# 2007 AMC 8 Problems/Problem 21

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

## Problem

Two cards are dealt from a deck of four red cards labeled $A$, $B$, $C$, $D$ and four green cards labeled $A$, $B$, $C$, $D$. A winning pair is two of the same color or two of the same letter. What is the probability of drawing a winning pair?

$\textbf{(A)}\ \frac{2}{7}\qquad\textbf{(B)}\ \frac{3}{8}\qquad\textbf{(C)}\ \frac{1}{2}\qquad\textbf{(D)}\ \frac{4}{7}\qquad\textbf{(E)}\ \frac{5}{8}$

~ pi_is_3.14

## Solution 2

Notice that, no matter which card you choose, there are exactly 4 cards that either has the same color or letter as it. Since there are 7 cards left to choose from, the probability is $\frac{4}{7}$. theepiccarrot7

## Solution 3

We can use casework to solve this.

Case $1$: Same letter

After choosing any letter, there are seven cards left, and only one of them will produce a winning pair. Therefore, the probability is $\frac17$.

Case $2$: Same color

After choosing any letter, there are seven cards left. Three of them will make a winning pair, so the probability is $\frac37$.

Now that we have the probability for both cases, we can add them: $\frac17+\frac37=\boxed{\textbf{(D)} \frac47}$.