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

Solution 1

There are 4 ways of choosing a winning pair of the same letter, 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 $\dfrac{4+12}{28} = \boxed{\textbf{(D)}\ \frac{4}{7}}$ .

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: $\frac17+\frac37=\boxed{\textbf{(D)} \frac47}$ .

2007 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
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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