Here is the question:
A bag contains only red marbles and green marbles, two of which are to be drawn without replacement. There are at least two marbles of each color in the bag. If the probability of both marbles being red is half the probability of both marbles being green, then what is the minimum possible number of marbles in the bag?
This is from a timed competition, fastest answers are best.
I tried setting up a system of equations $2(\frac{2}{x}\cdot\frac{1}{x-1})=\frac{x-2}{x} \cdot\frac{x-3}{x-1}$
This method is neither correct or fast. What am I doing wrong, and is there a better way?
Best Answer
There are $x$ red marbles and $y$ green ones. The probability of both draws returning red marbles is $\frac{x(x-1)}{(x+y)(x+y-1)}$; for green marbles it is $\frac{y(y-1)}{(x+y)(x+y-1)}$. So we need $2x(x-1)=y(y-1)$.
The numbers of the form $x(x-1)$ are called promic and begin $2, 6, 12\dots$ We see $6$ and $12$, so the minimum number of marbles needed is $x=3$ and $y=4$, for a total of $7$ marbles.