Question

8% of items in a shipment are known to be defective. If a sample of 5 items is randomly selected from this shipment, what is the probability that at least one defective item will be observed in this sample? Round your result to 2 significant places after the decimal (For example, 0.86732 should be entered as 0.87).

Answer 1

Answer: 0.34

Explanation:

Given : The proportion of items in a shipment are known to be defective : p=0.08

Number of items selected = 5

By using binomial probability distribution formula, the probability of getting success in x trials is given by :-

`P(x)=^nC_xp^x(1-p)^(n-x)`

Now, the probability that at least one defective item will be observed in this sample will be :-

`P(x\geq1)=1-P(x<1)\\\\=1-P(0)\\\\=1-^5C_0(0.08)^0(1-0.08)^(5)\\\\=1-(0.92)^5=0.3409184768\approx0.34`

Hence, the probability that at least one defective item will be observed in this sample is 0.34