Let $X_1 ,X_2,X_3$ be independent Poisson random variables with mean $1$ . Then $P(\max(X_1,X_2,X_3)=1)$ equals ?
$(A)1-e^{-3}$
$(B) e^{-3}$
$(C)1-8e^{-3}$
$(D)7e^{-3}$
I usually solve these types of problem when we are given $P(\max(X_1,X_2,X_3)\leq1)$(say) then i proceed in this way if they are independent $P(X_1\leq1)P( X_2\leq1)P(X_3\leq 1)$. I have reasoning in my head behind this $P(\max(X_1,X_2,X_3)\leq1)$ is that if maximum order statistic is $\leq1$ then other two random variables are also $\leq 1$ .
Thus in this question $P(X_1\leq1)P( X_2\leq1)P(X_3\leq 1)$ changes as $P(X_1=1)P( X_2=1)P(X_3= 1)=e^{-3}$
Someone tell me if I am on right track and give me some more knowledge on this topic.
\begin{align} P(\max(X_1, X_2, X_3)=1) &= P(\max(X_1,X_2,X_3) \leq 1) - P(\max(X_1,X_2,X_3) \leq 0) \\ &=P(X_1 \leq 1)^3-P(X_1 \leq 0)^3 \\ &=\left(\exp(-\lambda) +\exp(-\lambda) \lambda \right)^3 - \exp(-\lambda)^3 \\ &=(1+\lambda)^3 \exp(-3\lambda)- \exp(-3\lambda) \\ &=7\exp(-3) \end{align}
Alternative approach:
\begin{align} P(\max(X_1,X_2, X_3=1) &= 3P((1,0,0))+3P(1,1,0)+P(1,1,1)\\ &=3\exp(-3\lambda)\lambda+3\exp(-3\lambda)\lambda^2+\exp(-3\lambda) \lambda^3 \\ &=7\exp(-3) \end{align}
Remark about your approach:
We can't just assume $(X_1, X_2, X_3)=(1,1,1)$, don't forget cases such as $(X_1, X_2, X_3)=(1,0,0)$, we just need one of them to attain value $1$ and the other can take value either $0$ or $1$.