$X=Poi(\lambda)$ and $Y=Geom(p)$ and $Y$ is independent Geometric. Give a formula for $P(X=Y)$

Question

$X=Poi(\lambda)$ and $Y=Geom(p)$ and $Y$ is independent Geometric. Give a formula for $P(X=Y)$

No Idea how to do this $Y=(1-p)^kp$
$X=e^{-\lambda}(\lambda^n/n!)$

and I'm assuming it's asking for this?
$(1-p)^kp=e^{-\lambda}(\lambda^n/n!)$

Answer 1

$P(X=Y)= \sum\limits_{k=0}^{\infty} p(1-p)^{k} e^{-\lambda} \frac {\lambda ^{k}} {k!}$. This is nothing but $pe^{-\lambda} \sum\limits_{k=0}^{\infty} \frac {r^{k}} {k!}$ where $r=(1-p)\lambda$. Can you take it from here?