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(B) $A, B, C$ are pairwise independent events. $\Rightarrow P(A \cap B) = P(A) \cdot P(B) = P^2$ $P(B \cap C) = P(B) \cdot P(C) = P^2$ $P(C \cap A) =

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$3P(1-P)$

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(B) $A, B, C$ are pairwise independent events. $\Rightarrow P(A \cap B) = P(A) \cdot P(B) = P^2$ $P(B \cap C) = P(B) \cdot P(C) = P^2$ $P(C \cap A) = P(C) \cdot P(A) = P^2$ Given that all of them cannot occur simultaneously,$P(A \cap B \cap C) = 0$. Using the inclusion-exclusion principle: $P(A \cup B \cup C) = P(A) + P(B) + P(C) - [P(A \cap B) + P(B \cap C) + P(C \cap A)] + P(A \cap B \cap C)$ $P(A \cup B \cup C) = P + P + P - [P^2 + P^2 + P^2] + 0$ $P(A \cup B \cup C) = 3P - 3P^2 = 3P(1-P)$.

Let $A, B$ and $C$ be three events associated with sample space $S$. $A, B$ and $C$ are pairwise independent and $P(A)=P(B)=P(C)=P$. If all of them cannot occur simultaneously,then $P(A \cup B \cup C)$ is equal to

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