Illustrative example:
The pgf for $X \sim \mathrm{Bin}(m,p_1)$ is $G_1(z)=[(1-p_1)+p_1z]^m$. If $m \sim \mathrm{Bin} (n,p_2)$ the final distribution has pgf of $G_2(z)=[(1-p_1 p_2)+p_1 p_2z]^n$ representing $\mathrm{Bin}(n,p_1 p_2)$ distribution.
Mathematically the final answer is still a valid distribution for $s=p_1 p_2$ as long as $0<s <1$. This may be true even though $p_i<0$ or $p_i>1$ individually, as long as their product is positive and between 0 and 1.
For example, if $p_{\{1,2\}}=-\frac{1}{\sqrt 2}$ then both distributions alternate between positive and negative 'probabilities', yet their mixture is $s=\frac{1}{2}$; equivalent to flipping a coin $n$ times.
This reminds me of improper priors where the probabilities are positive but the sum does not equal 100%. Here we may have a "distribution" where the sum is 100% but the probabilities may be negative.
If improper priors are allowed if the posterior is valid, are negative probabilities allowed if the final answer is a valid distribution?