a shipment of 13 microwave ovens contains 3 defective units. a restaurant buys three of these units. what is…

a shipment of 13 microwave ovens contains 3 defective units. a restaurant buys three of these units. what is the probability of the restaurant buying at least two nondefective units?\nthe probability of the restaurant buying at least two nondefective units is \n(round to three decimal places as needed.)

a shipment of 13 microwave ovens contains 3 defective units. a restaurant buys three of these units. what is the probability of the restaurant buying at least two nondefective units?\nthe probability of the restaurant buying at least two nondefective units is \n(round to three decimal places as needed.)

Answer

Explanation:

Step1: Calculate the number of non - defective units

The total number of units is (n = 13), and the number of defective units is (d=3). So the number of non - defective units is (N=13 - 3=10).

Step2: Use the combination formula (C(n,k)=\frac{n!}{k!(n - k)!})

The probability of buying at least two non - defective units is (P(X\geq2)=P(X = 2)+P(X = 3)). For (P(X = 2)): We choose 2 non - defective units out of 10 and 1 defective unit out of 3. (C(10,2)=\frac{10!}{2!(10 - 2)!}=\frac{10\times9}{2\times1}=45), (C(3,1)=\frac{3!}{1!(3 - 1)!}=3) The number of ways to choose 3 units out of 13 is (C(13,3)=\frac{13!}{3!(13 - 3)!}=\frac{13\times12\times11}{3\times2\times1}=286) (P(X = 2)=\frac{C(10,2)\times C(3,1)}{C(13,3)}=\frac{45\times3}{286}=\frac{135}{286}) For (P(X = 3)): We choose 3 non - defective units out of 10 and 0 defective units out of 3. (C(10,3)=\frac{10!}{3!(10 - 3)!}=\frac{10\times9\times8}{3\times2\times1}=120), (C(3,0) = 1) (P(X = 3)=\frac{C(10,3)\times C(3,0)}{C(13,3)}=\frac{120\times1}{286}=\frac{120}{286})

Step3: Calculate (P(X\geq2))

(P(X\geq2)=\frac{135 + 120}{286}=\frac{255}{286}\approx0.891)

Answer:

(0.891)