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Suppose that independent Bernoulli trials with parameter $p$ are performed successively. Let $N$ be the number of trials needed to get $x$ successes, and $X$ be the number of successes in the first $n$ trials. Show $P(N=n) = \frac{x}n P(X=x)$. I tried to do something along the lines of the negative binomial distribution but I cant figure out how to get this.

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The event $\{N=n\}$ occurs exactly when the event $\{X=x\}$ occurs and the $n$-th trial is a success. There are $\binom{n}{x}$ equally likely ways that exactly $x$ successes occur on $n$ trials; each of these ways has probability $p^x(1-p)^{n-x}$. How many of these $\binom{n}{x}$ equally likely ways have a success on the $n$-th trial? Obviously $$\binom{n-1}{x-1} = \frac{(n-1)!}{(x-1)!(n-x)!} = \frac{x}{n}\times\frac{n!}{x!(n-x)!} = \frac{x}{n}\binom{n}{x}.$$ Can you take it from here?

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ah yes I see now.. It was the wording that got me off track. Thanks! –  TheHopefulActuary Nov 9 '12 at 3:38

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