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Question
There are 10% defective items in a large bulk of items. What is the probability that a sample of 4 items will include not more than one defective item?
Solution
Let X denote the number of defective items.
P(getting defective item) = p = `(10)/(100)` = 0.1 ...[Given]
∴ 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(sample will include not more than one defective item)
= P(X ≤ 1) = P(X = 0 or X = 1)
= P(X = 0) + P(X = 1)
= `""^4"C"_0 (0.1)^0 (0.9)^4 + ""^4"C"_1 (0.1)^1 (0.9)^3`
= (0.9)4 + 4 x 0.1 x (0.9)3
= (0.9)3 (0.9 + 0.4)
= 1.3 x (0.9)3.
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