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 has the same color or letter as it. Since there are 7 cards left to choose from, the probability is <math>\frac{4}{7}</math>. | + | 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 <math>\frac{4}{7}</math>. theepiccarrot7 |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2007|num-b=20|num-a=22}} | {{AMC8 box|year=2007|num-b=20|num-a=22}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 13:02, 3 November 2020
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
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
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. 
