Difference between revisions of "2007 AMC 8 Problems/Problem 21"

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Revision as of 13:51, 11 December 2010

There are 4 ways of choosing a winning pair of the same number, and $2 \left( \dbinom{4}{2} \right) = 12$ ways to choose a pair of the same color.

There's a total of $\dbinom{8}{2} = 28$ ways to choose a pair, so the probability is $\dfrac{4+12}{28} = \boxed{\dfrac{4}{7}}$.