Question

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?

Answer 1

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

Here, p=10%= $\frac{10}{100}=\frac{1}{10}$

and q=1 - p=1- $\frac{1}{10}=\frac{9}{10}$.

Clearly, X has a binomial distribution with n=12,

p=$\frac{1}{10}andq=\frac{9}{10}$

P(X=r) $={}^n{C}_r.p^r{q}^{n-r}={}^{12}{C}_r.{\left(\frac{1}{10}\right)}^r{\left(\frac{9}{10}\right)}^{12-r}$

Required probability = P (9 itmes are defective) = P(X=9) $={}^{19}{C}_9{p}^9{q}^3.$
$={}^{12}{C}_3{\left(\frac{1}{10}\right)}^9{\left(\frac{9}{10}\right)}^3$
= $\frac{12\times 11\times 10}{1\times 2\times 3}.=\frac{9^3}{{10}^{12}}=\frac{22\times 9^3}{{10}^{11}}$