<p><strong>For Problems 1–3:</strong> In a class of 10 students, probability of exactly <i>i</i> students passing an examination is directly proportional to <i>i</i><sup>2</sup>. Then answer the following questions:</p><p>If a student selected at random is found to have passed the examination, then the probability that he was the only student who has passed the examination is</p>
Step-by-Step Solution
Key Concept: Use Bayes' theorem: P(only 1 passed | student passed) = P(student passed | only 1 passed) × P(only 1 passed) / P(student passed). First establish P(i students pass) ∝ i², normalize it, then apply conditional probability.
<p><strong>Step 1: Find the probability distribution.</strong></p><p>Given: P(exactly i students pass) ∝ i²</p><p>So P(i) = ci² where i = 0, 1, 2, ..., 10</p><p>Since ∑P(i) = 1: c∑i² = c × [10(11)(21)/6] = 385c = 1</p><p>Therefore c = 1/385, and P(i) = i²/385</p><p><strong>Step 2: Find P(student passed).</strong></p><p>P(student passed) = ∑[i=1 to 10] P(i students pass) × P(selected student passed | i passed)</p><p>= ∑[i=1 to 10] (i²/385) × (i/10) = (1/3850)∑[i=1 to 10] i³</p><p>Since ∑i³ = [10(11)/2]² = 3025</p><p>P(student passed) = 3025/3850 = 11/14</p><p><strong>Step 3: Apply Bayes' theorem.</strong></p><p>P(only 1 passed | student passed) = P(student passed | only 1 passed) × P(only 1 passed) / P(student passed)</p><p>= (1/10) × (1/385) / (11/14) = (1/3850) × (14/11) = 14/42350 = 1/3025</p><p>∴ Answer: <strong>1/3025</strong> or equivalent form</p>
Correct Answer: B