$s = \{4, 4, 1, 5, 4 \}$ is an observed sample from a Poisson$(\lambda)$ distribution where $\lambda \in \{4,5,2 \}$ is unknown. Calculate the MLE of $\lambda$.
Attempt:
$L(\lambda \mid s) = e^{-n\lambda} \lambda^{n \bar{x}}$
$L(\lambda = 4\mid s) = e^{-5 \cdot 4} 4^{5 \cdot 3.6} \approx 141.641$
$L(\lambda = 5\mid s) = e^{-5 \cdot 5} 5^{5 \cdot 3.6} \approx 52.9783$
$L(\lambda = 2\mid s) = e^{-5 \cdot 2} 2^{5 \cdot 3.6} \approx 11.9013$
Therefore the MLE is $$\hat{\lambda} = 4$$
I can do this right? Just see which has the highest value and in this case it was $\lambda = 4?$
The likelihood is the probability you get all the data you find given a particular Poisson distribution having (unknown) $\lambda$.
This likelihood is the product of the individual probabilities you get the measured data:
$$P(4|\lambda)^3 P(1|\lambda) P(5|\lambda) = \frac{e^{-5 \lambda } \lambda ^{18}}{1658880} \propto e^{-5 \lambda } \lambda ^{18}.$$
You take the derivative of this with respect to $\lambda$, set it to zero (to find the maximum), which happens to be $3.6$... which makes eminent sense.
Here's the resulting distribution:
If you must select from the three $\lambda$s given, then compute the value of the expression above for each of the three candidate $\lambda$ values and choose the one that leads to the highest likelihood.
(Incidentally, perhaps you can see from this problem why we frequently work with the log likelihood...)