There are 5% defective items in a large bulk of items. What is the probability that a sample of 10 items will include not more than one defective item?

#### Solution

Let X = number of defective items

p = probability of defective item

∴ p = 5% = `5/100 = 1/20`

and q = 1 - p = `1 - 1/20 = 19/20`

∴ X ~ B `(10, 1/20)`

The p.m.f. of X is given by

P(X = x) = `"^nC_x p^x q^(n - x)`

i.e. p(x) = `"^10C_x (1/20)^x (19/20)^(10 - x)`, x = 0, 1, 2, ...,10

P(sample of 10 items will include not more than one defective item) = P[X ≤ 1]

= P(x = 0) + P(x = 1)

`= ""^10C_0 (1/20)^0(19/20)^(10 - 0) + "^10C_1 (1/20)^1 (19/20)^(10 - 1)`

`= 1*1*(19/20)^10 + 10 xx (1/20) xx (19/20)^9`

`= (19/20)^9 [19/20 + 10/20]`

`= (19/20)^9 (29/20) = 29(19^9/20^10)`

Hence, the probability that a sample of 10 items will include not more than one defective item = `29 (19^9/20^10)`.