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

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== Problem 2 ==
 
== Problem 2 ==
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An urn contains <math>4</math> green balls and <math>6</math> blue balls. A second urn contains <math>16</math> green balls and <math>N</math> blue balls. A single ball is drawn at random from each urn. The probability that both balls are of the same color is 0.58. Find <math>N</math>.
  
 
== 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>.
 
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 15:56, 14 March 2014

Problem 2

An urn contains $4$ green balls and $6$ blue balls. A second urn contains $16$ green balls and $N$ blue balls. A single ball is drawn at random from each urn. The probability that both balls are of the same color is 0.58. Find $N$.

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 $N=144$.

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