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A shipment of 40 parts contains 12 defective parts. Suppose 3 parts are selected at random...

A shipment of 40 parts contains 12 defective parts. Suppose 3 parts are selected at random from the shipment. What is the probability that all 3 parts are defective?

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Answer #1

probability that all 3 parts are defective =P(first defective)*P(second defective|first defective)*P(third defective|first two defective)=(12/40)*(11/39)*(10/38)=0.0223

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