A shipment of 12 microwave ovens contains 2 defective units. A restaurant buys 4 of these units. What is the probability of the
restaurant buying at least 3 nondefective units? The probability of the restaurant buying at least 3 nondefective units is
1 answer:
Answer:
91%
Step-by-step explanation:
we can use combinations:
Total probabilities = 12! / (4!8!) = 479,001,600 / 967,680 = 495
Probability of no defective unit = ₁₀C₄ = 10!/4!6! = 3,628,800 / 17,280 = 210
Probability of one defective unit = ₁₀C₃ · ₂C₁ = 10!/3!7! · 2!/1! = 120 · 2 = 240
Probability of no more than one defective unit = (210 + 240) / 495 = 450 / 495 = 91%
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Best of Luck!
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