Let $X_1$ and $X_2$ be independent Poisson variables, such be $X_1 \sim \operatorname{Poisson}(\lambda_1)$ and $X_2 ∼ \operatorname{Poisson}(\lambda_2)$.
What will be the probability distribution of $P(X_1 X_2 \leq 1)$?
My attempt:
$P(X_1 X_2 \leq 1) = P(X_1 X_2 \leq 1|X_1> 1)P(X_1>1) + P(X_1 X_2 \leq 1|X_1 \leq 1)P(X_1\leq 1) $
Is this a good direction? Maybe it's easier to show?
$P(X_1X_2\leq 1)=P(X_1=0)+P(X_1=1, X_2 = 1)+P(X_2=0)-P(X_1=X_2=0)$ since $X_1X_2 \leq 1$ only when one of the variables is $0$ or both of them are $1$. I will let you finish the computation using indepedence of $X_1$ and $X_2$.