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(D) Let $X$ denote the number of defective bulbs out of $20$ bulbs selected.Given that the total number of bulbs is $100$ and $10$ are defective,the p

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$\left(\frac{9}{10}\right)^{20}$

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(D) Let $X$ denote the number of defective bulbs out of $20$ bulbs selected.Given that the total number of bulbs is $100$ and $10$ are defective,the probability $p$ of selecting a defective bulb is $p = \frac{10}{100} = \frac{1}{10}$.Consequently,the probability $q$ of selecting a non-defective bulb is $q = 1 - p = 1 - \frac{1}{10} = \frac{9}{10}$.We are selecting $n = 20$ bulbs. The probability of getting no defective bulb is given by the binomial distribution formula $P(X = k) = {}^{n}C_{k} \cdot p^{k} \cdot q^{n-k}$.For $k = 0$,$P(X = 0) = {}^{20}C_{0} \cdot \left(\frac{1}{10}\right)^{0} \cdot \left(\frac{9}{10}\right)^{20-0}$.$P(X = 0) = 1 \cdot 1 \cdot \left(\frac{9}{10}\right)^{20} = \left(\frac{9}{10}\right)^{20}$.

In a box containing $100$ bulbs,$10$ are defective. The probability that out of $20$ bulbs selected at random,none is defective is

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