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=\boxed{144}</math>.
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First, we find the probability both are green, then the probability both are blue, and add the two probabilities. The sum should be equal to <math>0.58</math>.  
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The probability both are green is <math>\frac{4}{10}\cdot\frac{16}{16+N}</math>, and the probability both are blue is <math>\frac{6}{10}\cdot\frac{N}{16+N}</math>, so  
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<cmath> \frac{4}{10}\cdot\frac{16}{16+N}+\frac{6}{10}\cdot\frac{N}{16+N}=\frac{29}{50}</cmath>
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Solving this equation,
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<cmath>20\left(\frac{16}{16+N}\right)+30\left(\frac{N}{16+N}\right)=29</cmath>
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Multiplying both sides by <math>16+N</math>, we get
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<cmath>20\cdot 16 + 30\cdot N = 29(16+n)\Rightarrow 320+30N=464+29N \Rightarrow N = \boxed{144}</cmath>
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2014|n=I|num-b=1|num-a=3}}
 
{{AIME box|year=2014|n=I|num-b=1|num-a=3}}
 
{{MAA Notice}}
 
{{MAA Notice}}
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[[Category:Intermediate Probability Problems]]

Revision as of 15:09, 28 November 2019

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 green, then the probability both are blue, and add the two probabilities. The sum should be equal to $0.58$.

The probability both are green is $\frac{4}{10}\cdot\frac{16}{16+N}$, and the probability both are blue 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, \[20\left(\frac{16}{16+N}\right)+30\left(\frac{N}{16+N}\right)=29\] Multiplying both sides by $16+N$, we get \[20\cdot 16 + 30\cdot N = 29(16+n)\Rightarrow 320+30N=464+29N \Rightarrow N = \boxed{144}\]

See also

2014 AIME I (ProblemsAnswer KeyResources)
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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