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

$$\space\space\space$$:
$$\space\space\spacea$$. $$a-1$$ 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.
• I haven't checked your work so I don't know if it's right or wrong. However, seeing that the given answer is so simple, you might try and see if you can finf a correspondingly simple proof.
– bof
Jul 23, 2020 at 5:56
• What other (related) problems is $\frac a{a+b}$ the answer to, and can you see any reason why the answers should be the same? Well, what is the probability that the first ball out of the box is white? Assuming you keep drawing until there are no balls left, what's the probability that the last ball out of the box is white?
– bof
Jul 23, 2020 at 6:01
• 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 not working. Jul 23, 2020 at 6:14
• @bof a very simple proof is here---->math.stackexchange.com/q/3085480/415432 one related problem whose answer is $\frac{ a}{a+b}$ is "$n$ balls are successively drawn without replacement from the bag. what is the probability that the $j^{th}$ ball drawn is white". Jul 23, 2020 at 6:15

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 $${a\over a+b}$$.

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 $$a/(a+b)$$

• tell me if I am wrong, but in case 1 is the answer is $a$ because the white balls are not identical. and thus there are $a$ ways in which the last remaining white ball can be chosen. similarly for case2 the black balls are also not identical Jul 23, 2020 at 6:41
• why do u think they r unidentical? it will be 'a' ways for the first case in this manner: first way is to take out all black balls, then another way will be to take out (all black balls+1 white ball), another way will be to take out (all black balls+ 2 white balls) and so on.. Jul 23, 2020 at 6:47
• but is that equal to the number of ways in which one (white)ball is left at last? Jul 23, 2020 at 7:00
• I dont think the questions asks for number of ways in which only one white ball is left at last. Jul 23, 2020 at 7:05
• no it doesn't but, what you wrote is "The number of ways in which you can do this such that one ball is left at last will be 'a'" Jul 23, 2020 at 7:07