A shipment of 21 computers contains 5 defective units. If 3 units are selected at random what is the probability that: a. All 3 are defective? (Round answer to 4 decimal places) Answer: (4 marks) b. All 3 are not defective? (Round answer to 4 decimal places) Answer: (4 marks)
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- Probability of first pick being defective: 5/21 - Probability of second pick being defective: 4/20 - Probability of third pick being defective: 3/19 - Multiply these probabilities together: (5/21) * (4/20) * (3/19) = 0.007519 Show more…
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