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(C) Step $1$: Define the variables. Total articles $= 15$. Defective articles $= 3$. Non-defective articles $= 15 - 3 = 12$. Sample size $= 5$. We nee

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$\frac{220}{1001}$

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(C) Step $1$: Define the variables. Total articles $= 15$. Defective articles $= 3$. Non-defective articles $= 15 - 3 = 12$. Sample size $= 5$. We need the probability of choosing exactly $2$ defective articles in this sample. Step $2$: Use combinations to calculate possible outcomes. The probability is calculated using the hypergeometric distribution formula: $P(X = k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}}$ $1$. Number of ways to choose $2$ defective articles out of $3$: $\binom{3}{2} = \frac{3!}{2!1!} = 3$. $2$. Number of ways to choose $3$ non-defective articles out of $12$: $\binom{12}{3} = \frac{12 	imes 11 	imes 10}{3 	imes 2 	imes 1} = 220$. $3$. Total number of ways to choose $5$ articles out of $15$: $\binom{15}{5} = \frac{15 	imes 14 	imes 13 	imes 12 	imes 11}{5 	imes 4 	imes 3 	imes 2 	imes 1} = 3003$. Step $3$: Calculate the probability. $P = \frac{3 	imes 220}{3003} = \frac{660}{3003} = \frac{220}{1001}$. Thus,the correct answer is $\frac{220}{1001}$.

In a consignment of $15$ articles,it is found that $3$ are defective. If a sample of $5$ articles is chosen at random from it,then the probability of having $2$ defective articles is

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