On this problem how might we use the binomial distribution? I know the solution is using the NB but cant you use (x/n)(ncx)(p^x)(q^nx) somehow?
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Originally posted by claytonmccandless View PostOn this problem how might we use the binomial distribution? I know the solution is using the NB but cant you use (x/n)(ncx)(p^x)(q^nx) somehow?
I thought that
fb = (nCx)(p^x)(q^(nx))
and
fnb = (x/n)(nCx)(p^x)(q^(nx))
so fnb = (x/n)fnb
So I would think that the answer to your question is that yes, you can use a binomial distribution, but in order to use it in this case, you have to multiply it be the coefficient (x/n). And by doing so, you actually end up turning it into the negative binomial distribution.
I believe that you can think of the negative binomial as just a more restrictive case of the binomial distribution. For the Binomial distribution, x is the random variable and the PDF gives the probability of x successes in a fixed number of n trials. For the Neg Binomial distribution, n is the random variable and the PDF gives the probability that n trials are required for a fixed number of x successes to occur. So in the B.D. we just count the # of successes in n trials, but in the N.B.D. we count the number of trials it takes to get to x successes. So even if you use the same values for x and n in the B.D. and the N.B.D. you get different probabilities because the B.D. counts all the chances of having x success in n trials (the xth success can be before the last trial) but the N.B.D. only counts the chances of having the x successes where the last success occurs on the last trial. So you have x1 successes in the first n1 trials, and the xth success occurs on the nth trial. That is a subset of the chances of any occurrence of x successes in n trials. So that is why the N.B.D. is a fraction of the B.D.
I actually show further detail on how to use the N.B.D. for this question in another thread here:
http://www.actuary.com/actuarialdis...ad.php?p=65709
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