# Why is this adjustment of the negative binomial distribution true?

SECOND-YEAR PROBABILITY: For iid trials with events $$F,S$$ and $$P(S)=p$$, what is the probability $$y$$ trials will occur before the $$r$$th success?

This is a question from Wackerly, Mathematical Stats 7ed. I tried to reason this out and arrived at this answer: $$p(x)=\begin{pmatrix}y\\r\end{pmatrix}p^{r-1}q^{y-r+1}$$ Where the correct answer was: $$p(x)=\begin{pmatrix}y\\r-1\end{pmatrix}p^{r}q^{y-r+1}$$ Can someone walk me through why this is? Especially the nCr part.

My reasoning was based off an arbitrary example of FFFS. The "new distribution" would be looking for FFFF, so I needed one extra fail (hence y-r+1) and there would be one less success but apparently it doesn't decrease p. I think it's because I misunderstood the question, my hypothetical scenario should be FFFFS under the new distribution. Can someone confirm this? As for nCr part I really don't know what's going on around $$\begin{pmatrix}y\\r-1\end{pmatrix}$$.

I think you are asking for the probability that $$y$$ trials will occur before the $$r$$th "$$S$$" trial.
Take for example $$y=5$$ and $$r=3$$. Situations you would consider are FSSFS, SFSFS, SSFFS, etc. Each is a sequence of $$r$$ "S" events and $$y-r+1$$ "F" events, so each sequence occurs with probability $$p^r q^{y-r+1}$$. To count the number of valid sequences, note that you just need the last trial (the $$(y+1)$$st trial) to be S, and the first $$y$$ trials can be any arbitrary arrangement of $$r-1$$ "S" events and $$y-r+1$$ "F" events. So you multiply by $$\binom{y}{r-1}$$.