When is $1 = P(a)+P(b)+P(c)+...+P(z)$ in $S = \{a,b,c,...,z\}$ in probability? I know this is true for independent events because if the probablity of one doesn't affect the probability of the others then the sum should be 1 or 100%. How about the others?
I figure the same applies to mutually exclusive events but I am not sure.
 A: If the given set $S$ is the sample space, then it always holds, by the definition of a probability distribution. i.e., 
$$\sum_{x\in S} P(x) =1.$$
Also, if that is the sample space, the events are by definition dependent. If any of these occur, none of the others have. For example, $P(\{a\}|\{b\})=0$.
A: Assuming $S$ is the sample space (the set of possible outcomes), this is a consequence of the following axiom:

If $A$ and $B$ are mutually exclusive events, then $P(A) + P(B) = P(A\cup B)$.

That is: the probability of at least one of these events is the same as the sum of their probabilities. This axiom applies to any set of pairwise disjoint events which is finite or countably infinite.
You're slightly abusing notating by writing $P(a)$ since $a$ is an element of a sample space, where an event is a subset of the sample space - so you likely would formally write $P(\{a\})$. This aside, since $\{a\}$ and $\{b\}$ and so on are all disjoint events, you can apply the axiom to see that
$$P(\{a\})+P(\{b\})+P(\{c\})+\ldots+P(\{z\})=P(\{a,b,c\ldots,z\})=P(S)$$
and then $P(S)=1$ by definition, since $S$ is the entire sample space.
Note that there is not an analog for this in terms of independent events: summing up the probabilities of independent events doesn't result in a meaningful probability the way that summing up mutually exclusive events does and there's no way to consider events covering all the possibilities in a uniform way and still being independent. Summing probabilities of independent events doesn't have any special meaning - except possibly "expected number of events that happen" though this interpretation doesn't rely on independence.
