A large chain retailer purchases a certain kind of electronic device from a manufacturer. The manufacturer indicates that the defective rate of the device is 3%. The inspector of the retailer picks 20 items from a shipment. What is the probability that the store will receive at most one defective item?
Solution
Let X = number of a defective electronic device.
p = probability that a device is defective
∴ p = 3% = `3/100`
∴ q = 1 - p = `1 - 3/100 = 97/100`
Given: n = 20
∴ X ~ B `(20, 3/100)`
The p.m.f. of X is given as :
P[X = x] = `"^nC_x p^x q^(n - x)`
i.e. p(x) = `"^20C_x (3/100)^x (97/100)^(20 - x)`
P (store will receive at most one defective item)
= P[X ≤ 1] = p[X = 0] + P[X = 1]
= p(0) + p(1)
`= ""^20C_0 (3/100)^0 (97/100)^(20 - 0) + "^20C_1 (3/100)^1 (97/100)^(20 - 1)`
`= 1 xx 1 xx (0.97)^20 + 20xx (0.03) xx (0.97)^19`
`= (0.97 + 0.6)(0.97)^19`
`= (1.57)(0.97)^19`
Hence, the probability that the store will receive at most one defective item `= (1.57)(0.97)^19`
[Note: Answer in the textbook is incorrect.]
