Suppose $X$ and $Y$ are independent Poisson-distributed random variables with parameters $1$ and $2$ respectively. Find $\mathbb{P}(\frac{X + Y}{ 2} \geq 1)$.
The way I attempted to solve this problem is by saying that $$ \mathbb{P}\Big(\frac{X + Y}{2} \geq 1\Big) = \mathbb{P}\Big(\frac{X}{2} + \frac{Y}{2}\geq 1\Big) = 1 - \mathbb{P}\Big(\frac{X}{2} + \frac{Y}{2} = 0\Big)$$ but this does not seem to be the correct answer. I am wondering where I went wrong.
The event $\{\frac{X+Y}{2}=0\}$ is the same as the event $\{X=0,Y=0\}$ since $X$ and $Y$ are non-negative. This means that you're missing the events $\{X=1,Y=0\}$ and $\{X=0,Y=1\}$ in your final expression, because in these cases $\frac{X+Y}{2}=\frac{1}{2}<1$.