Poisson Conditional Probability - Very lost! A discrete random variable is said to have a Poisson distribution if its possible values are the non-negative integers and if, for any non-negative integer $k$,
$$P(X=k)=e^{-\lambda}\frac{\lambda^k}{k!}$$
where $\lambda>0$. It turns out that $E(X)=\lambda$.
Minitab has a calculator for calculating Poisson probabilities, which is very similar to the calculator for Binomial probabilities.
The Poisson distribution model is widely used for modeling the number of "rare" events.
Suppose we have a Poisson random variable $X$ with mean (or expected value) equal to $2$ and another Poisson random variable $Y$ with mean $3$.
Suppose $X$ and $Y$ are independent random variables, in which case $W = X+Y$ will be a Poisson random variable with mean equal to $5 (= 2+3)$.
Find the conditional probability that $X = 5$ given that $W = 10$.

The distribution of the conditional probability is $$f_{X\mid W}(x\mid w)f_W(w)=f_{X,W}(x,w).$$
The probability that $X=5$ and $W=10$ is the same as the probability that $X=5$ and $Y=5$, so the joint probability on the right is the same as $f_{X,Y}(x,y)$, where $Y=W-X$. (Explicitly, this is a change of coordinates, but it has unit Jacobian). Since $X$ and $Y$ are independent, the joint probability is easy to understand. Dividing it by the distribution of $W$, which we also know, gives us the conditional probability.
