# Why does negative binomial random variable uses ${n-1 \choose r-1}$ instead of ${n \choose r}$ as coefficient?

negative binomial random variable works like this, but can you explain why we use ${n-1 \choose r-1}$ instead of ${n \choose r}$?

$$P(X=x) = {x-1\choose r-1}(1-p)^{x-r}p^r$$

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Basic reason the use of ${n-1} \choose {r-1}$ lies in the description of the situation where negative binomial occurs.We want rth success in nth trial.Hence,$r-1$ success in $n-1$ trial and nth trial being a success.Hence $n-1 \choose r-1$.

Now you can redefine the situation in which $n \choose r$ makes sense but try doing it,it is not very easy to perceive.

Lets start from scratch: We want first $n-1$ trials result in $r-1$ successes and $(n-1)-(r-1)=n-r$ failures.Any of $r-1$ out of $n-1$ trials can be success,and remaining $n-r$ should be be failures (note that choosing $r-1$ success also fixes $n-r$ failures.Hence ${n-1 \choose r-1} \times p^{r-1} \times {(1-p)}^{n-r}$ . Now the last trial must be success,probability of which is $p$,hence we multiply by $p$ and get required answer.

Hope this helps.

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i understand r-1 success in n-1 trial part, but the last trial must be a success then that makes it ${n \choose r}$... if we do ${n-1 \choose r-1}$ we lose the last trial! –  user133466 Oct 4 '12 at 19:39
Consider again,if we use $n \choose r$ then it includes the possibility that first r were successes and remaining were failures,then we don't get nth trial as success.For nth trial to be rth success as we want,we only have freedom in choosing $r-1$ successes from $n-1$ trials as rth success is to occur at nth trial. –  TheJoker Oct 4 '12 at 19:44
This is a good answer, Thank you! however what I don't understand is that with ${n-1 \choose r-1}$ aren't we skipping the last trial? It's a guaranteed success, but we shouldn't ignore it –  user133466 Oct 4 '12 at 20:47
No we are not skipping the last trial.However note that last trial is not guaranteed success,it is success with probability $p$.Hence we multiplied by $p$,otherwise we'ld have got ${n-1 \choose r-1} {(1-p)}^{(n-r)}p^{(r-1)}$ –  TheJoker Oct 4 '12 at 22:33

Dunno, maybe for the sum of independent NB$(r,p)$ and NB$(s,p)$ to be NB$(r+s,p)$.

Slightly less convincing: for NB$(r,p)$ to be geometric when $r=1$.

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