Difference between revisions of "2014 AIME I Problems/Problem 2"
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== Solution == | == Solution == | ||
| − | First, we find the probability both are | + | First, we find the probability both are green, then the probability both are blue, and add the two probabilities which equals <math>0.58</math>. The probability both are green is <math>\frac{4}{10}\cdot\frac{16}{16+N}</math>, and the probability both are blue is <math>\frac{6}{10}\cdot\frac{N}{16+N}</math>, so <cmath> \frac{4}{10}\cdot\frac{16}{16+N}+\frac{6}{10}\cdot\frac{N}{16+N}=\frac{29}{50}. </cmath> Solving this equation, we get <math>N=\boxed{144}</math>. |
== See also == | == See also == | ||
{{AIME box|year=2014|n=I|num-b=1|num-a=3}} | {{AIME box|year=2014|n=I|num-b=1|num-a=3}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 10:29, 27 March 2014
Problem 2
An urn contains 
green balls and 
blue balls. A second urn contains 
green balls and 
blue balls. A single ball is drawn at random from each urn. The probability that both balls are of the same color is 
. Find .
Solution
First, we find the probability both are green, then the probability both are blue, and add the two probabilities which equals . The probability both are green is 
, and the probability both are blue is 
, so ![\[\frac{4}{10}\cdot\frac{16}{16+N}+\frac{6}{10}\cdot\frac{N}{16+N}=\frac{29}{50}.\]](http://latex.artofproblemsolving.com/d/8/f/d8f005768975d07178f8959f1405f3ab149361a5.png)
Solving this equation, we get 
.
See also
| 2014 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
