2014 AIME I Problems/Problem 2: Difference between revisions
No edit summary |
|||
| Line 2: | Line 2: | ||
== Solution == | == Solution == | ||
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 13:24, 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 
.
