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Question
A lot of 100 items contain 10 defective items. Five items are selected at random from the lot and sent to the retail store. What is the probability that the store will receive at most one defective item?
Solution
Let X = number of defective items.
p = probability that item is defective
∴ p = `10/100 = 1/10`
∴ q = `1 - "p" = 1 - 1/10 = 9/10`
Given: n = 5
∴ X ~ B `(5, 1/10)`
The p.m.f. of X is given as:
P[X = x] = `"^nC_x p^x q^(n - x)`
i.e. p(x) = `"^5C_x (1/10)^x (9/10)^(5 - x)`
P (store will receive at most one defective item)
= P[X ≤ 1] = P[X = 0] + P[X = 1]
= p(0) + p(1)
`= ""^5C_0 (1/10)^0 (9/10)^(5 - 0) + "^5C_1 (1/10)^1 (9/10)^(5 - 1)`
`= 1 xx 1 xx (9/10)^5 + 5 xx 1/10 xx (9/10)^4`
`= (0.9)^5 + (0.05)(0.9)^4`
`= (0.9 + 0.5)(0.9)^4`
= (1.4)(0.9)4
Hence, the probability that the store will receive at most one defective item is (1.4)(0.9)4
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