For three non-impossible events $A$,$B$,and $C$,$P(A \cap B \cap C) = 0$,$P(A \cup B \cup C) = \frac{3}{4}$,$P(A \cap B) = \frac{1}{3}$,and $P(C) = \frac{1}{6}$. The probability that exactly one of $A$ or $B$ occurs but $C$ does not occur is:

  • A
    $\frac{1}{12}$
  • B
    $\frac{5}{6}$
  • C
    $\frac{1}{4}$
  • D
    $\frac{2}{3}$

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