Difference between revisions of "2007 AMC 8 Problems/Problem 21"
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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>. | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC8 box|year=2007|num-b=20|num-a=22}} | ||
Revision as of 22:44, 9 December 2012
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 number, 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 | ||
