Probability
Independent Events
Grade 12
Question:
<p>Event 'A' is independent of event \(B\), \(B \cup C\) and \(B \cap C\). If \(P(A) = \frac{1}{2}\), \(P(B) = \frac{1}{3}\) and \(P(C) = \frac{1}{4}\). Then:</p>
<p>\(P\left(\frac{A}{C}\right) = \frac{1}{2}\)</p>
<p>\(P\left(\frac{\overline{B \cup C}}{A}\right) = \frac{11}{12}\) (where \(B\) and \(C\) are independent events)</p>
<p>\(P\left(\frac{\bar{A}}{B \cap C}\right) = \frac{1}{2}\)</p>
<p>\(A\) and \(C\) are not independent events</p>
Step-by-Step Solution
Key Concept: If event A is independent of B, B∪C, and B∩C simultaneously, then A must be independent of C as well. This is because independence from both the union and intersection of two events forces independence from each individual event.
<p><strong>Step 1:</strong> Given A is independent of B, B∪C, and B∩C.</p><p><strong>Step 2:</strong> If A is independent of both B∪C and B∩C, then:<br>P(A∩(B∪C)) = P(A)·P(B∪C) and P(A∩(B∩C)) = P(A)·P(B∩C)</p><p><strong>Step 3:</strong> Expanding P(A∩(B∪C)) = P((A∩B)∪(A∩C)) = P(A∩B) + P(A∩C) - P(A∩B∩C)</p><p>Since A is independent of B: P(A∩B) = P(A)·P(B) = ½·⅓ = 1/6</p><p><strong>Step 4:</strong> For A independent of B and A independent of B∪C simultaneously, A must be independent of C as well. Therefore P(A∩C) = P(A)·P(C) = ½·¼ = 1/8</p><p><strong>Step 5:</strong> Since A is independent of B∩C: P(A∩B∩C) = P(A)·P(B∩C)</p><p>If B and C are independent: P(B∩C) = P(B)·P(C) = ⅓·¼ = 1/12<br>Thus P(A∩B∩C) = ½·1/12 = 1/24</p><p><strong>Step 6:</strong> Verify: P(A∩(B∪C)) = 1/6 + 1/8 - 1/24 = 4/24 + 3/24 - 1/24 = 6/24 = 1/4<br>And P(A)·P(B∪C) = ½·(⅓ + ¼ - 1/12) = ½·(4+3-1)/12 = ½·½ = 1/4 ✓</p><p>∴ Answer: P(A∩B) = 1/6, P(A∩C) = 1/8, P(A∩B∩C) = 1/24 (and related independence statements)</p>
Correct Answer: A,B,C