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Hi folks this is a self learn probability (counting) question from DeGroot. The question is:

Suppose that a box contains r red balls and w white balls. Suppose also that the balls are drawn out of the box one at a time at random without replacement. What is is the probability that all r red balls will be obtained before two white balls are obtained?

Now my reasoning behind this is that the sample space is the first r+1 balls drawn. The r+2nd ball is not important because you have to draw all the red balls plus one white ball to satisfy the conditions of the problem. Therefore for the entire sample space the number of combinations $\dbinom{r+w}{r+1}$. The number of combinations that would satisfy the conditions is $\dbinom{r+1}{r}$. Therefore the probability is $\frac{\dbinom{r+1}{r}}{\dbinom{r+w}{r+1}}$.

The answer provided in the back of the book is $\frac{\dbinom{r+1}{r}}{\dbinom{r+w}{r}}$.

I can't for the life of me see why the entire sample space is $\dbinom{r+w}{r}$ instead of $\dbinom{r+w}{r+1}$. Can anyone please provide an explanation?

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The number of combinations that satisfy the conditions is $w$, not $ r+1\choose r $ (there are $w$ ways to choose the one white ball, and there is one way to choose the $n$ red balls). This leads to same answer as that in the back of the book. I don't see offhand how the book arrived at their particular expression. –  David Mitra Jun 19 '13 at 0:12

1 Answer 1

The entire sample space is the combinations that include r red balls - this gives ${r+w} \choose r$ as the sample space - note that this includes all states with $r$ red balls and from 0 to $w$ white balls.

Of those, the ones that meet the condition are those that have 0 white balls and $r$ red balls and 1 white ball and $r$ red balls - hence ${r+1}\choose r$

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Yes that is correct and it makes sense. Thank you very much. –  panos Jun 19 '13 at 2:08

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