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(D) The total number of apples is $N = 100$. The number of defective apples is $M = 10$,and the number of non-defective apples is $N - M = 90$. We sel

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$0.01458$

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(D) The total number of apples is $N = 100$. The number of defective apples is $M = 10$,and the number of non-defective apples is $N - M = 90$. We select a sample of $n = 6$ apples. We want to find the probability that exactly $k = 3$ apples are defective. This follows a hypergeometric distribution. The probability is given by: $P(X = k) = \frac{\binom{M}{k} \binom{N-M}{n-k}}{\binom{N}{n}}$ Substituting the values: $P(X = 3) = \frac{\binom{10}{3} \binom{90}{3}}{\binom{100}{6}}$ Calculating the combinations: $\binom{10}{3} = \frac{10 	imes 9 	imes 8}{3 	imes 2 	imes 1} = 120$ $\binom{90}{3} = \frac{90 	imes 89 	imes 88}{3 	imes 2 	imes 1} = 117480$ $\binom{100}{6} = \frac{100 	imes 99 	imes 98 	imes 97 	imes 96 	imes 95}{6 	imes 5 	imes 4 	imes 3 	imes 2 	imes 1} = 1192052400$ $P(X = 3) = \frac{120 	imes 117480}{1192052400} = \frac{14097600}{1192052400} \approx 0.0118$ However,using the binomial approximation (since $n/N = 0.06 $P(X = 3) = \binom{6}{3} (0.1)^3 (0.9)^3 = 20 	imes 0.001 	imes 0.729 = 0.01458$. Thus,the correct option is $D$.

In a box containing $100$ apples,$10$ are defective. The probability that in a sample of $6$ apples,$3$ are defective is

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