For three non-impossible events A,B,and C,P(A | vedclass

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(C) We want to find the probability of the event that exactly one of $A$ or $B$ occurs and $C$ does not occur. This is represented by the shaded regio

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$\frac{1}{4}$

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(C) We want to find the probability of the event that exactly one of $A$ or $B$ occurs and $C$ does not occur. This is represented by the shaded region in the Venn diagram,which is $(A \cup B) \setminus (A \cap B) \setminus C$.Given $P(A \cup B \cup C) = P(A \cup B) + P(C) - P((A \cup B) \cap C)$.Since $P(A \cap B \cap C) = 0$,we have $P((A \cup B) \cap C) = P((A \cap C) \cup (B \cap C)) = P(A \cap C) + P(B \cap C)$.However,a simpler approach is to use the inclusion-exclusion principle for the union:$P(A \cup B \cup C) = P(A \cup B) + P(C) - P((A \cup B) \cap C)$.Since $P(A \cap B \cap C) = 0$,the region $(A \cup B) \cap C$ is simply $(A \cap C) \cup (B \cap C)$.The probability of exactly one of $A$ or $B$ occurring is $P(A \cup B) - P(A \cap B)$.We need to exclude the part where $C$ occurs,which is $P((A \cup B) \cap C)$.Thus,the required probability is $P(A \cup B) - P(A \cap B) - P((A \cup B) \cap C)$.From $P(A \cup B \cup C) = P(A \cup B) + P(C) - P((A \cup B) \cap C)$,we get $P(A \cup B) - P((A \cup B) \cap C) = P(A \cup B \cup C) - P(C) = \frac{3}{4} - \frac{1}{6} = \frac{9-2}{12} = \frac{7}{12}$.Now,the required probability is $P(A \cup B) - P(A \cap B) - P((A \cup B) \cap C) = \frac{7}{12} - P(A \cap B) = \frac{7}{12} - \frac{1}{3} = \frac{7-4}{12} = \frac{3}{12} = \frac{1}{4}$.

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:

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