Given a set $C$ composed by three distinguishable kinds of elements, the probability of the event $I_n^C$, defined as "to get at least one element of each kind, performing $n>0$ trials of one element at a time with replacement from $C$", is $$ P(I_n^C)=1-\left(\frac{\alpha+\gamma}{c}\right)^n-\left(\frac{\beta+\gamma}{c}\right)^n-\left(\frac{\alpha+\beta}{c}\right)^n+\left(\frac{\alpha}{c}\right)^n+\left(\frac{\beta}{c}\right)^n+\left(\frac{\gamma}{c}\right)^n, $$ where $\alpha,\beta,\gamma>0$ are the integers accounting for the numbers of elements of the three kinds, and $c=\alpha+\beta+\gamma>2$.
This result can be easily proved by means of the principle of inclusion-exclusion together with Bayes' theorem.
A trivial property of the event $I_n^C$ is that, for any $\alpha,\beta,\gamma,n>0$, it must be $P(I_n^C)=0\iff n\leq 2$, and $P(I_n^C)>0\iff n> 2$.
If we impose the relation $1-\left(\frac{\alpha+\gamma}{c}\right)^n-\left(\frac{\beta+\gamma}{c}\right)^n=0$, it must be either $$ -\left(\frac{\alpha+\beta}{c}\right)^n+\left(\frac{\alpha}{c}\right)^n+\left(\frac{\beta}{c}\right)^n+\left(\frac{\gamma}{c}\right)^n=0 \iff n\leq 2, $$ in case $P(I_n^C)=0$, or $$ -\left(\frac{\alpha+\beta}{c}\right)^n+\left(\frac{\alpha}{c}\right)^n+\left(\frac{\beta}{c}\right)^n+\left(\frac{\gamma}{c}\right)^n>0 \iff n> 2, $$ in case $P(I_n^C)>0$.
Therefore, to prove the statement
$$1-\left(\frac{\alpha+\gamma}{c}\right)^n-\left(\frac{\beta+\gamma}{c}\right)^n=0\Rightarrow n\leq 2, $$
which corresponds to Fermat's last theorem, is fully equivalent to prove the statement
$$
1-\left(\frac{\alpha+\gamma}{c}\right)^n-\left(\frac{\beta+\gamma}{c}\right)^n=0\Rightarrow -\left(\frac{\alpha+\beta}{c}\right)^n+\left(\frac{\alpha}{c}\right)^n+\left(\frac{\beta}{c}\right)^n+\left(\frac{\gamma}{c}\right)^n=0,
$$
since $P(I_n^C)$ cannot be negative.
In other words, the new statement assess that the imposition $1-\left(\frac{\alpha+\gamma}{c}\right)^n-\left(\frac{\beta+\gamma}{c}\right)^n=0$ (which corresponds to ask that the probability to get at least one element of kind $\alpha$ is equal to the probability not to get any element of kind $\beta$, in $n$ trials) implies that the event $I_n^C$ cannot take place.
The new statement can be written in the form
$$ (\alpha+\beta+\gamma)^n=(\alpha+\gamma)^n+(\beta+\gamma)^n \Rightarrow \alpha^n+\beta^n+\gamma^n=(\alpha+\beta)^n, $$
which therefore represents an equivalent formulation of Fermat's last theorem (i.e. if we prove the above implication, we prove Fermat's theorem).
Is this reasoning correct?
I apologize in case of naivety and silly mistakes! Thanks however for your help!
NOTE: This post is a wrap of many others, among which An urn with three kinds of balls... and a weird constraint!, and A conjecture involving the equation $x^n+y^n+z^n= (x+y)^n$.