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Let A ,B ,C be three events such that P(...

Let `A ,B ,C` be three events such that `P(A)=0. 3 ,P(B)=0. 4 ,P(C)=0. 8 ,P(AnnB)=0. 88 ,P(AnnC)=0. 28 ,P(AnnBnnC)=0. 09.` If `P(AuuBuuC)geq0. 75 ,` then show that `0. 23lt=P(BnnC)lt=0. 48.`

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To solve the problem step-by-step, we will use the principle of inclusion-exclusion for probabilities. We are given the probabilities of events A, B, and C, as well as their intersections. We need to show that \(0.23 < P(B \cap C) < 0.48\) given that \(P(A \cup B \cup C) \geq 0.75\). ### Step 1: Write the formula for \(P(A \cup B \cup C)\) Using the principle of inclusion-exclusion, we can express \(P(A \cup B \cup C)\) as follows: \[ P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C) ...
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