Probability
Classical Probability
Grade 12
Question:
<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 13:</strong> The probability that the chessboard contains equal number of red and black squares is</p>
<p>\(\dfrac{{}^{64}C_{32}}{2^{64}}\)</p>
<p>\(\dfrac{64!}{32! \cdot 2^{64}}\)</p>
<p>\(\dfrac{2^{32}-1}{2^{64}}\)</p>
<p>none of these</p>
Step-by-Step Solution
Key Concept: An 8×8 chessboard has 64 squares total. For equal red and black squares, we need exactly 32 red and 32 black squares. The probability is the ratio of favorable outcomes (ways to choose 32 squares from 64) to total outcomes (2^64 ways to color all squares).
<p><strong>Step 1:</strong> Count total squares: 8 × 8 = 64 squares</p><p><strong>Step 2:</strong> For equal coloring: need 32 red and 32 black squares</p><p><strong>Step 3:</strong> Total possible colorings = 2^64 (each square independently red or black)</p><p><strong>Step 4:</strong> Favorable outcomes = C(64, 32) (ways to choose which 32 squares are red)</p><p><strong>Step 5:</strong> Probability = C(64, 32)/2^64</p><p><strong>Step 6:</strong> Simplify using C(64, 32)/2^64 = C(64, 32)/2^64</p><p>∴ Answer: A</p><p><em>Note: This can also be expressed as </em><strong>C(64, 32)/2^64</strong><em> or approximately </em><strong>√(2/(64π)) ≈ 0.0996</strong><em> using Stirling's approximation.</em></p>
Correct Answer: A