What is the <strong>MINIMUM</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.

  1. A. 0.1
  2. B. 0.2 ✓
  3. C. 0.35
  4. D. 0.45

Correct Answer: B. 0.2

Explanation

First, find P(A \cap B) using P(A \cup B) = P(A) + P(B) - P(A \cap B) \implies 0.8 = 0.6 + 0.4 - P(A \cap B) \implies P(A \cap B) = 0.2. Using the formula for union of three events: P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(A \cap C) + P(A \cap B \cap C). Substitute the values: P(A \cup B \cup C) = 0.6 + 0.4 + 0.5 - 0.2 - P(B \cap C) - 0.3 + 0.2 = 1.2 - P(B \cap C). Since probability cannot exceed 1, P(A \cup B \cup C) \leq 1, meaning 1.2 - P(B \cap C) \leq 1 \implies P(B \cap C) \geq 0.2. The minimum value is 0.2.

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