Computing probabilities involving increasing event outcomes Suppose I have an unfair die with three outcomes: $A,B,C$
\begin{align*}
A &\sim 1/512 \\
B &\sim 211/512 \\
C &\sim 300/512 \\
\end{align*}
The case of one dice is trivial. However, if I roll two of these dice, what is the probability of at least one of the dice having outcome $A$? 
I do not know how to tackle this problem since the outcomes in the sample space do not all have the same probability of occurring.
Thanks!

There are 4 ways your two dice can end up:
1.) First is $A$ and second is $A$, 
2.) First is not $A$ and second is $A$,
3.) First is $A$ and second is not $A$,
4.) First is not $A$ and second is not $A$.
Each die has $P(A)=1/512$ and $P(not A)=511/512$ so
1.) First is $A$ and second is $A$: $P(A)\cdot P(A)=(1/512)(1/512)=\frac{1}{512^2}$.
2.) First is not $A$ and second is $A$: $P(not A)\cdot P(A)=(511/512)(1/512)=\frac{511}{512^2}$.
3.) First is $A$ and second is not $A$: $P(A)\cdot P(notA)=(1/512)(511/512)=\frac{511}{512^2}$.
4.) First is not $A$ and second is not $A$: $P(not A)\cdot P(not A)=(511/512)(511/512)=\frac{511^2}{512^2}$.
The probability of at least one outcome being $A$ is the sum of the probabilities that satisfy the description. Thus:
$$P(\text{at least one }A)=\frac{511}{512^2}+\frac{511}{512^2}+\frac{511^2}{512^2}$$
Or equivalent it is the complement of everything that doesn't satisfy your description:
$$P(\text{at least one }A)=1-P(\text{no }A)=1-\frac{1}{512^2}$$
