2014 AIME I Problems/Problem 2

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

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

2014 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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2014 AIME I Problems/Problem 2

Problem 2

Solution

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