It is known that 10% of certain articles manufactured are defective. What is the probability that in a random sample of 12 such articles, 9 are defective?

#### Solution

The repeated selections of articles in a random sample space are Bernoulli trails. Let X denote the number of times of selecting defective articles in a random sample space of 12 articles.

Clearly, X has a binomial distribution with n = 12 and p = 10% = `10/100 = 1/10`

q = 1 - p = `1 - 1/10 = 9/10`

Given: n = 12

∴ X ~ B `(12, 1/10)`

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

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

i.e. p(x) = `"^12C_x (1/10)^x (9/10)^(12 - x)`, x = 1, 2, 3,...,12

P(9 defective articles) = P[X = 9]

= p(9) = `"^12C_9 (1/10)^9 (9/10)^(12 - 9)`

`= (12!)/(9! 3!) (1/10)^9 (9/10)^3`

`= (12 xx 11 xx 10 xx 9!)/(9! xx 3 xx 2 xx 1) xx 1/10^9 xx 9^3/10^3`

`= 2 xx 11 xx 10 * 9^3/10^12 = 22 (9^3/10^11)`

Hence, the probability of getting 9 defective articles `=22 (9^3/10^11)`.