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?

Answer 1

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(n, p) with n = 4, p = 0.1 and q = 0.9

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

P(X = x) = `""^n"C"_x (p)^x (q)^(n - 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)  (0.9)^3`

= 1 × (0.9)⁴ + 4 × 0.1 × (0.9)³

= (0.9)⁴ + 4 × 0.1 × (0.9)³

= (0.9)³ [0.9 + 0.4]

= (0.9)³ (1.3)

= 1.3 × (0.9)³.