What is the probability that balls left are white?

From a bag containing $b$ black balls and $a$ white balls, balls are successively drawn without replacement until only those of the same colour are left. What is the probability that balls left are white?

the favourable event occurs when the balls drawn consist of

1. $0$ white and $b$ black balls
2. $1$ white and $b$ black balls
3. $2$ white and $b$ black balls

$spacespacespace$:
$spacespacespace$$a$. $a-1$ white and $b$ black balls

the last ball drawn cannot be white. it has to be black.

the probability of drawing $i$ white and $b$ black balls, successively without replacement from a bag containing $b$ black balls and $a$ white balls, such that the last ball drawn is black is $$p_i = frac{binom{a}{i}binom{b}{1}binom{b-1}{b-1}(b-1+i)!}{binom{a+b}{b+i}(b+i)!} = frac{frac{a!}{(a-i)!i!} cdot b cdot 1 cdot (b-1+i)!}{frac{(a+b)!}{(b+i)!(a-i)!}(b+i)!} = frac{a! cdot b cdot (b-1+i)!}{(a+b)! cdot i!}$$

$$text{the probability that the balls left are white } = sum_{i=0}^{a-1}p_i = frac{a! cdot b}{(a+b)!}sum_{i=0}^{a-1}frac{(b-1+i)!}{i!}$$

the given answer is $frac{a}{a+b}$

---

my question is:

is what I have done correct? if yes, how do I proceed further? if no, then what is wrong in my approach?

while I know that there is a very elegant proof for this question, I am curious as to why a very simple approach to solving this question is not working.