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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?

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  • $\begingroup$ $$P(Y \geq 1) = 1 - P(Y<1)= 1-P(Y=0)$$ $\endgroup$ Jun 18 '18 at 20:41
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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?

You don't multiply by two because they are symmetric. That makes no sense. Instead use the principle of inclusion and exclusion.

$$\begin{split}\mathsf P(A\text{ or } B)~&=\mathsf P(A)+\mathsf P(B)-\mathsf P(A\text{ and } B)\\\mathsf P(X_1{=}0\text{ or }X_2{=}0) & = \tfrac {1^0}{0!}e^{-1}+\tfrac {1^0}{0!}e^{-1}-\big(\tfrac {1^0}{0!}\big)^2e^{-2}\end{split}$$

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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.

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