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

(Undo revision 60871 by Agrgordon (talk))
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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 bals 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 <math>0.58</math>. Find <math>N</math>.
 
An urn contains <math>4</math> green balls and <math>6</math> blue balls. A second urn contains <math>16</math> green bals 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 <math>0.58</math>. Find <math>N</math>.
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'''Solution 1:'''The probability of drawing balls of the same color is the sum of the probabilities of drawing two green balls and of drawing two blue balls. This give us <math>\frac{29}{50} = \frac{2}{5} \cdot \frac{16}{16+N} + \frac{3}{5} \cdot \frac{N}{16 + N}</math>

Revision as of 08:05, 14 March 2014

An urn contains $4$ green balls and $6$ blue balls. A second urn contains $16$ green bals 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 1:The probability of drawing balls of the same color is the sum of the probabilities of drawing two green balls and of drawing two blue balls. This give us $\frac{29}{50} = \frac{2}{5} \cdot \frac{16}{16+N} + \frac{3}{5} \cdot \frac{N}{16 + N}$

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