Sum of Poisson and geometric distribution

Question

Suppose that $X$ has Poisson distribution with parameter $λ$ and that $Y$ has geometric distribution with parameter $p$ and is independent of $X$. Find a simple formula in terms of $λ$ and $p$ for $P(X + Y = 2)$

This is what I have so far:

Based on $$P(X + Y = n) = \sum_{k=0}^nP(X=k)P(Y=n-k) $$

so

$$P(X + Y = 2)=\sum_{k=0}^2{{λ^ke^{-λ}}\over k!}p(1-p)^{2-k-1} $$

$${e^{-λ}}p(1-p)\sum_{k=0}^2{{λ^k}\over k!}(1-p)^{-k} $$

$${e^{-λ}}p(1-p)\sum_{k=0}^2{\left({{λ}\over (1-p)}\right)^k {1\over k!}} $$

At this point I do not know how to proceed, or were my previous steps incorrect?

[The correct solution is ${e^{-λ}}p(1-p)+{e^{-λ}}λp$]

Answer 1

The permissible values for $X$ and $Y$ are limited, so the sum has very few terms: $$\Pr[X + Y = 2]$$ allows only the outcomes $$(X,Y) \in \{(1,1), (0,2)\}.$$ This is for the parametrization of $Y$ that excludes $0$, namely $$\Pr[Y = y] = p(1-p)^{y-1}, \quad y \in \{1, 2, \ldots\}.$$ So it follows that $$\begin{align*} \Pr[X + Y = 2] &= \Pr[(X = 1) \cap (Y = 1)] + \Pr[(X = 0) \cap (Y = 2)] \\ &\overset{\text{ind}}{=} \Pr[X = 1]\Pr[Y = 1] + \Pr[X = 0]\Pr[Y = 2] \\ &= e^{-\lambda}\frac{\lambda^1}{1!} \cdot p(1-p)^{1-1} + e^{-\lambda}\frac{\lambda^0}{0!} \cdot p(1-p)^{2-1} \\ &= \lambda e^{-\lambda} p + e^{-\lambda} p(1-p), \end{align*}$$ as claimed.

Answer 2

Just expand your sum, when $k=0$ the summand is $1$ so you get $e^{-\lambda}p(1-p)$, when $k=1$, you get $e^{-\lambda}p \lambda$ and when $k=2$ you have $$e^{-\lambda}\frac{p}{2(1-p)}\lambda^2$$ and now add and simplify...