# If A ~ Poisson(a), then what is joint probability mass function of male and female fish?

Suppose that the number of fish in a big sea $A$ adheres to a Poisson distribution. What is joint probability mass function of having $x$ male and $y$ female fish?

I thought that I could just divide A by $\frac{1}{2}$ and add the two together... but that didn't make very much sense on second thought...:

$$M \rightarrow \text{ the number of male fish.} \\ F \rightarrow \text{ the number of female fish.} \\ P(M = x, Y = y) = 2\frac{e^{-a}a^{x+y}}{2(x+y!)}$$

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Note that the probability of having $x+y$ fish (in presumably a small defined area) is $$e^{-\lambda}\frac{\lambda^{x+y}}{(x+y)!},\tag{1}$$ where $\lambda$ is the parameter you called $a$.
Given that there are $x+y$ fish in the area, the probability that $x$ are male and $y$ are female is equal to $$\binom{x+y}{x}\left(\frac{1}{2}\right)^{x+y}.\tag{2}$$ (The "fact" that the conditional distribution is binomial is based on a not necessarily reasonable independence assumption.)
Multiply $(1)$ and $(2)$. The expression can be simplified somewhat.
It is easy to modify the second expression if instead of equidistribution we assume that a fish is male with probability $p$, and female with probability $q=1-p$.