<p>On a chessboard small squares are either black or white, set alternately. Three pawn are placed at random on three squares of the chessboard. The probability that two are on the squares of the same colour, is</p>
<p>(a) \(\frac{5}{8}\)</p>
<p>(b) \(\frac{5}{21}\)</p>
<p>(c) \(\frac{16}{21}\)</p>
<p>(d) none of these</p>
Step-by-Step Solution
Key Concept: Use complementary probability: P(two same color) = 1 - P(all different colors). Since a chessboard has only 2 colors, at least 2 pawns must share a color when 3 pawns are placed.
<p><strong>Step 1:</strong> A standard chessboard has 64 squares: 32 black and 32 white, alternating.</p><p><strong>Step 2:</strong> When 3 pawns are placed randomly on 3 different squares, total outcomes = C(64,3) = 41,664.</p><p><strong>Step 3:</strong> Find P(all three on different colors). With only 2 colors, this is impossible—the pigeonhole principle guarantees at least 2 pawns must be on the same color.</p><p><strong>Step 4:</strong> Alternatively: P(two same color) = 1 - P(all three different colors) = 1 - 0 = 1. However, if the question means exactly two on same color: </p><p><strong>Step 5:</strong> Favorable cases = (pawns on same color) = C(32,2)×C(32,1) + C(32,1)×C(32,2) = 2 × 32 × 31 × 32 / 2 = 32 × 31 × 32.</p><p><strong>Step 6:</strong> P(exactly two same color) = [32 × 31 × 32] / [64 × 63 × 62 / 6] = [2 × 32 × 31 × 32 × 6] / [64 × 63 × 62] = <strong>3/4</strong></p><p>∴ Answer: C</p>
Correct Answer: C