<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 2:</b> The probability that \(X \geq 3\) equals</p>
Step-by-Step Solution
Key Concept: P(X ≥ 3) means we fail to get a 6 in the first two tosses, then anything can happen. Use complementary counting: P(X ≥ 3) = P(first two tosses are not 6) = (5/6)².
<p><strong>Step 1:</strong> Recognize that X ≥ 3 means the first 6 appears on the 3rd toss or later.</p><p><strong>Step 2:</strong> This is equivalent to: the first two tosses are NOT 6.</p><p><strong>Step 3:</strong> P(not getting 6 on toss 1) = 5/6</p><p><strong>Step 4:</strong> P(not getting 6 on toss 2) = 5/6</p><p><strong>Step 5:</strong> Since tosses are independent: P(X ≥ 3) = (5/6) × (5/6) = 25/36</p><p>∴ Answer: B (assuming B = 25/36)</p>
Correct Answer: B