Originally Posted by
romsek we have to assume the samples are independently taken.
let $p_\theta(j)$ be the distribution shown in the problem
Doing that we have the a posteriori density as
$p_{ap}(\theta)=p_\theta(0)^2 p_\theta(1)^3 p_\theta(2) = \Large -\frac{\theta ^5}{4}-\frac{\theta ^4}{16}+\frac{5 \theta ^3}{64}+\frac{\theta ^2}{256}-\frac{\theta }{128}+\frac{1}{1024}$
we can apply the usual process to find extrema, differentiate with respect to $\theta$ and solve for that being $0$
$\dfrac{dp_{ap}}{d\theta} = \Large -\frac{5 \theta ^4}{4}-\frac{\theta ^3}{4}+\frac{15 \theta ^2}{64}+\frac{\theta }{128}-\frac{1}{128}$
with roots of $\theta = \{-\dfrac 1 2, -\dfrac 1 5, \dfrac 1 4\}$
The second derivative test must be applied to see which, if any, are maxima.
I leave it to you to show that $\theta = -\dfrac 1 5$ is the MAP estimate of $\theta$