Question

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?
the probability of the restaurant buying at least two nondefective units is
(round to three decimal places as needed.)

Explanation:

Step1: Calculate the number of non - defective units

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

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 \(3\). The number of ways to choose \(3\) units out of \(13\) is \(C(13,3)=\frac{13!}{3!(13 - 3)!}=\frac{13\times12\times11}{3\times2\times1}=286\)
The number of ways to choose \(2\) non - defective and \(1\) defective is \(C(10,2)\times C(3,1)=\frac{10!}{2!(10 - 2)!}\times\frac{3!}{1!(3 - 1)!}=\frac{10\times9}{2\times1}\times3 = 135\)
So \(P(X = 2)=\frac{C(10,2)\times C(3,1)}{C(13,3)}=\frac{135}{286}\)

• For \(P(X = 3)\):

We choose \(3\) non - defective units out of \(10\). The number of ways is \(C(10,3)=\frac{10!}{3!(10 - 3)!}=\frac{10\times9\times8}{3\times2\times1}=120\)
So \(P(X = 3)=\frac{C(10,3)}{C(13,3)}=\frac{120}{286}\)

Step3: Calculate \(P(X\geq2)\)

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

Answer:

\(0.891\)