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?

Lets look at another perpective .  Assume that we are arranging all balls in line . Moreover , this line is constructed according to selection order of the balls.
Then, how many ways are there to align all of the balls ? . The answer is $\frac{36!}{20! \times 16!}$.
Now assume that the last ball is red and we placed one red ball in the last place , so we need to align $19$ red balls and $16$ blue balls in a straight line (their positons are determined by selection order).Then , we can arrange them by $\frac{35!}{19! \times 16!}$ ways.
As a result , $\frac{\frac{35!}{19! \times 16!}}{\frac{36!}{20! \times 16!}}=\frac{20}{36} = \frac{5}{9}$
