# 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 $$20\cdot 16 + 30\cdot N = 29(16+n)\Rightarrow 320+30N=464+29N \Rightarrow N = \boxed{144}$$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 