Probability
Addition Theorem of Probability
Grade 12
Question:
<p>A, B, C are events such that \(P(A) = 0.3\), \(P(B) = 0.4\), \(P(C) = 0.8\), \(P(AB) = 0.08\), \(P(AC) = 0.28\) and \(P(ABC) = 0.09\). If \(P(A \cup B \cup C) \geq 0.75\), then show that \(P(BC)\) lies in the interval \(0.23 \leq x \leq 0.48\).</p>
<p>\([0.23, 0.48]\)</p>
<p>\([0.25, 0.50]\)</p>
<p>\([0.20, 0.45]\)</p>
<p>\([0.30, 0.48]\)</p>
Step-by-Step Solution
Key Concept: Use the inclusion-exclusion principle P(A∪B∪C) = P(A) + P(B) + P(C) - P(AB) - P(AC) - P(BC) + P(ABC) and the constraint P(A∪B∪C) ≥ 0.75 to establish bounds on P(BC). The key is recognizing that P(BC) ≥ P(ABC) (since ABC ⊆ BC) provides the lower bound.
<p><strong>Step 1: Apply Inclusion-Exclusion Principle</strong></p><p>P(A∪B∪C) = P(A) + P(B) + P(C) - P(AB) - P(AC) - P(BC) + P(ABC)</p><p>Substituting known values:</p><p>P(A∪B∪C) = 0.3 + 0.4 + 0.8 - 0.08 - 0.28 - P(BC) + 0.09</p><p>P(A∪B∪C) = 1.23 - P(BC)</p><p><strong>Step 2: Find Upper Bound on P(BC)</strong></p><p>Given P(A∪B∪C) ≥ 0.75:</p><p>1.23 - P(BC) ≥ 0.75</p><p>P(BC) ≤ 1.23 - 0.75 = 0.48</p><p><strong>Step 3: Find Lower Bound on P(BC)</strong></p><p>Since ABC ⊆ BC, we must have:</p><p>P(BC) ≥ P(ABC) = 0.09</p><p>However, we also need P(BC) ≥ P(AC) = 0.28 because P(ABC) = 0.09 and P(AC) = 0.28 implies the remaining part of AC (which must intersect with B at least partially or not at all). Actually, since P(ABC) = 0.09 < P(AC) = 0.28, part of C is in A but not B. For consistency: P(BC) ≥ P(ABC) = 0.09. But examining the union constraint more carefully: since P(B) = 0.4 and P(AB) = 0.08, and we need valid probabilities, P(BC) must be at least 0.23 to ensure P(A∪B∪C) doesn't exceed 1 while satisfying all constraints.</p><p><strong>Step 4: Verify Bounds</strong></p><p>Lower bound: P(BC) ≥ 0.23 (ensures consistency with all given probabilities and union constraint)</p><p>Upper bound: P(BC) ≤ 0.48 (from P(A∪B∪C) ≥ 0.75)</p><p>∴ <strong>0.23 ≤ P(BC) ≤ 0.48</strong></p>
Correct Answer: A