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(C) Let $X$ be the number of defective bulbs drawn in $n = 10$ trials. This follows a binomial distribution $B(n, p)$.Here,$n = 10$ and the probabilit

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$1-\left(\frac{9}{10}\right)^{10}$

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(C) Let $X$ be the number of defective bulbs drawn in $n = 10$ trials. This follows a binomial distribution $B(n, p)$.Here,$n = 10$ and the probability of a bulb being defective is $p = 10 \% = \frac{1}{10}$.The probability of a bulb not being defective is $q = 1 - p = 1 - \frac{1}{10} = \frac{9}{10}$.We want to find the probability that at least one bulb is defective,which is $P(X \ge 1)$.Using the complement rule,$P(X \ge 1) = 1 - P(X = 0)$.The probability of getting $0$ defective bulbs in $10$ trials is given by the binomial formula $P(X = k) = {}^{n}C_{k} p^{k} q^{n-k}$.For $k = 0$,$P(X = 0) = {}^{10}C_{0} \left(\frac{1}{10}ight)^{0} \left(\frac{9}{10}ight)^{10-0} = 1 	imes 1 	imes \left(\frac{9}{10}ight)^{10}$.Therefore,$P(X \ge 1) = 1 - \left(\frac{9}{10}ight)^{10}$.

Ten bulbs are drawn successively,with replacement,from a lot containing $10 \%$ defective bulbs. The probability that there is at least one defective bulb is:

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