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Question
Solve the following problem :
A large chain retailer purchases an electric device from the manufacturer. The manufacturer indicates that the defective rate of the device is 10%. The inspector of the retailer randomly selects 4 items from a shipment. Find the probability that the inspector finds at most one defective item in the 4 selected items.
Solution
Let X be the number of defective items.
P(an item is defective) = p = `(10)/(100)` = 0.1
∴ q = 1 – p = 1 – 0.1 = 0.9
Given, n = 4
∴ X ~ B(4, 0.1)
The p.m.f. of X is given by
P(X = x) = `""^4"C"_x (0.1)^x (0.9)^(4 - x), x` = 0, 1, ...,4
P(at most one defective item)
= P(X ≤ 1)
= P(X = 0 or = 1)
= P(X = 0) + P(X = 1)
= `""^4"C"_0 (0.1)^0 (0.9)^4 + ""^4"C"_1 (0.1) (0.9)^3`
= (0.9)4 + 4(0.1) (0.9)3
= 0.9477
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