Let $X_1$ and $X_2$ be two random independent variables with Poisson distribution $\lambda = 1$ Let $X_1$ and $X_2$ be two random independent variables with Poisson distribution $\lambda = 1$. Denoting by $Y = min\{X_1,X_2\}$, I want to calculate $P[Y \geq 1]$.
This is what I did:
$$ P[Y \geq 1] = 1 - P[Y \leq 1] $$
I now calculate $P[Y \leq 1]$. The minum between $X_1$ and $X_2$ is less or equal than one in the following cases: 
1) $X_1$ zero or one, and $X_2$ whatever.
2) $X_2$ zero or one, and $X_1$ whatever. 
I calculate 1) and then I multiply by two because they are symmetric. 
P[$X_1$ = 0 or $X_2$ = 0] = $e^{-1}*\frac{1^0}{0!}*e^{-1}\frac{1^1}{1!}$
My final answer is then $1 - 2e^{-2}$, however, it is wrong. Where did I make a mistake?

The quickest solution is $P(Y\ge 1)=P(X_1,\,X_2\ge 1)=(1-e^{-1})^2$. Note that $P(Y\ge 1)=1-P(Y=0)$, because the distribution of $Y$ is discrete.
