Difference between revisions of "2014 AIME I Problems/Problem 2"

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== Solution ==
 
== Solution ==
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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>n=144</math>.

Revision as of 14:24, 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 $0.58$. The probability both are blue is $\frac{4}{10}\cdot\frac{16}{16+N}$, and the probability both are green is $\frac{6}{10}\cdot\frac{N}{16+N}$, so \[\frac{4}{10}\cdot\frac{16}{16+N}+\frac{6}{10}\cdot\frac{N}{16+N}=\frac{29}{50}.\] Solving this equation, we get $n=144$.