2007 AMC 8 Problems/Problem 21

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

See Also

2007 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
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All AJHSME/AMC 8 Problems and Solutions

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