2007 AMC 8 Problems/Problem 21
Revision as of 13:51, 11 December 2010 by Thegoldenratio (talk | contribs) (Created page with 'There are 4 ways of choosing a winning pair of the same number, and <math>2 \left( \dbinom{4}{2} \right) = 12</math> ways to choose a pair of the same color. There's a total of …')
There are 4 ways of choosing a winning pair of the same number, and 
ways to choose a pair of the same color.
There's a total of 
ways to choose a pair, so the probability is 
.
