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

Given: Sample space =12
Let X be the number of defected articles in a sample space of 12 articles.

Picking articles from a random sample space are the Bernoulli trials
So, X has binomial distribution.
$P(X=x){=}^n{C}_x{q}^{n-x}{p}^x$
Probablity that certain article that is manufactured is defective is $p=10\%=\frac{1}{10}$
Here$n=12,p=10\%=\frac{1}{10}$
$\Rightarrow q=1-p=1-\frac{1}{10}=\frac{9}{10}$
Putting the value of p,q & n
$P(X=x){=}^{12}{C}_x{\left(\frac{9}{10}\right)}^{12-x}\cdot {\left(\frac{1}{10}\right)}^x...\left(1\right)$
Probability of selecting 9 defective articles =P(X=x)
Putting the value of x=9 in (1) $\Rightarrow P(X=x){=}^{12}{C}_9{\left(\frac{9}{10}\right)}^3{\left(\frac{1}{10}\right)}^9$

$\Rightarrow P(X=x)=220\cdot {\frac{9}{10}}^3\cdot {\frac{1}{10}}^9$

$\Rightarrow P(X=x)=\frac{22\times 9^3}{{10}^{11}}$