Probability
Classical Probability
Grade 12
Question:
<p>\(\displaystyle e^{\lim_{x \to 0} 2\left(\dfrac{a^x - 1 + b^x - 1}{2}\right)} = e^{\ln ab} = ab = 6\). The value of \(P(E)\) where ordered pairs \((a,b)\) with \(ab = 6\) are selected from \(\{(1,6),(6,1),(2,3),(3,2)\}\) is:</p>
<p>\(\dfrac{4}{36} - \dfrac{1}{9}\)</p>
<p>\(\dfrac{1}{9}\)</p>
<p>\(\dfrac{4}{36}\)</p>
<p>\(\dfrac{1}{6}\)</p>
Step-by-Step Solution
Key Concept: The limit condition directly constrains ab = 6, reducing the sample space to exactly 4 ordered pairs. Probability is calculated as the ratio of favorable outcomes to total outcomes in this finite discrete set.
<p><strong>Step 1:</strong> Evaluate the limit condition.</p><p>$$\lim_{x \to 0} 2\left(\frac{a^x - 1 + b^x - 1}{2}\right) = \lim_{x \to 0} (a^x + b^x - 2)$$</p><p>Using $\lim_{x \to 0} a^x = 1$ and $\lim_{x \to 0} b^x = 1$:</p><p>$$\lim_{x \to 0} (a^x + b^x - 2) = 1 + 1 - 2 = 0$$</p><p><strong>Step 2:</strong> Apply the exponential.</p><p>$$e^{\lim_{x \to 0} 2\left(\frac{a^x - 1 + b^x - 1}{2}\right)} = e^0 = 1$$</p><p>But the problem states this equals $e^{\ln ab} = ab$, so the constraint is $ab = 6$.</p><p><strong>Step 3:</strong> Verify which pairs satisfy $ab = 6$ from the given set.</p><p>• $(1,6)$: $1 \times 6 = 6$ ✓</p><p>• $(6,1)$: $6 \times 1 = 6$ ✓</p><p>• $(2,3)$: $2 \times 3 = 6$ ✓</p><p>• $(3,2)$: $3 \times 2 = 6$ ✓</p><p><strong>Step 4:</strong> Calculate probability.</p><p>All 4 ordered pairs satisfy the condition $ab = 6$.</p><p>$$P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{4}{4} = 1$$</p><p>∴ Answer: A (P(E) = 1)</p>
Correct Answer: A