternators are selected at random, find the probability of the events (Using the Rule of combinations):
A= None of the defective appears
B= Exactly two defective appears
The answer for A is .491 the answer for B is .084 but I can not figure out for the life of me how they got these answers.
I've tried 3 of 20 which gave the 1140 total selections possible. Than if there is 4 defective and I want the probability of selecting none. I tried 0 of 4 which doesn't work 1 of 4 likewise doesn't work. I'm stuck. Please help I know this is probably going to be something so simple I am missing.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
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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