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(D) Let the probabilities of $A$, $B$, and $C$ passing the test be $P(T_A) = k$, $P(T_B) = 2k$, and $P(T_C) = 3k$ respectively, based on the ratio $1:

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$\frac{1}{5}$

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(D) Let the probabilities of $A$, $B$, and $C$ passing the test be $P(T_A) = k$, $P(T_B) = 2k$, and $P(T_C) = 3k$ respectively, based on the ratio $1:2:3$. Let $I$ be the event that a person faces the interview successfully. The conditional probabilities are $P(I|T_A) = 0.8$, $P(I|T_B) = 0.7$, and $P(I|T_C) = 0.6$. We need to find the probability that $A$ is selected, given that one of them is selected. This is calculated using Bayes' Theorem: $P(A|I) = \frac{P(T_A) 	imes P(I|T_A)}{P(T_A) 	imes P(I|T_A) + P(T_B) 	imes P(I|T_B) + P(T_C) 	imes P(I|T_C)}$ $P(A|I) = \frac{k 	imes 0.8}{k 	imes 0.8 + 2k 	imes 0.7 + 3k 	imes 0.6}$ $P(A|I) = \frac{0.8k}{0.8k + 1.4k + 1.8k} = \frac{0.8k}{4.0k} = \frac{0.8}{4} = \frac{1}{5}$.

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