How do I find the marginal distribution with this summation/series?

Question

Let $X$ be a Poisson(2) and $Y$ be Binomial(10,3/4) random variables, If $X$ and $Y$ are independent, then $P(XY=0)$ is

I thought of using transformation to find the distribution of $XY$ so I let $U=XY, V=Y$ with$|j|=\frac{1}{v}$ hence the transformed bivariate distribution from $$f(x,y)=\frac{e^{-2}2^x}{x!}.\binom{10}{y}\left(\frac{3}{4}\right)^y\left(\frac{1}{4}\right)^{10-y} \text{is} f(u,v)=\frac{e^{-2}2^{\frac{u}{v}}}{\left(\frac{u}{v}\right)!}.\binom{10}{v}\left(\frac{3}{4}\right)^v\left(\frac{1}{4}\right)^{10-v}\frac{1}{v}$$ now I don't know how to find distribution of $U$ with it as unable to solve this $$\sum_{v=1}^{10}\frac{e^{-2}2^{\frac{u}{v}}}{\left(\frac{u}{v}\right)!}.\binom{10}{v}\left(\frac{3}{4}\right)^v\left(\frac{1}{4}\right)^{10-v}\frac{1}{v}$$ any suggestions? May be there is some other method which I don't know about.

Answer 1

There is no need for a change of variables; we have $$\{XY=0\} = \{X=0\}\cup\{X\ne0,Y=0\}, $$ and so by independence, \begin{align} \mathbb P(XY=0) &= \mathbb P(X=0) + \mathbb P(X\ne0)\mathbb P(Y=0)\\ &= e^{-2} + (1-e^{-2})\left(\frac14\right)^{10}\\ &= e^{-2} + (1-e^{-2})2^{-20}. \end{align}