Difference between revisions of "2007 AMC 8 Problems/Problem 21"
m (→Solution) |
|||
| Line 4: | Line 4: | ||
==Solution== | ==Solution== | ||
| − | There are 4 ways of choosing a winning pair of the same | + | There are 4 ways of choosing a winning pair of the same letter, and <math>2 \left( \dbinom{4}{2} \right) = 12</math> ways to choose a pair of the same color. |
There's a total of <math>\dbinom{8}{2} = 28</math> ways to choose a pair, so the probability is <math>\dfrac{4+12}{28} = \boxed{\textbf{(D)}\ \frac{4}{7}}</math>. | There's a total of <math>\dbinom{8}{2} = 28</math> ways to choose a pair, so the probability is <math>\dfrac{4+12}{28} = \boxed{\textbf{(D)}\ \frac{4}{7}}</math>. | ||
Revision as of 11:14, 25 July 2013
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?
Solution
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 
.
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. 
