probability question on drawing ball 
I have a bag containing $20$ red balls and $16$ blue balls. I uniformly randomly take balls out from the bag without replacement until all balls of both colors have been removed. If the probability that the last ball I took was red can be represented as $\frac{p}{q}$, where $p$ and $q$ are coprime positive integers. Find $p+q$

My solution: as the last ball is red, that means we have drawn $19$ red balls and $16$ blue balls so far. The total no of ways of doing it is $35 \choose19$. Now total no of ways of drawing all the balls is $36 \choose 20$. The probability is
$$\frac{35\choose19}{36\choose20}=\frac{5}{9}$$
But the answer is $\frac{4}{9}$. Can you please correct my solution?

I'm going to add another answer since my other one is still correct, but the original question is slightly different.
The way you phrased it, $\frac{5}{9}$ is definitely correct. The original question says, "all balls of a color have been removed." From that perspective, it's asking what is the probability of there being at least one blue ball when all red balls have been drawn.
The "trick" is to argue that the last ball (it you kept going) will either be red or blue. If it's red, then all blue balls will have already been drawn. If it's blue, all red balls will have been drawn. Since those outcomes are disjoint, the probability of all red balls being drawn is equal to the probability that the final ball drawn is blue which is $1-\frac{5}{9}=\frac{4}{9}$.
