So I have a mathproblem where the Poisson distribution of the number of flowers X in one area is given by $\lambda_1=4$. The Poisson distribution for the numbers of flowers Y in the other area is given by $\lambda_2=6$. I am then asked to find an expression for $P(X=Y)$, and explain what condition that must be satisfied to calculate this. I'm not that good at probability problems, so what does $P(X=Y)$ mean, and how should I find this expression? I know that
$P(X)=\frac{4^X}{X!}e^{-4}$ and $P(Y)=\frac{6^Y}{Y!}e^{-6}$
You can use $P(X=Y)=P(\bigcup_n \{ X= n\} \cap \{Y=n\})=\sum_nP(\{X=n\}\cap\{Y=n\})$. Now all you need is a condition to compute $P(\{X=n\}\cap\{Y=n\})$. The easiest would be independence.
Edit: Regarding your notation. As $X$ is a random variable and $P$ a probability measure, $\{X=n\}$ denotes the set of all $\omega\in\Omega$ such that $X(\omega)=n$. So writing $P(X)$ is abuse of notation, you should better write $P(X=n)=\frac{4^n}{n!}e^{-4}$. By $P(X=Y)$ we are looking for the probability of the set $\{\omega\in\Omega: X(\omega)=Y(\omega)\}$ which are indeed those $\omega$, such that there exists $n$ with $X(\omega)=n$ and $Y(\omega)=n$ which leads to the probability above.