I'm unable to think through this question please help.
Suppose our sample space distinguishes points with different orders of selection. For example suppose that $S =\{SSSSFFF\ldots\}$ consists of all words of length n where letters are drawn without replacement from a total of $r$ $S$’s and $N$ � $r$ $F$’s. Derive a formula for the probability that the word contains exactly $X$ $S$’s. In other words, determine the hypergeometric probability function using a sample space in which order of selection is considered.
The answer is: $ f(x)= \dfrac{\binom{n}{x} r^x (N-r)^{n-x}}{N^n}$
with $x$ ranging from $\max(0, n-(N-r))$ to $\min(n,r)$
Thanks in advance!
For your sampling without replacement problem, there are $\binom{N}{n}$ equally likely choices. (Imagine that the letters have ID numbers written on them with invisible ink.) Whether we grab the $n$ items all at once or sequentially makes no difference to the probability we have $x$ S's.
There are $\binom{r}{x}$ ways to choose $x$ S's, and for each of these ways there are $\binom{N-r}{n-x}$ to choose $n-x$ F's. So our probability is $$\frac{\binom{r}{x}\binom{N-r}{n-x}}{\binom{N}{r}}.$$
Remark: For sampling with replacement, at each pick our probability of success S is $\frac{r}{N}$, and the probability of failure is $1-\frac{r}{N}$, which is $\frac{N-r}{N}$. Thus the probability of $x$ successes in $n$ trials is $$\binom{n}{x}\left(\frac{r}{N}\right)^x \left(1-\frac{r}{N}\right)^{n-x}.$$ This is the $f(x)$ of the OP. When we sample with replacement, the distribution is binomial, not hypergeometric.