# Probability - R successes before the kth failure

1. Feb 27, 2013

### dkotschessaa

1. The problem statement, all variables and given/known data

In a sequence of independent trials with probability of success p, what is the probability that there are r successes before the k-th failure?

2. Relevant equations

Binomial distribution

$f(x;n,p) = {{n}\choose{r}} p^{x} (1-p)^{n-r}$

3. The attempt at a solution

I know that the answer is

${{k+r+1}\choose{r}} p^{r} (1-p)^{k}$

This almost makes sense to me.

Since x = number of successes = r
and I know n represents the number of trials - but I'm not sure how this amounts to k+r+1.

and if n is k+r+1, then the exponent of (1-p), which is n-x, should be

(k+r+1) - r = k+1

But here it is just k.

why?

-Dave K

2. Feb 27, 2013

### Ray Vickson

The "answer" you give above is not quite correct. Look at it for the case of k = 1; it does not work properly in that case. (Note: your formula would be the probability of having r successes and 1 failure in (r+1) trials, but that would include all outcomes of the form FSS...S, SFSS...S, ... SS...SF, which is NOT what you want. You want the probability of the single point SSS...SF.)

The number of tosses X until the first failure is geometrically distributed with parameter 1-p; note that this count *includes* the failure itself, so the number *before* the first failure is Y = X-1 (which is in {0, 1, 2, ...}). After each failure the counting process starts again, so you want the distribution Zk = Y1 + Y2 + ... + Yk, where the Yi are iid copies of Y. You are asking for P{Zk = r}. You can get a closed-form formula for this using generating-function methods, or else by getting a recursion for P{Z(k-1)=s} and P{Y=t}. The answer will be almost what you wrote above, but with a slight difference.

Last edited: Feb 27, 2013
3. Feb 28, 2013

### dkotschessaa

Thanks. I suspected the answer was wrong, but it was what the professor gave us. He is giving us wrong answers, and sometimes even wrong questions. We are all baffled. We have a test tomorrow.

Thanks!