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 $p$ , so the probability that a girl is chosen is $1-p$ (as the probabilities must sum to $1$ ). Therefore $\begin{align*}\frac{2}{3}(1-p) = p &\iff 2(1-p) = 3p \\ &\iff 2-2p = 3p \\&\iff 5p = 2 \\&\iff p = \frac{2}{5}.\end{align*}$

Now simply notice that the probability a boy is chosen is precisely the proportion of the number of boys out of the total number of students (boys and girls), so the answer is precisely $\boxed{\text{(B)} \ \frac{2}{5}}$ .

1985 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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1985 AHSME Problems/Problem 6

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