For three events A, B and C,P(text{Exactly on | vedclass

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(C) Let $P(A), P(B), P(C)$ be the probabilities of events $A, B, C$. Given: $P(A) + P(B) - 2P(A \cap B) = \frac{1}{4}$ ... $(1)$ $P(B) + P(C) - 2P(B \

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$\frac{7}{16}$

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(C) Let $P(A), P(B), P(C)$ be the probabilities of events $A, B, C$. Given: $P(A) + P(B) - 2P(A \cap B) = \frac{1}{4}$ ... $(1)$ $P(B) + P(C) - 2P(B \cap C) = \frac{1}{4}$ ... $(2)$ $P(C) + P(A) - 2P(C \cap A) = \frac{1}{4}$ ... $(3)$ Adding $(1), (2),$ and $(3)$,we get: $2[P(A) + P(B) + P(C)] - 2[P(A \cap B) + P(B \cap C) + P(C \cap A)] = \frac{3}{4}$ Dividing by $2$: $[P(A) + P(B) + P(C)] - [P(A \cap B) + P(B \cap C) + P(C \cap A)] = \frac{3}{8}$ We know that $P(A \cup B \cup C) = [P(A) + P(B) + P(C)] - [P(A \cap B) + P(B \cap C) + P(C \cap A)] + P(A \cap B \cap C)$. 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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