In a shipment of 20 computers, 2 are defective. Three computers are randomly selected and tested. What is the probability that all 2 are defective if the first one is not replaced after being tested?
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Expert's answer
2020-12-15T07:04:56-0500
Probabilities of being defective in case the 1st is not replaced (2nd is replaced)
First one: "\\frac{2}{20}"
Second one: "\\frac{1}{19}"
Third one: no more defective
Therefore we have: "\\frac{2}{20}\\cdot\\frac{1}{19}=\\frac{1}{190}"
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