A box contains 100 bulbs,out of which 10 are | vedclass

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(D) Total number of bulbs $N = 100$.Number of defective bulbs $D = 10$.Number of non-defective bulbs $G = 100 - 10 = 90$.We are drawing a sample of $n

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$(\frac{9}{10})^{5}$

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(D) Total number of bulbs $N = 100$.Number of defective bulbs $D = 10$.Number of non-defective bulbs $G = 100 - 10 = 90$.We are drawing a sample of $n = 5$ bulbs without replacement.The probability that none is defective is given by the hypergeometric distribution formula:$P(X = 0) = \frac{\binom{90}{5} \binom{10}{0}}{\binom{100}{5}}$$P(X = 0) = \frac{\frac{90 	imes 89 	imes 88 	imes 87 	imes 86}{5 	imes 4 	imes 3 	imes 2 	imes 1}}{\frac{100 	imes 99 	imes 98 	imes 97 	imes 96}{5 	imes 4 	imes 3 	imes 2 	imes 1}}$$P(X = 0) = \frac{90 	imes 89 	imes 88 	imes 87 	imes 86}{100 	imes 99 	imes 98 	imes 97 	imes 96} \approx (0.9)^{5}$.Given the options provided,the intended model is binomial approximation (sampling with replacement),where $p = \frac{90}{100} = 0.9$ is the probability of picking a non-defective bulb.Thus,$P(X = 0) = (0.9)^{5} = (\frac{9}{10})^{5}$.

$A$ box contains $100$ bulbs,out of which $10$ are defective. $A$ sample of $5$ bulbs is drawn. The probability that none is defective is

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