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I am solving a probability question from my math course, and I have gotten to the following expression:

$$\left(\sum_{i=1}^n \frac{(x+y+i-2)!}{y ! (x-1)! (i-1)!}\right) \frac{n! x! y!}{(n + x + y)!}$$

However, the final solution is such:

$$\frac{n x}{(x + y) (x + y + n) }$$

Using Wolfram Mathematica, I confirmed that my equation and the final answer are equivalent, but how would I go about simplifying my answer to the get final answer by hand?

In case anyone is curious, the question asks that if there are $x$ jelly donuts, $y$ chocolate donuts, and $n$ vanilla donuts in a box, what is the probability that if they are taken out and subsequently eaten one by one uniformly at random, that the chocolate donuts will be the first type of donut to be completely finished, the jelly donuts will be the second type to be finished, and the vanilla donuts will be the last to be finished?

Any alternate solutions are also appreciated, thanks for the help.

Rajat Mittal
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2 Answers2

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By removing the donuts randomly one at a time, you're simply putting them in a random order. The last donut is equally likely to be any one of the $n+x+y$ donuts, so the probability that it is vanilla is $n/(n+x+y)$. Now given that the last donut is vanilla, the non-vanilla donuts are still in a random order. The conditional probability that the last non-vanilla donut is jelly, given that the last donut is vanilla, is $x/(x+y)$. Thus the probability of your event is $$\frac{n}{n+x+y} \cdot \frac{x}{x+y} = \frac{nx}{(n+x+y)(x+y)}$$

Robert Israel
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  • Thanks a lot. That really helps. But do you happen to know of an algebraic way to simplify the expression I had gotten to in my work? – Rajat Mittal Nov 27 '16 at 19:48
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About the algebraic simplification, first note that :

$$\sum_{i=1}^{n}\frac{(x+y+i-2)!}{y!(x-1)!(i-1)!}\times \frac{n!x!y!}{(n+x+y)!}=\sum_{i=0}^{n-1}\frac{(x+y+i-1)!}{i!}\times \frac{x\times n!}{(n+x+y)!}$$

Here, I have just extracted the terms that do not depend on $i$ from the sum, and made the change of variable "$i=i-1$". Then you can make appear a coefficient binomial in the sum by dividing by $(x+y-1)!$ to get :

$$\sum_{i=0}^{n-1}\frac{(x+y+i-1)!}{i!}=\sum_{i=0}^{n-1}\binom{x+y+i-1}{i}\times (x+y-1)!$$

Then apply the formula

$$\sum_{i=0}^{n-1} \binom{X+i}{i}=\binom{X+n}{n-1}$$

(you can find a proof of this formula here : Prove that $C(n,0) + C(n+1,1) +\dots+ C(n+r,r) = C(n+r+1,r).$ )

where $X=x+y-1$ to get :

$$\sum_{i=0}^{n-1}\frac{(x+y+i-1)!}{i!}\times \frac{x\times n!}{(n+x+y)!} =\binom{n+x+y-1}{n-1}\times \frac{x\times (x+y-1)!n!}{(n+x+y)!}$$

Simplify and you get the result.

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