Difference between revisions of "2007 AMC 8 Problems/Problem 21"
(Created page with 'There are 4 ways of choosing a winning pair of the same number, and <math>2 \left( \dbinom{4}{2} \right) = 12</math> ways to choose a pair of the same color. There's a total of …') |
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| + | ==Problem== | ||
| + | Two cards are dealt from a deck of four red cards labeled <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math> and four green cards labeled <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math>. A winning pair is two of the same color or two of the same letter. What is the probability of drawing a winning pair? | ||
| + | <math> \textbf{(A)}\ \frac{2}{7}\qquad\textbf{(B)}\ \frac{3}{8}\qquad\textbf{(C)}\ \frac{1}{2}\qquad\textbf{(D)}\ \frac{4}{7}\qquad\textbf{(E)}\ \frac{5}{8} </math> | ||
| + | |||
| + | ==Solution== | ||
There are 4 ways of choosing a winning pair of the same number, and <math>2 \left( \dbinom{4}{2} \right) = 12</math> ways to choose a pair of the same color. | There are 4 ways of choosing a winning pair of the same number, 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{\ | + | 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 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 
.
