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!

The previous answers flesh out the standard take quite well. Here's another slant on these kinds of problems.
Treat the die faces as a polynomial:
$(\frac{a}{512}+\frac{211 b}{512}+\frac{300 c}{512})$
Squaring this represents rolling two (or twice), cubing it three, etc.
Squared, we get:
$(\frac{a^2}{262144}+\frac{211 a b}{131072}+\frac{75 a c}{32768}+\frac{44521 b^2}{262144}+\frac{15825 b c}{32768}+\frac{5625 c^2}{16384})$
This "encodes" the possible outcomes. The coefficients (like $\frac{211}{131072}$) are the probabilities of the different possible outcomes, and the variables and exponents encode the two faces seen and their multiplicity (e.g., $a^2$ means both were $a$).
So, adding up all the coefficients for where there is an $a$ component gives you the probability of getting at least one $a$.
Nicely, if your problem were two die (or more) with differing face probabilities, or differing number of faces, you simple multiply the polynomials representing each die as needed for the rolls, and same applies.
