10% of certain articles manufactured are defective. We need to determine the probability that in a random sample of $(n = 12)$ such articles, 9 are defective.
Let X be the number of times we select a defective article out of 12 articles. It is aBernoulli trial as they satisfy the conditions (i) finite number of trials, (ii) independent trials, (iii) there is a definite outcome and (iv) the probability of success does not change for each trial..
P (getting a defective article) $ = p = $10%$\; = \large\frac{1}{10}$$ \rightarrow q = 1 - p =\large \frac{9}{10}$
Since X has a bionomial distribution, the probability of x success in n-Bernoulli trials, $P (X = x) = \large^{n}C_x. p^x.q^{n–x}$ where $x = 0, 1, 2,...,n$ and $(q = 1 – p)$
We need to calculate the probability of 9 items being defective, i.e, $P (X = 9)$
$\Rightarrow P (X=9) =\; ^{12}C_9 \times \large\frac{1}{10}^9 $$\times \large\frac{9}{10}^{12-9}$$ =\; \large\frac{12 \times\ 11 \times 10 \times 9!}{9! \times3 \times 2 \times 1}( \frac{1}{10})^9 (\frac{9}{10})^3$
$ P (X = 9) = 22\large\frac{9^3}{10^{11}}$