Probability (bag containing red and white balls)

*A bag contains r red and w white balls. A ball is selected at random, is removed, without replacement, until those of the same colour are left in the bag.*

Find the probability that exactly one white ball left remains in the bag when the removal stops.

The solution says the (w-1) white balls have to be selected from (w+r-2), and thus the required probability is

( (r+w-2) choose (w-1) ) / ( (r+w) choose w )

Can anyone explain?

Thanks!

Re: Probability (bag containing red and white balls)

Think of it terms of probability = successes / total possible outcomes.

The total possible number of outcomes is equal to the total number of ways to arrange two types of indistinguishable objects in a line (Imagine all the balls lined up in a random order inside the container, and you draw them out sequentially. It's the same situation, just a different way of thinking about it)

for example, f I have 2 r's and 1 w, possible outcomes are

rrw

rwr

wrr

which equals (r+w) choose w = (r+w)!/(r!*w!) = (r+w) choose r

This is the denominator in the probability, it represents all possible outcomes.

The only way I can be left with a single white ball at the end is if I have a red and white ball in the next to the last step. So my sequence of balls in the container must look like

RWRRWRW......RW

the last two balls MUST be red, and then white, in that order. Everything before that doesn't matter. So I don't care about the order of the first (r+w-2) balls because as long as I have one white and one red ball in the next to the last step, I'm ok.

Another way to look at it is (r+w-2) choose (w-1) = [(r-1) + (w-1)] choose (w-1) = [(r-1)+(w-1)]!/(r-1)! * (w-1)!

In english, the answer is (ways to arrange red and white balls so that I am guaranteed to end with one red and one white)/(total ways to arrange red and white balls)