<p><b>For Problems 1–3:</b> A fair die is tossed repeatedly until a 6 is obtained. Let \(X\) denote the number of tosses required.</p><p><b>Problem 3:</b> The conditional probability that \(X \geq 6\) given \(X > 3\) equals</p>
Step-by-Step Solution
Key Concept: Use the memoryless property of geometric distribution: P(X ≥ 6 | X > 3) = P(X ≥ 3) since the process 'restarts' after 3 failed tosses. Alternatively, condition on X > 3 and count remaining tosses needed.
<p><strong>Step 1:</strong> Identify the distribution. X ~ Geometric(p) where p = 1/6 (probability of rolling a 6).</p><p><strong>Step 2:</strong> Apply the memoryless property: P(X ≥ m + n | X > m) = P(X ≥ n) for geometric distributions.</p><p><strong>Step 3:</strong> Here, P(X ≥ 6 | X > 3) = P(X ≥ 3) since 6 = 3 + 3.</p><p><strong>Step 4:</strong> Calculate P(X ≥ 3) = 1 - P(X < 3) = 1 - [P(X=1) + P(X=2)]</p><p>P(X=1) = (1/6), P(X=2) = (5/6)(1/6)</p><p>P(X ≥ 3) = 1 - (1/6) - (5/36) = 1 - (6/36) - (5/36) = 25/36</p><p><strong>Alternatively:</strong> P(X ≥ 3) = (5/6)² = 25/36 (probability that first 2 tosses are not 6)</p><p>∴ Answer: <strong>25/36</strong></p>
Correct Answer: B