Question

suppose a shipment of 130 electronic components contains 3 defective components. to determine whether the shipment should be accepted, a quality - control engineer randomly selects 3 of the components and tests them. if 1 or more of the components is defective, the shipment is rejected. what is the probability that the shipment is rejected?
the probability that the shipment is rejected is
(round to four decimal places as needed.)

Explanation:

Step1: Calculate the total number of ways to select 3 components

The total number of ways to select 3 components out of 130 is given by the combination formula \(C(n,r)=\frac{n!}{r!(n - r)!}\), where \(n = 130\) and \(r=3\). So \(C(130,3)=\frac{130!}{3!(130 - 3)!}=\frac{130\times129\times128}{3\times2\times1}= 353720\)

Step2: Calculate the number of ways to select 3 non - defective components

The number of non - defective components is \(130 - 3=127\). The number of ways to select 3 non - defective components out of 127 is \(C(127,3)=\frac{127!}{3!(127 - 3)!}=\frac{127\times126\times125}{3\times2\times1}=333500\)

Step3: Calculate the probability of accepting the shipment

The probability of accepting the shipment (i.e., all 3 components are non - defective) is \(P(\text{accept})=\frac{C(127,3)}{C(130,3)}=\frac{333500}{353720}\approx0.9429\)

Step4: Calculate the probability of rejecting the shipment

The probability of rejecting the shipment is \(P(\text{reject})=1 - P(\text{accept})\)
\(P(\text{reject})=1-\frac{333500}{353720}=\frac{353720 - 333500}{353720}=\frac{20220}{353720}\approx0.0572\)

Answer:

\(0.0572\)