Talk:2011 AMC 12B Problems/Problem 9

Is the problem flawed? The range of numbers [-20, 10] includes 31 total numbers: [-1, -20], 0, and [1, 10]. Which makes the probabilities drastically different as we're searching for a product that is greater than zero. For example the probability for two consecutive negative numbers being chosen becomes $\frac{20}{31} \cdot \frac{20}{31} = \frac{400}{961}$

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