What is the probability that balls left are white?

Question

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
   

$\space\space\space$:
$\space\space\space$$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 $$p_i = \frac{\binom{a}{i}\binom{b}{1}\binom{b-1}{b-1}(b-1+i)!}{\binom{a+b}{b+i}(b+i)!} = \frac{\frac{a!}{(a-i)!i!} \cdot b \cdot 1 \cdot (b-1+i)!}{\frac{(a+b)!}{(b+i)!(a-i)!}(b+i)!} = \frac{a! \cdot b \cdot (b-1+i)!}{(a+b)! \cdot i!}$$

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

$$ = \frac{a! \cdot b!}{(a+b)!}\sum_{i=0}^{a-1} {b-1+i \choose i} = \frac{a! \cdot b!}{(a+b)!}\sum_{i=0}^{a-1} {b-1+i \choose i}{1 \choose 0}{a-i \choose a-i} $$

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.

Note to myself: for the summation $\sum_{i=0}^{a-1} {b-1+i \choose i}{1 \choose 0}{a-i \choose a-i}$

1. The vandermonde identity unfortunately does not apply
2. since, the summation is from $0$ to $a-1$, the last slot is always chosen. i.e. the last ball chosen is always white.

Answer 1

Lets take two cases-

1. where all ball left are white
2. where all balls left are black

Case 1:

You can take out all the black balls at one go and then successively pick out a white ball one by one. The number of ways in which you can do this such that the bag doesn't get emptied will be 'a'(This works since all the white balls are identical.).

Case 2:

Similarly, you can take out all the white balls at one go and then successively pick out a black ball one by one. The number of ways in which you can do this such that the bag doesn't get emptied will be 'b'(This works since all the black balls are identical.).

So the probablity will be simply $a/(a+b)$

Answer 2

It seems that your calculation is correct. Using your final formula Mathematica obtained for all $(a,b)\in[100]^2$ the official value ${a\over a+b}$.

In reality it is expected that you solve this problem without any calculation. Just don't stop drawing until you have drawn all $a+b$ balls. The color of the last ball is the "remaining" color you are interested in. If you agree that the probability for the first ball being white is ${a\over a+b}$ then you should also agree that the probability for the last ball being white is ${a\over a+b}$.