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(A) Let $X$ be the number of defective eggs in the $10$ eggs drawn. Since the drawing is done with replacement,the trials are Bernoulli trials. Here,$

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

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(A) Let $X$ be the number of defective eggs in the $10$ eggs drawn. Since the drawing is done with replacement,the trials are Bernoulli trials. Here,$n = 10$ and the probability of a defective egg is $p = \frac{10}{100} = \frac{1}{10}$. Thus,the probability of a non-defective egg is $q = 1 - p = 1 - \frac{1}{10} = \frac{9}{10}$. We need to find the probability of at least one defective egg,which is $P(X \geq 1)$. Using the complement rule,$P(X \geq 1) = 1 - P(X = 0)$. For a binomial distribution,$P(X = k) = ^{n}C_{k} p^{k} q^{n-k}$. So,$P(X = 0) = ^{10}C_{0} (\frac{1}{10})^{0} (\frac{9}{10})^{10} = 1 	imes 1 	imes (\frac{9}{10})^{10} = (\frac{9}{10})^{10}$. Therefore,$P(X \geq 1) = 1 - (\frac{9}{10})^{10}$.

Ten eggs are drawn successively with replacement from a lot containing $10 \%$ defective eggs. Find the probability that there is at least one defective egg.

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