Probability
Classical Probability
Grade 12
Question:
<p>The decimal parts of the logarithms of two numbers taken at random are found to six places of decimal. What is the chance that the second can be subtracted from the first without 'borrowing'?</p><p>For each column of the two numbers, <em>n(S)</em> = number of ways to fill the two places by the digits 0, 1, 2, ..., 9 = 10 × 10 = 100.</p>
<p>\(\left(\dfrac{55}{100}\right)^6\)</p>
<p>\(\left(\dfrac{11}{20}\right)^6\)</p>
<p>\(\left(\dfrac{55}{100}\right)^6 \cdot \left(\dfrac{11}{20}\right)^6\)</p>
<p>\(\left(\dfrac{11}{20}\right)^{12}\)</p>
Step-by-Step Solution
Key Concept: The decimal part of logarithms is uniformly distributed on [0,1). For subtraction without borrowing in a decimal representation, each digit of the first number must be ≥ the corresponding digit of the second number. Since digits are independent and uniformly distributed, calculate P(digit₁ ≥ digit₂) for each position and multiply.
<p><strong>Step 1:</strong> The decimal parts of logarithms are uniformly distributed on [0,1). Each digit (0-9) appears with equal probability 1/10 in each decimal place.</p><p><strong>Step 2:</strong> For subtraction without borrowing, at each decimal place we need: digit₁ ≥ digit₂. For a single position, count favorable outcomes: if digit₂ = k, then digit₁ can be k, k+1, ..., 9 (giving 10-k choices).</p><p><strong>Step 3:</strong> P(no borrowing in one position) = Σ(k=0 to 9) [P(digit₂ = k) × P(digit₁ ≥ k)] = Σ(k=0 to 9) [(1/10) × (10-k)/10] = (1/100)Σ(10-k) = (1/100)(10+9+8+...+1) = 55/100 = 11/20.</p><p><strong>Step 4:</strong> For all six decimal places independently: P(all positions no borrowing) = (11/20)⁶. However, for practical six-place decimals, this simplifies to checking if the probability at each place is 11/20, giving final answer (11/20) or equivalently 0.55.</p><p>∴ Answer: B</p>
Correct Answer: B