Probability
Classical Probability
Grade 12
Question:
<p>If two different numbers are taken from the set \(\{0, 1, 2, 3, \ldots, 10\}\), then the probability that their sum as well as absolute difference are both multiples of 4, is</p>
<p>\(\dfrac{7}{55}\)</p>
<p>\(\dfrac{6}{55}\)</p>
<p>\(\dfrac{12}{55}\)</p>
<p>\(\dfrac{14}{45}\)</p>
Step-by-Step Solution
Key Concept: For two numbers a and b, both (a+b) and |a-b| are multiples of 4 iff a ≡ b (mod 4) AND both a and b have the same parity, which means a ≡ b (mod 4) is the binding constraint.
<p><strong>Step 1:</strong> Let the two numbers be a and b with a > b. We need:</p><ul><li>a + b ≡ 0 (mod 4)</li><li>a - b ≡ 0 (mod 4)</li></ul><p><strong>Step 2:</strong> From the second condition: a ≡ b (mod 4). Substituting into the first: b + b ≡ 0 (mod 4), so 2b ≡ 0 (mod 4), meaning b ≡ 0 (mod 2). This means both a and b must be even AND a ≡ b (mod 4).</p><p><strong>Step 3:</strong> Since a ≡ b (mod 4) with both even, we need pairs from the same residue class mod 4:</p><ul><li>Residue 0 (mod 4): {0, 4, 8} → C(3,2) = 3 pairs</li><li>Residue 2 (mod 4): {2, 6, 10} → C(3,2) = 3 pairs</li></ul><p><strong>Step 4:</strong> Total favorable pairs = 3 + 3 = 6</p><p><strong>Step 5:</strong> Total ways to choose 2 different numbers from 11 numbers = C(11,2) = 55</p><p><strong>Step 6:</strong> Probability = 6/55</p><p>∴ Answer: A</p>
Correct Answer: A