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2007 AMC 8 Problems/Problem 19

Revision as of 02:19, 16 November 2010 by Thealexrider (talk | contribs) (Created page with 'Suppose you take out a card: Green, <math> A </math>. There are <math> 7 </math> cards left in the deck. There are three cards with the same color as the first card: Green <math>…')
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Suppose you take out a card: Green, $A$. There are $7$ cards left in the deck. There are three cards with the same color as the first card: Green $B$, Green $C$, Green $D$. There is only $1$ card with the matching alphabet: Red $A$.

Since this is an "or" condition, that means you add the possibilities because you can win either way (same color or same matching alphabet)...

$\frac{3}{7} + \frac{1}{7}$ = $\frac{4}{7}$ $\Longrightarrow$ $\boxed{D}$

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