What is the <strong>MAXIMUM</strong> value of P(B\cap C)?
Consider the following for the next two (02) items that follow : A, B and C are three events such that P(A)=0.6, P(B)=0.4, P(C)=0.5, P(A\cup B)=0.8, P(A\cap C)=0.3, and P(A\cap B\cap C)=0.2 and P(A\cup B\cup C)\geq 0.85.
- A. 0.1
- B. 0.2
- C. 0.35 ✓
- D. 0.45
Correct Answer: C. 0.35
Explanation
From the previous solution, we derived the relation P(A \cup B \cup C) = 1.2 - P(B \cap C). We are given that P(A \cup B \cup C) \geq 0.85. Substituting this into the inequality gives 1.2 - P(B \cap C) \geq 0.85, which simplifies to P(B \cap C) \leq 1.2 - 0.85 = 0.35. Thus, the maximum value of P(B \cap C) is 0.35.
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