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

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== 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 <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>
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== Solution ==
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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>\begin{align*}
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20\cdot16+30\cdot N&=29(16+N)\\
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320+30N&=464+29N\\
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N&=\boxed{144}
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\end{align*}</cmath>
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== See also ==
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{{AIME box|year=2014|n=I|num-b=1|num-a=3}}
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{{MAA Notice}}
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[[Category:Intermediate Probability Problems]]

Latest revision as of 19:12, 15 January 2022

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

\begin{align*} 20\cdot16+30\cdot N&=29(16+N)\\ 320+30N&=464+29N\\ N&=\boxed{144} \end{align*}

See also

2014 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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