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 non - defective units?
the probability of the restaurant buying at least two non - defective 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 \(d=3\). 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.
\(P(X = 2)=\frac{C(10,2)\times C(3,1)}{C(13,3)}\)

$$
LATEXBLOCK0
$$

\(P(X = 2)=\frac{45\times3}{286}=\frac{135}{286}\)

• For \(P(X = 3)\):

We choose 3 non - defective units out of 10 and 0 defective units out of 3.
\(P(X = 3)=\frac{C(10,3)\times C(3,0)}{C(13,3)}\)

$$
LATEXBLOCK1
$$

\(P(X = 3)=\frac{120\times1}{286}=\frac{120}{286}\)

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

\(P(X\geq2)=\frac{135 + 120}{286}=\frac{255}{286}\approx0.892\)

Answer:

\(0.892\)