Probability
Classical Probability with Limits
Grade 12
Question:
<p>If \(a\) and \(b\) are chosen randomly by throwing a pair of fair dice, then the probability that \(\lim_{x\to 0}\left(\dfrac{a^x + b^x}{2}\right)^{\frac{2}{x}} = 6\) equals:</p>
<p>(a) \(\dfrac{4}{9}\)</p>
<p>(b) \(\dfrac{2}{9}\)</p>
<p>(c) \(\dfrac{3}{9}\)</p>
<p>(d) \(\dfrac{1}{9}\)</p>
Step-by-Step Solution
Key Concept: The limit $\lim_{x\to 0}\left(\frac{a^x + b^x}{2}\right)^{\frac{2}{x}}$ evaluates to $e^{\ln(ab)}=(ab)^1=ab$ using the standard form $\lim_{x\to 0}(1+f(x))^{1/f(x)}=e$. Therefore we need $ab=6$.
<p><strong>Step 1: Evaluate the limit.</strong></p><p>Let $L = \lim_{x\to 0}\left(\frac{a^x + b^x}{2}\right)^{\frac{2}{x}}$</p><p>Taking natural logarithm: $\ln L = \lim_{x\to 0}\frac{2}{x}\ln\left(\frac{a^x + b^x}{2}\right)$</p><p><strong>Step 2: Apply L'Hôpital's rule.</strong></p><p>As $x\to 0$: $a^x + b^x \to 2$, so the argument approaches 1 (form $0/0$)</p><p>$\ln L = 2\lim_{x\to 0}\frac{\ln\left(\frac{a^x + b^x}{2}\right)}{x}$</p><p>Using L'Hôpital: $\ln L = 2\lim_{x\to 0}\frac{a^x\ln a + b^x\ln b}{a^x + b^x} = 2\cdot\frac{\ln a + \ln b}{2} = \ln(ab)$</p><p><strong>Step 3: Solve for the condition.</strong></p><p>Therefore $L = ab$. Given $L = 6$, we need $ab = 6$</p><p><strong>Step 4: Count favorable outcomes.</strong></p><p>Pairs $(a,b)$ from two fair dice with product 6:</p><p>$(1,6), (2,3), (3,2), (6,1)$ — exactly 4 favorable outcomes</p><p><strong>Step 5: Calculate probability.</strong></p><p>Total outcomes = $6 \times 6 = 36$</p><p>Probability = $\frac{4}{36} = \frac{1}{9}$</p><p>∴ Answer: B</p>
Correct Answer: B