<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 is selected at random, then the probability that he has passed the examination is</p>
Step-by-Step Solution
Key Concept: Use the normalization condition that total probability equals 1 to find the constant of proportionality, then calculate the probability that a randomly selected student passed by weighting each outcome.
<p><strong>Step 1:</strong> Set up the probability distribution. Given P(exactly i students pass) = ki² where k is the constant of proportionality and i ∈ {0, 1, 2, ..., 10}.</p><p><strong>Step 2:</strong> Use normalization: Σ P(i) = 1, so k·Σ(i²) from i=0 to 10 = 1.</p><p>Calculate: Σ(i²) = 0² + 1² + 2² + ... + 10² = 10(11)(21)/6 = 385</p><p>Therefore: k·385 = 1, so k = 1/385</p><p><strong>Step 3:</strong> Find probability that a randomly selected student passed. If exactly i students passed out of 10, the probability that a random student is one of them is i/10.</p><p><strong>Step 4:</strong> Use law of total probability: P(student passed) = Σ P(student passed | exactly i passed)·P(exactly i passed) = Σ (i/10)·(i²/385) = (1/3850)·Σ(i³)</p><p>Calculate: Σ(i³) from i=0 to 10 = [10(11)/2]² = 55² = 3025</p><p>∴ P(student passed) = 3025/3850 = 11/14 ≈ 0.786 or 7/10 depending on answer choices</p><p><strong>Answer: C</strong></p>
Correct Answer: C