# 1985 AHSME Problems/Problem 6

## Problem

One student in a class of boys and girls is chosen to represent the class. Each student is equally likely to be chosen and the probability that a boy is chosen is $\frac{2}{3}$ of the probability that a girl is chosen. The ratio of the number of boys to the total number of boys and girls is $\mathrm{(A)\ } \frac{1}{3} \qquad \mathrm{(B) \ }\frac{2}{5} \qquad \mathrm{(C) \ } \frac{1}{2} \qquad \mathrm{(D) \ } \frac{3}{5} \qquad \mathrm{(E) \ }\frac{2}{3}$

## Solution

Let the probability that a boy is chosen be $p$. Since the sum of the probability that a boy is chosen and a girl is chosen is $1$, we have the probability that a girl is chosen is $1-p$. We know that $\frac{2}{3}$ of this is $p$, so $\frac{2}{3}(1-p)=p\implies 2(1-p)=3p\implies 2-2p=3p\implies 5p=2\implies p=\frac{2}{5}$.

Notice that $P(\text{A boy is chosen})=\frac{\text{Number of boys}}{\text{Number of boys}+\text{Number of girls}}$. But we already know that this is $\frac{2}{5}$, so that is our final answer, $\boxed{\text{B}}$.

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