2011 AMC 12B Problems/Problem 9

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For the product to be greater than zero, we must have either both numbers negative or both positive.

Both numbers are negative with a $\frac{2}{3}*\frac{2}{3}=\frac{4}{9}$ chance.

Both numbers are positive with a $\frac{1}{3}*\frac{1}{2}=\frac{1}{9}$ chance.

Therefore, the total probability is $\frac{4}{9}+\frac{1}{9}=\frac{5}{9}$ and we are done. $\Box$

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