Probability
Classical Probability
Grade 12
Question:
<p>Out of <strong>(2n + 1)</strong> tickets numbered consecutively, three numbers on them are in AP. Find the chance that the numbers drawn are in AP.</p>
<p>\(\dfrac{n}{2(2n-1)(4n^2-1)}\)</p>
<p>\(\dfrac{3n}{(2n-1)(4n^2-1)}\)</p>
<p>\(\dfrac{3n}{(2n+1)(2n-1)}\)</p>
<p>\(\dfrac{n^2}{(2n+1)(2n-1)}\)</p>
Step-by-Step Solution
Key Concept: The total ways to choose 3 tickets from (2n+1) tickets is C(2n+1,3). To find favorable outcomes, count all 3-element arithmetic progressions where the middle term lies strictly between the extremes in the range [1, 2n+1].
<p><strong>Step 1:</strong> Total ways to choose 3 numbers from (2n+1) tickets:</p><p>Total outcomes = C(2n+1, 3) = (2n+1)(2n)(2n-1)/6</p><p><strong>Step 2:</strong> Count favorable outcomes (3 numbers in AP).</p><p>Let the AP be: a, a+d, a+2d where a ≥ 1, d ≥ 1, and a+2d ≤ 2n+1</p><p>For fixed d: a can range from 1 to (2n+1-2d)</p><p>Number of valid values of a = (2n+1-2d)</p><p><strong>Step 3:</strong> Sum over all possible values of d.</p><p>d ranges from 1 to n (since 2d ≤ 2n, we need d ≤ n)</p><p>Favorable outcomes = Σ(d=1 to n) (2n+1-2d)</p><p>= Σ(d=1 to n) 2(n+1-d) = 2[n + (n-1) + ... + 1]</p><p>= 2 × n(n+1)/2 = n(n+1)</p><p><strong>Step 4:</strong> Calculate probability.</p><p>P(AP) = n(n+1) / [(2n+1)(2n)(2n-1)/6]</p><p>= 6n(n+1) / [(2n+1)(2n)(2n-1)]</p><p>= 3(n+1) / [(2n+1)(2n-1)]</p><p>= 3(n+1) / (4n² - 1)</p><p>∴ Answer: C</p>
Correct Answer: C