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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?

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  • $\begingroup$ This appears to be a question about a negative binomial random variable. There several versions of negative binomial distributions. Please describe the random variable and carefully define $m, n, p.$ $\endgroup$ – BruceET Apr 2 '18 at 2:00
  • $\begingroup$ I have added some of these details. Please see if you can answer it now? $\endgroup$ – toing Apr 2 '18 at 2:11
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The book answer (or your transcription of it) appears to be incorrect

By the way, it is simpler to count failures, and look at the results in reverse.
With $0 \le k < m$, the last trial must be a success, and the $k$ failures can be distributed any which way in the remaining $(k+n-1)$ trials, thus

$$P = \sum_{k=0}^{m} \binom{k+n-1}k p^n (1-p)^k$$

If you want to count successes (as the book has done), you should now be able to correct the formula you have transcribed.

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