problem of n success before m failures where binomial probability of success is p has a standard textbook solution as follows $$P = \sum_{k=n}^{m+n-1} \binom{m+n-1}k p^k (1-p)^{m+n-1-k}$$
I am however unable to come up with this solution and i am not sure where i deviate and how?
I start with combinatorial logic that for any t trials
(a) last trial has to be a success => probability of p
(b) in t-1 trials before last trial, n-1 have to be success and (t-1)-(n-1) failures => probability of $\binom{t-1}{n-1} p^{n-1} (1-p)^{t-n}$
Combining (a) and (b) gives me probability that t trials have t-n failures before n success $P_{t} = \binom{t-1}{n-1} p^{n} (1-p)^{t-n}$ I understand that this is negative binomial pmf as well.
From here on, i say that total probability is sum of all the probabilities with various t i.e. $P = \sum_{n}^{m+n-1} P_{t}$ that can be written as $$P = \sum_{t=n}^{m+n-1} \binom{t-1}{n-1} p^{n-1} (1-p)^{t-n} $$
I believe something is off with my last step.. may be these events of different $t$ trials are not mutually exclusive but i am not able to see it.
For example: n = 3 success and m = 2 failures; p = binomial probability
(a) I can have a 3 trial solution SSS with probability $p^3$
(b) 4 trial solution {SSFS,SFSS,FSSS} with probability $ \binom{3}{2} p^{3} (1-p)^1$
(c) 5 trial solution is not possible since n=2 would have happened
I would therefore add probabilities from (a) and (b) as solution but that would give $$ P = p^3 + \binom{3}{2} p^{3} (1-p)^1 $$
and of course, this is not right. Can someone help and point out why this is not correct and what can i fix here?