MLE of poisson random variable. 
Let $x_1=x_2=x_3=1, x_4=x_5=x_6=2\ $ be a random sample from a Poisson random
  variable with mean $\theta$, where $\theta\in \{1,2\}$. Then, the maximum likelihood estimator of $\theta$ is equal to...

What I know is that, the MLE of poisson distribution is given by $$ \hat{\theta}_{MLE}=\sum_{i=1}^n\frac{X_i}{n} .$$ If we evaluate here then $\hat{\theta}_{MLE}$ is coming $1.5$, which is not there in the range of $\theta$. Then how will I find the MLE in this case?

You have to decide under which one of the two possible parameters your sample is more probable (literally). Under $\theta = 1$ you have 
$$
\prod_{i=1}^3P(X_i=1)\prod_{i=4}^6P(X_i=2)=(e^{-1}/1!)^3(e^{-1}/2!)^3.
$$
Under $\theta = 2$ you have 
$$
\prod_{i=1}^3P(X_i=1)\prod_{i=4}^6P(X_i=2)=(e^{-2}2^1/1!)^3\times (e^{-2}2^2/2!)^3
$$
so check which expression is greater.
