A Math League Probability Problem

We can solve for $P_n$ by writing it in terms of g. Because there are $g+1$ marbles in the bag, with one removed after every drawing, $P_6=\frac{g}{g+1}\cdot\frac{g-1}{g}...\frac{g-5}{g-4}\cdot\frac{1}{g-5}$ . The $\frac{g}{g+1}\cdot\frac{g-1}{g}...\frac{g-5}{g-4}$ represents the probability that the first 6 drawings result in green marbles, while the $\frac{1}{g-5}$ represents the probability that the seventh drawing results in a red marble. In a similar way, $P_5=\frac{g}{g+1}\cdot\frac{g-1}{g}...\frac{g-4}{g-3}\cdot\frac{1}{g-4}$ .

When we simplify by cross cancelling, it turns out that both $P_5$ and $P_6$ equal $\frac{1}{g+1}$ . $P_5+P_6=\frac{g+3}{g^2-37}$ now turns to $\frac{2}{g+1}=\frac{g+3}{g^2-37}$ .

We can now solve for g:

$\frac{2}{g+1}=\frac{g+3}{g^2-37}$

1. Multiply both sides by $g+1$ : $2=\frac{g^2+4g+3}{g^2-37}$

2. Multiply both sides by $g^2-37$ : $2{g^2}-74=g^2+4g+3$

3. Move everything over to the left side: $g^2-4g-77=0$

4. Factorize the left side: $(g-11)(g+7)=0$

5. Solve for g: From $(g-11)(g+7)=0$ , we get $g=11$ or $g=-7$ ; however, since there can't be a negative number of green marbles, $g=11$ is the only possible value for g.

Now, we know that there are 11 green marbles and 1 red marble, so there are $\boxed{12}$ marbles in the bag.