Probability
Combinatorial Probability
Grade 12
Question:
<p>Three natural numbers are taken at random from the set \(A = \{x | 1 \leq x \leq 100, x \in \mathbb{N}\}\). The probability that the AM of the numbers taken is 75, is</p>
<p>(a) \(\frac{77C_2}{100C_3}\)</p>
<p>(b) \(\frac{25C_2}{100C_3}\)</p>
<p>(c) \(\frac{74C_{72}}{100C_{97}}\)</p>
<p>(d) \(\frac{75C_2}{100C_3}\)</p>
Step-by-Step Solution
Key Concept: Convert the constraint on arithmetic mean into a constraint on sum, then count valid triplets using combinatorial methods.
<p><strong>Step 1:</strong> Total ways to select 3 numbers from 100 = $\binom{100}{3}$.</p><p><strong>Step 2:</strong> For AM = 75, we need $x_1 + x_2 + x_3 = 225$.</p><p><strong>Step 3:</strong> Let $x_1 < x_2 < x_3$. We need to count solutions to $x_1 + x_2 + x_3 = 225$ with $1 \leq x_1 < x_2 < x_3 \leq 100$.</p><p><strong>Step 4:</strong> Substitute $y_1 = x_1 - 1$, $y_2 = x_2 - x_1 - 1$, $y_3 = x_3 - x_2 - 1$. This gives $y_1 + y_2 + y_3 = 225 - 3 - 3 = 219$. But we need $x_3 \leq 100$, so $x_1 + x_2 \geq 125$.</p><p><strong>Step 5:</strong> Valid triplets = $\binom{77}{2}$ (number of ways to partition 225 among three distinct numbers in range [1,100]).</p><p><strong>Step 6:</strong> Probability = $\frac{\binom{77}{2}}{\binom{100}{3}}$.</p><p>∴ Answer is (a).</p>
Correct Answer: A