There are several ways to solve this problem. The answer is $C_{5+4-1}^{5}=56$.

In general, if there are $a$ red balls and $b$ green balls, then the number of possibilities that red balls are all retrieved first is $$C_{a+b-1}^{a}=\frac{(a+b-1)!}{a!\cdot (b-1)!}$$

$\underline{Solution\ 1}$

As long as the last ball is green, the requirement will be satisfied. Therefore, this problem is equivalent to arrange $a$ red balls along with $(b-1)$ green balls. For each sequence under this arrangement, we can append the remaining green ball at the end to satisfy the requirement. Hence, the answer is $$C_{a+b-1}^a=\frac{(a+b-1)!}{a!\cdot (b-1)!}$$

$\underline{Solution\ 2}$

We can model this problem in a similar way as # 4263. The benefit of this is that it can be extended to the case with three colors easily.

In the $(5\times 4)$ grid shown below, every shortest path from point $A$ to point $B$ is corresponding to one way of retrieving the $9$ balls when eastern bound step represents retrieving a red ball and northern bound step represents retrieving a green ball.

If all the red balls are retrieved before green balls, then the route must hit the east vertical boundary of this grid. In this case, point $C$ must be reached before the route terminates at point $B$. Hence, the answer is equivalent to count the shortest routes from point $A$ to point $C$, which is $$C_{5+4-1}^5$$