Difference between revisions of "2014 AIME I Problems/Problem 2"
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| − | First, we find the probability both are blue, then the probability both are green, and add the two probabilities which equaling <math>0.58</math>. The probability both are blue is <math>\frac{4}{10}\cdot\frac{16}{16+N}</math>, and the probability both are green 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> | + | First, we find the probability both are blue, then the probability both are green, and add the two probabilities which equaling <math>0.58</math>. The probability both are blue is <math>\frac{4}{10}\cdot\frac{16}{16+N}</math>, and the probability both are green 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=144</math>. |
Revision as of 13:25, 14 March 2014
Problem 2
Solution
First, we find the probability both are blue, then the probability both are green, and add the two probabilities which equaling 
. The probability both are blue is 
, and the probability both are green 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 
.
