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(D) Let $E_1, E_2, E_3, E_4$ be the events of selecting boxes $A, B, C, D$ respectively. Let $F$ be the event that the selected fuse is defective.$P(E

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(D) Let $E_1, E_2, E_3, E_4$ be the events of selecting boxes $A, B, C, D$ respectively. Let $F$ be the event that the selected fuse is defective.$P(E_1) = \frac{5000}{11000} = \frac{5}{11}, P(E_2) = \frac{3000}{11000} = \frac{3}{11}, P(E_3) = \frac{2000}{11000} = \frac{2}{11}, P(E_4) = \frac{1000}{11000} = \frac{1}{11}$.The conditional probabilities of selecting a defective fuse are:$P(F|E_1) = \frac{3}{100}, P(F|E_2) = \frac{2}{100}, P(F|E_3) = \frac{1}{100}, P(F|E_4) = \frac{0.5}{100} = \frac{1}{200}$.Using Bayes' Theorem, the probability that the defective fuse came from box $D$ is $P(E_4|F) = \frac{P(E_4)P(F|E_4)}{\sum_{i=1}^{4} P(E_i)P(F|E_i)}$.$P(E_4|F) = \frac{\frac{1}{11} 	imes \frac{1}{200}}{\frac{5}{11} 	imes \frac{3}{100} + \frac{3}{11} 	imes \frac{2}{100} + \frac{2}{11} 	imes \frac{1}{100} + \frac{1}{11} 	imes \frac{1}{200}}$.$P(E_4|F) = \frac{\frac{1}{2200}}{\frac{15}{1100} + \frac{6}{1100} + \frac{2}{1100} + \frac{1}{2200}} = \frac{\frac{1}{2200}}{\frac{30+12+4+1}{2200}} = \frac{1}{47}$.

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