Question

It is known that $10\%$ of certain articles manufactured are defective. The probability that in a random sample of $12$ such articles, $9$ are defective is:

• A. $\frac{22}{(10{)}^{11}}$
• B. $\frac{22\times 9^3}{(10{)}^{11}}$
• C. $\frac{9^3}{{10}^{11}}$
• D. $\frac{{11}^3}{{10}^{11}}$

Answer 1

The correct option is B $\frac{22\times 9^3}{(10{)}^{11}}$

Let us assume $X$ represent the number of times of selecting defected articles in a random sample space of given $12$ articles.
Clearly, we have $X$ has the binomial distribution where $n=12$ and $p=10\%=1/10$
And, $q=1-p=1-\frac{1}{10}=\frac{9}{10}$
$\therefore P(X=x)={}^n{C}_x{q}^{n-x}{p}^x$
$={}^{12}{C}_x\times {\left(\frac{9}{10}\right)}^{12-x}\times {\left(\frac{1}{10}\right)}^x$
Hence, probability of selecting $9$ defective articles ${=}^{12}{C}_9{\left(\frac{9}{10}\right)}^3.{\left(\frac{1}{10}\right)}^9=220.\frac{9^3}{{10}^3}.\frac{1}{{10}^9}=\frac{22\times 9^3}{(10{)}^{11}}$