$X$ Poisson distribution, $Y$ geometric distribution - how to find $P(Y>X)$?

Question

Suppose that $X$ has Poisson distribution with parameter $\lambda$ and that $Y$ has geometric distribution with parameter $p$ and is independent of $X$.

What is the $P(Y>X)$ ? (the final formula should not contain any infinite sum).

$X$ and $Y$ represent two different distributions and it is not a standard case with a sum of two independent variables. I am confused here. Does the convolution apply to this case? What should be the steps to solve this problem?

Answer 1

for the record, there is the full solution to this problem

$P(X=n)= e^{-\lambda} \frac{\lambda^n}{n!} \\ P(Y=n+1)=(1-p)^{(n+1)-1} \\$

$P(Y>X)=\sum_n P(Y>X,X=n)= \sum_n P(Y>n,X=n)= \sum_n P(Y>n)P(X=n) = \sum_n P(Y\ge n+1) P(X=n)= \\ = \sum_n (1-p)^{(n+1)-1} * e^{-\lambda} \frac{\lambda^n}{n!} = \sum_n e^{-\lambda} * \frac{(\lambda(1-p))^n}{n!} = e^{-\lambda}*e^{\lambda(1-p)}=e^{-\lambda p}$