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

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%