Quantity I: Probability of drawing both balls red or both balls white = $\frac{\binom{4}{2} + \binom{6}{2}}{\binom{20}{2}}$ Quantity II: Probability of drawing balls of different colors = \(\frac{4\times 6\times 2 + 4\times 6\times 8 + 6\times 2\times 8 + 4\times 2\times 8}{\binom{20}{2}

Question

Quantity I: Probability of drawing both balls red or both balls white = $\frac{\binom{4}{2} + \binom{6}{2}}{\binom{20}{2}}$ Quantity II: Probability of drawing balls of different colors = $\frac{4\times 6\times 2 + 4\times 6\times 8 + 6\times 2\times 8 + 4\times 2\times 8}{\binom{20}{2}}$ Compare Quantity I and Quantity II.

Accepted Answer

Quantity I < Quantity II. Quantity I counts only same-color pairs of red-red or white-white, while Quantity II counts all pairs of different colors. Since the number of different-color combinations is much larger, Quantity II is greater. Therefore, Quantity I < Quantity II.

Solution

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