2007 AMC 8 Problems/Problem 21
Contents
Problem
Two cards are dealt from a deck of four red cards labeled , , , and four green cards labeled , , , . A winning pair is two of the same color or two of the same letter. What is the probability of drawing a winning pair?
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 ways to choose a pair of the same color.
There's a total of ways to choose a pair, so the probability is .
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 . theepiccarrot7
Solution 3
We can use casework to solve this.
Case : 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 .
Case : Same color
After choosing any letter, there are seven cards left. Three of them will make a winning pair, so the probability is .
Now that we have the probability for both cases, we can add them: .
See Also
2007 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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