Difference between revisions of "2007 AMC 8 Problems/Problem 21"
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==Solution 2== | ==Solution 2== | ||
| − | Notice that, no matter which card you choose, there are exactly 4 cards that either | + | Notice that, no matter which card you choose, there are exactly <math>4</math> cards that either have the same color or letter as it. Since there are <math>7</math> cards left to choose from, the probability is <math>\boxed{\textbf{(D)}\frac47}</math>. |
| + | -theepiccarrot7 | ||
==Solution 3== | ==Solution 3== | ||
Latest revision as of 12:05, 24 December 2024
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
Video Solution by WhyMath
Solution 2
Notice that, no matter which card you choose, there are exactly 
cards that either have the same color or letter as it. Since there are 
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 | |
| 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. 
