2007 AMC 8 Problems/Problem 21

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?

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

Video Solution by OmegaLearn

https://youtu.be/OOdK-nOzaII?t=1712

~ pi_is_3.14

Video Solution

https://youtu.be/OOdK-nOzaII?t=1698

Video Solution by WhyMath

https://youtu.be/HZ0lxSAujrI

Solution 2

Notice that, no matter which card you choose, there are exactly $4$ cards that either have the same color or letter as it. Since there are $7$ cards left to choose from, the probability is $\boxed{\textbf{(D)}\frac47}$ . -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: $\frac17+\frac37=\boxed{\textbf{(D)} \frac47}$ .

2007 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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