From: "r.e.s." Subject: Re: modified geometrical distribution? Date: Sun, 27 Jun 1999 15:18:07 -0700 Newsgroups: sci.math Keywords: negative binomial distribution Yes, it makes sense to consider the probability that, in independent repeated trials, the n-th success occurs on the k-th trial (k=n, n+1, n+2, ...), where in each trial p=pr(success). pr(n-th success on the k-th trial) =pr(n-th trial is a success, and among the (k-1) preceding trials, there are (n-1) successes) =pr(n-th trial is a success) * pr(among the (k-1) preceding trials, there are (n-1) successes| n-th trial is a success) = p * pr(among the (k-1) preceding trials, there are (n-1) successes), because of independence = p * C(k-1, n-1) p^(n-1) (1-p)^(k-n), applying the usual binomial distribution, C()=binomial coeff. = C(k-1, n-1) p^n (1-p)^(k-n), k=n, n+1, n+2,... This is called the "negative binomial distribution" or "Pascal distribution". (It was found by Pascal while answering the questions of a nobleman whose gaming interest was in the occurrence of a 6-6 when repeatedly throwing a pair of dice.) -- r.e.s. (Spam-block=XX) James wrote ... : Given p as a probability of a sucess. : The question is the waiting time distribution : until n successes. : If n=1, it is a geometric distribtution. : But I don't know how to solve this for n>1. : For example, n=2, how can I solve this? : Or it doesn't make a sense at all.