Consider an urn containing $c$ elements, $b$ of which are black.
If we perform $n$ trials with replacement of one element at a time from the urn, the probability to get $n$ times a black ball is $\left(\frac{b}{c}\right)^n$.
Since $c, b$ and $n$ are integer numbers, at each trial corresponds one and only one extracted ball. Therefore, I guess we can represent the $n$ trials as follows:
In this scheme, $c=10$, $b=4$ and $n=14$.
The blue path represents a sequence of trials in which the extracted balls are both black and not-black, whereas the black path represents a sequence of trials in which all the extracted balls are black.
Since at each trial corresponds one and only one extracted ball, all these possible paths (blue and black) are discrete functions of $n$.
Let $N_{c,n}$ and $N_{b,n}$ the number of possible functions/paths that is possible to build in such scheme.
My conjecture is that
$$ \frac{N_{b,n}}{N_{c,n}}=\left(\frac{b}{c}\right)^n. $$
However, I am not sure how to prove or disprove such claim.
My question, strongly related to the conjecture above, is
$N_{b,n}=b^n$?
If yes, how to prove it? If not, how to correctly evaluate $N_{b,n}$?
I apologize in case this is an obvious problem.
Thanks for your help!
Your paths seem to be records of which particular ball is drawn. For example, your blue path starts $10,5,7,3,5,\ldots$ while your black path starts $2,3,3,4,2,\ldots$ restricted to the black numbers $\{1,2,3,4\}$
So the number of possible paths overall is $N_{c,n} = c^n$ since each of the $n$ steps can be any of the $c$ numbers independently of the others
and the number of possible paths just using black numbers is similarly $N_{b,n} = b^n$
So the proportion of paths just using black numbers is $$\dfrac{N_{b,n}}{N_{c,n}} = \dfrac{b^n}{c^n} = \left(\dfrac{b}{c} \right)^n$$ and since each path is equally likely, this also gives the probability of all black draws