, 2013

First, we consider the first bag, which contains a total of $2+2=4$ marbles.

There are 4 possible marbles that can be drawn first, leaving 3 possible marbles that can be drawn second. This gives a total of $4\times 3 = 12$ ways of drawing two marbles.

For both marbles to be red, there are 2 possible marbles (either red marble) that can be drawn first, and 1 marble that must be drawn second (the remaining red marble). This gives a total of $2\times 1=2$ ways of drawing two red marbles.

For both marbles to be blue, there are 2 possible marbles that can be drawn first, and 1 marble that must be drawn second. This gives a total of $2\times 1=2$ ways of drawing two blue marbles.

Therefore, the probability of drawing two marbles of the same colour from the first bag is the total number of ways of drawing two marbles of the same colour ($2+2=4$) divided by the total number of ways of drawing two marbles (12), or $\dfrac{4}{12}=\dfrac{1}{3}$.

Second, we consider the second bag, which contains a total of $2+2+g=g+4$ marbles.

There are $g+4$ possible marbles that can be drawn first, leaving $g+3$ possible marbles that can be drawn second. This gives a total of $(g+4)(g+3)$ ways of drawing two marbles.

As with the first bag, there are $2\times 1=2$ ways of drawing two red marbles.

As with the first bag, there are $2\times 1=2$ ways of drawing two blue marbles.

For both marbles to be green, there are $g$ possible marbles that can be drawn first, and $g-1$ marbles that must be drawn second. This gives a total of $g(g-1)$ ways of drawing two green marbles.

Therefore, the probability of drawing two marbles of the same colour from the second bag is the total number of ways of drawing two marbles of the same colour ($2+2+g(g-1)=g^2-g+4$) divided by the total number of ways of drawing two marbles ($(g+4)(g+3)$), or $\dfrac{g^2-g+4}{(g+4)(g+3)}$.

Since the two probabilities that we have calculated are to be equal and $g \neq 0$, then $$\begin{aligned} \dfrac{1}{3} & = \dfrac{g^2-g+4}{(g+4)(g+3)} \\ (g+4)(g+3) & = 3g^2 - 3g+12 \\ g^2 + 7g + 12 & = 3g^2 - 3g + 12 \\ 10g & = 2g^2 \\ 0 & = 2g^2-10g \\ 0 & = 2g(g-5)\end{aligned}$$ Therefore, $g=0$ or $g=5$. Since $g \neq 0$, then $g=5$.