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(C) Let $P(A) = x$, $P(B) = y$, $P(C) = z$, $P(A \cap B) = p$, $P(B \cap C) = q$, $P(C \cap A) = r$, and $P(A \cap B \cap C) = k = \frac{1}{16}$.Given

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$\frac{7}{16}$

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(C) Let $P(A) = x$, $P(B) = y$, $P(C) = z$, $P(A \cap B) = p$, $P(B \cap C) = q$, $P(C \cap A) = r$, and $P(A \cap B \cap C) = k = \frac{1}{16}$.Given $P(	ext{exactly one of } A 	ext{ or } B) = P(A) + P(B) - 2P(A \cap B) = x + y - 2p = \frac{1}{4}$.Similarly, $y + z - 2q = \frac{1}{4}$ and $z + x - 2r = \frac{1}{4}$.Adding these three equations: $2(x + y + z) - 2(p + q + r) = \frac{3}{4} \implies x + y + z - (p + q + r) = \frac{3}{8}$.The probability that at least one event occurs is $P(A \cup B \cup C) = (x + y + z) - (p + q + r) + k$.Substituting the values: $P(A \cup B \cup C) = \frac{3}{8} + \frac{1}{16} = \frac{6+1}{16} = \frac{7}{16}$.

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