$\mathbb{P}(X>Y)$ for $X,Y$ two Poisson

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Two teams have to play a final of a tournament. Team A score a number of goals that can be shaped like a random variable $X \sim Poi(\lambda_{A}=2.5)$. Team B score a number of goals that can be shaped like a random variable $Y \sim Poi(\lambda_{B}=2)$. The variables $X$ and $Y$ are independent.

  • 1) What is the probability that A win?

  • 2) What is the probability of a tie?

  • 3) What is the probability of an over (that is the sum of goals of two teams is greater than $4$)?


I think I've solved only the point 2).

Knowing that $H=${total number of goals}$\sim Poi(4.5)\rightarrow \mathbb{P}(H=h)=\frac{e^{-4.5}4.5^h}{h!}$ and setting $T=X-Y$, I wrote:

$$\begin{align*} \mathbb{P}(T=0)&=\sum_{s=0}^{\infty}\mathbb{P}(X=s,Y=s)\\&=\sum_{s=0}^{\infty}\mathbb{P}(X=s,Y=s,H=2s)\\&=\sum_{s=0}^{\infty}\mathbb{P}(H=2s)\mathbb{P}(X=s,Y=s|H=2s)\\&=\sum_{s=0}^{\infty}\mathbb{P}(H=2s)\mathbb{P}(X=s|H=2s)\\&=\sum_{s=0}^{\infty}\frac{e^{-4.5}4.5^{2s}}{2s!}\binom{2s}{s}p^s(1-p)^{2s-s}\\&=\sum_{s=0}^{\infty}\frac{e^{-4.5}4.5^{2s}[p(1-p)]^s}{s!s!} \end{align*}$$

Is it correct?

Instead, I find it difficult to formalize $\mathbb{P}(X>Y)$ and $\mathbb{P}(X+Y>4)$ when $X$ and $Y$ are discrete variables. Can you help me? Thanks in advance.

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For 1), given the condition you gave, you need to consider outcomes (4,0), (3,1)

For 2), again, given the condition (2,2), because you can't get a draw otherwise.

For 3), an integer $n$ can be a sum of two integers $\leq n$ in $n+1$ ways, so given independence you need the probability $$ P(X+Y=n) = \sum_{k=0}^{n}P(X=k)P(Y=n-k) $$ EDIT: for 3), also consider $X+Y=Z$, which is also Poisson with rate $\lambda_1 + \lambda_2$.