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

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== Solution ==
 
== Solution ==
 
First, we find the probability both are blue, then the probability both are green, and add the two probabilities which equals <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 equals <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>.
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== See also ==
 +
{{AIME box|year=2013|n=I|num-b=1|num-a=3}}
 +
{{MAA Notice}}

Revision as of 19:37, 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 equals $0.58$. 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 $N=144$.

See also

2013 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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