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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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up vote 2 down vote accepted

You are probably intended to make unreasonable assumptions. Like equal probability of male and female. And to forget about the fact that males like to hang around females, or the other way around.

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 $q=1-p$.

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