Question 751081: A container has 50 electronic components, of which 10 are defective.
If 8 components are drawn at random from the container, what is the probability that exactly 3 of them are defective?
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! A container has 50 electronic components, of which 10 are defective.
If 8 components are drawn at random from the container, what is the probability that exactly 3 of them are defective?
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Binomial with n = 8 and p(defective) = 10/50 = 1/5
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P(x = 3) = 8C3*(1/5)^3*(4/5)^5 = binompdf(8,1/5,3) = 0.1468
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I used a TI-84 to get the numerical value.
Cheers,
Stan H.
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