Probability
Independent Events
Grade 12
Question:
<p>Let \(A\) and \(B\) be two events such that \(P(\overline{A \cup B}) = \dfrac{1}{6}\), \(P(A \cap B) = \dfrac{1}{4}\) and \(P(\overline{A}) = \dfrac{1}{4}\), where \(\overline{A}\) stands for the complement of the event \(A\). Then the events \(A\) and \(B\) are</p>
<p>mutually exclusive and independent</p>
<p>equally likely but not independent</p>
<p>independent but not equally likely</p>
<p>independent and equally likely</p>
Step-by-Step Solution
Key Concept: Use the complement rule P(A∪B) = 1 - P(A̅∪B̅), then apply P(A∪B) = P(A) + P(B) - P(A∩B) to find P(B), and check if P(A∩B) = P(A)·P(B) for independence.
<p><strong>Step 1:</strong> From P(A̅∪B̅) = 1/6, use De Morgan's Law: A̅∪B̅ = (A∩B)̅</p><p>Therefore: P((A∩B)̅) = 1/6, which gives P(A∩B) = 1 - 1/6 = 5/6</p><p><strong>Step 2:</strong> But wait—we're given P(A∩B) = 1/4. This means P(A̅∪B̅) corresponds to the complement differently. Actually, P(A̅∪B̅) = 1 - P(A∪B) is wrong. Correctly: P(A̅∪B̅) = P((A∩B)̅) only if we use complements properly.</p><p>Given directly: P(A̅) = 1/4, so P(A) = 3/4 and P(A∩B) = 1/4</p><p><strong>Step 3:</strong> From P(A̅∪B̅) = 1/6: Using P(A̅∪B̅) = P(A̅) + P(B̅) - P(A̅∩B̅)</p><p>We have: 1/6 = 1/4 + P(B̅) - P(A̅∩B̅)</p><p><strong>Step 4:</strong> Alternatively, 1 - P(A∪B) = 1/6 gives P(A∪B) = 5/6</p><p>Using P(A∪B) = P(A) + P(B) - P(A∩B):</p><p>5/6 = 3/4 + P(B) - 1/4</p><p>5/6 = 1/2 + P(B), so P(B) = 5/6 - 1/2 = 1/3</p><p><strong>Step 5:</strong> Check independence: P(A)·P(B) = (3/4)·(1/3) = 1/4 = P(A∩B) ✓</p><p>∴ Answer: D (Events A and B are <strong>independent</strong>)</p>
Correct Answer: D