I want to calculate:
$$ \mathbb{P}[ 2Y = 3X ~\vert~ \text{$X$ is even} ]$$
given this joint pmf:
$$ \mathbb{P}(X = k, Y = n) = e^{-2} \frac{(1-e^{-1})^k}{(n-k)!}$$
From the joint I found the density of $X$ ( by summing from $n = k$ to infinity) : $$ f_X(x) = e^{-1}(1-e^{-1})^k $$ So $ X + 1 \sim \text{Geometric}(e^{-1}) $
Then I found the density of $Y$ conditional $X = x$ by dividing the joint by density of $X$, I found the density $f_{Y \vert X = k}(n) = \frac{ e^{-1}}{(n-k)!}$, where $n \geq k$. I hope my calculations are all right until now.
But I think I messed up somewhere, because $$ \mathbb{P}\left[ Y = \frac{3}{2}X ~\middle|~ X = 2c \right] = \mathbb{P}[ Y = 3c ~\vert~ X = 2c ] = \sum_{c=0}^\infty \frac{ e^{-1}}{(3c-2c)!}$$ adds up to one which makes no sense.
How do I find $ \mathbb{P}[ 2Y = 3X ~\vert~ \text{$X$ is even} ]$ ?
Given the joint pmf $$\mathbb{P}(X = k, Y = n) = e^{-2} \frac{(1-e^{-1})^k}{(n-k)!}~, \qquad n\geq k\geq 0 \label{eq_joint_pmf} \tag*{Eq.(1)}$$ where $k = 0$ and $n = k$ are explicitly included, indeed we have the marginal for $X$ as Geometric $$f_X(k) = \frac1{e} \bigl( 1 - \frac1{e}\bigr)^k \label{eq_x-marginal} \tag*{Eq.(2)}$$ The desired conditional probability is $$\mathbb{P}\bigl[\, 2Y = 3X ~\vert~ \text{$X$ is even} \,\bigr] = \frac{ \mathbb{P}\bigl[\, 2Y = 3X ~~\&~~\text{$X$ is even} \,\bigr] }{ \mathbb{P}\bigl[\, \text{$X$ is even} \,\bigr] } \label{eq_target} \tag*{Eq.(3)}$$
Let me assume that the event $\{ X ~\text{is even}\}$ means $X = 0,2,4,6,\ldots$ with zero included. First we obtain the denominator in \ref{eq_target} by summing \ref{eq_x-marginal} over the desired subset \begin{align} \mathbb{P}\bigl[\, \text{$X$ is even} \,\bigr] = \sum_{c = 0}^{\infty} f_X(2c) &= \sum_{c = 0}^{\infty} \frac1{e} \bigl( 1 - \frac1{e}\bigr)^{2c} \\ &= \frac1{e} \frac1{1 - \bigl( 1 - \frac1{e}\bigr)^2 } =\frac1{ 2 - \frac1{e} } \approx 0.6127 \label{eq_pr_x-even} \tag*{Eq.(4)} \end{align} First we obtain the numerator in \ref{eq_target} by summing \ref{eq_joint_pmf}, again, over the desired subset where $X = k = 2c$ and $Y = n = 3c$. \begin{align} \mathbb{P}\bigl[\, 2Y = 3X ~~\&~~\text{$X$ is even} \,\bigr] &= \sum_{c = 0}^{\infty} e^{-2} \frac{(1-e^{-1})^{2c} }{ (3c - 2c)! } && \text{, denote $\lambda \equiv \bigl( 1 - \frac1{e} \bigr)^2$} \\ &= \frac1{ e^2 } \sum_{c = 0}^{\infty} \frac{ \lambda^c }{ c! } && \\ &= \frac1{ e^2 } e^\lambda && \label{eq_pr_x-even_y-3} \tag*{Eq.(5a)} \end{align} To divide by \ref{eq_pr_x-even} one might consider rewriting \ref{eq_pr_x-even_y-3} as $$\frac1{ e^2 } e^\lambda = \frac1{ e } \exp\left[ -1 + 1 - \frac2{e} + \frac1{ e^2} \right] = \frac1{ e } \exp\left[ \frac{-1}{e} \bigl( 2 - \frac1{e} \bigr) \right] \label{eq_pr_x-even_y-3_alt} \tag*{Eq.(5b)} $$ so that one can have a nice form for the desired conditional probability \ref{eq_target} with $\rho \equiv 2 - e^{-1}$
The procedure is similar if one wants to exclude zero from the set of even numbers. The result \begin{align} &\phantom{{}={}}\mathbb{P}\bigl[\, 2Y = 3X ~~\&~~\text{$X$ is even} \,\bigr] \cdot \mathbb{P}\bigl[\, \text{$X$ is even} \,\bigr] ^{-1} \\ &= \frac{ \displaystyle \exp\left[ \bigl( \frac1{e} - 1 \bigr)^2 \right] -1 }{ e^2 } \cdot \left( \frac{ \displaystyle\bigl( \frac1{e} - 1 \bigr)^2 }{ 2 - e^{-1} } \right)^{-1} \approx 0.271529 \end{align}