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(B) The repeated selections of articles in a random sample are Bernoulli trials. Let $X$ denote the number of defective articles in a random sample of

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$\frac{22 	imes 9^{3}}{10^{11}}$

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(B) The repeated selections of articles in a random sample are Bernoulli trials. Let $X$ denote the number of defective articles in a random sample of $12$ articles.Clearly,$X$ follows a binomial distribution with $n=12$ and $p=10 \% = \frac{10}{100} = \frac{1}{10}$.Therefore,$q = 1 - p = 1 - \frac{1}{10} = \frac{9}{10}$.The probability mass function is given by $P(X=x) = ^{n}C_{x} p^{x} q^{n-x}$.Substituting the values,we get $P(X=9) = ^{12}C_{9} \left(\frac{1}{10}ight)^{9} \left(\frac{9}{10}ight)^{12-9}$.$P(X=9) = ^{12}C_{3} \left(\frac{1}{10}ight)^{9} \left(\frac{9}{10}ight)^{3}$.Since $^{12}C_{3} = \frac{12 	imes 11 	imes 10}{3 	imes 2 	imes 1} = 220$,we have:$P(X=9) = 220 	imes \frac{1}{10^{9}} 	imes \frac{9^{3}}{10^{3}} = 220 	imes \frac{9^{3}}{10^{12}} = \frac{22 	imes 9^{3}}{10^{11}}$.

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?

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