The problem statement is:
Given the R.V. $Y \sim \operatorname{Poisson}(2)$ and the R.V. $X$ with conditional probability function
$$P(X=x \mid Y=y) = \begin{cases} \frac{1}{5} & \text{if } x = y + 1 \\ \frac{3}{5} & \text{if } x = y + 2 \\ \frac{1}{5} & \text{if } x = y + 3 \\ \end{cases}$$
What is $P(X=3)$?
I don't know how to approach for this exercise. I was trying to compute the probability P(X=3) by the joint probability of $X$ and $Y$, but I don't know how to get $P(Y=y)$ and find the joint probability function using this conditional probability function in this particular case.
The answer is $\frac{9e^{-2}}{5} \ = 0.244$
$\begin{align*} &\quad P(X=3)\\[1ex] &={{P(X=3|Y=0)P(Y=0)} + {P(X=3|Y=1)P(Y=1)} + {P(X=3|Y=2)P(Y=2)}}\\[1ex] &=\frac{1}{5}\frac{2^0e^{-2}}{0!}+\frac{3}{5}\frac{2^1e^{-2}}{1!}+\frac{1}{5}\frac{2^2e^{-2}}{2!}\\[1ex]&=\frac{9e^{-2}}{5} \end{align*}$