<p><strong>For Problems 12–14</strong><br>If the squares of a \(8 \times 8\) chessboard are painted either red or black at random.</p><p><strong>Problem 12:</strong> The probability that not all the squares in any column are alternating in color is</p>
Step-by-Step Solution
Key Concept: Calculate the probability that at least one column does NOT have alternating colors, using the complement: P(at least one non-alternating) = 1 - P(all columns alternating). For each column to alternate, adjacent squares must differ in color, giving 2 choices per column (start with red or black).
<p><strong>Step 1:</strong> Find probability that a single column has alternating colors.</p><p>For one column (8 squares): adjacent squares must alternate. First square has 2 choices (R or B), then each subsequent square is forced. So 2 favorable outcomes out of 2^8 = 256 total colorings.</p><p>P(one column alternates) = 2/2^8 = 1/128</p><p><strong>Step 2:</strong> Find probability that ALL 8 columns alternate simultaneously.</p><p>Each of 8 columns independently alternates with probability 1/128.</p><p>P(all 8 columns alternate) = (1/128)^8 = 1/2^56</p><p><strong>Step 3:</strong> Apply complement for 'not all columns are alternating'.</p><p>P(not all columns alternate) = 1 - P(all columns alternate)</p><p>= 1 - 1/2^56 = (2^56 - 1)/2^56</p><p><strong>Simplified form:</strong> The answer is <strong>1 - 1/2^56</strong> or equivalently <strong>(2^56 - 1)/2^56</strong>, which for practical purposes ≈ 1.</p><p>∴ Answer: A</p>
Correct Answer: A