mutually exclusive definition I know that if two events are mutually exclusive it means that at most one is true. 
However, what does this mean in terms of probability theory? Is it that for a set of mutually exclusive possibilities, their corresponding probabilities always sum to 1? Or does only one event have probability of 1?
I am studying mutually exclusive events versus collectively exhaustive, and I'm not sure if I'm confusing them.
 A: Events $A_1, A_2, A_3, \ldots$ are said to be mutually exclusive if the intersection
of any pair of distinct events is empty, that is,
$$(A_i \cap A_j) = \emptyset ~~\mathrm{for~all~} i\neq j.$$
Since the empty set has probability $0$, this implies that $P(A_i \cap A_j) = 0$.
The third axiom of probability then tells us that
$$P(A_1 \cup A_2\cup \cdots) = P(A_1) + P(A_2) + \cdots$$
and since $A_1 \cup A_2\cup \cdots \subset \Omega$, we have that the
probability of the union cannot exceed $P(\Omega)=1$.  Thus,
$$P(A_1) + P(A_2) + \cdots \leq 1 ~\mathrm{for~mutually~exclusive~events~}
A_1, A_2, A_3, \ldots$$
On the other hand, the collection of events
$\{A_1, A_2, A_3, \ldots\}$ is said to be collectively exhaustive 
if 
$$A_1 \cup A_2\cup \cdots = \Omega,$$ that is,
their union is the entire sample space.  Neither of these properties
implies the other.  When a collection of events has both properties,
it is said to be a partition of the sample space: we have
partitioned (meaning divided up) the entire sample space
into mutually exclusive events and so every outcome $\omega \in \Omega$
is a member of exactly one event in the partition.
Example: If $\Omega = \{1,2,3,4\}$, then 


*

*$A_1 = \{1,2\}$ and $A_2=\{3\}$ are mutually exclusive but not collectively exhaustive.

*$B_1 = \{1,2,3\}$ and $B_2 = \{3,4\}$ are collectively exhaustive
but not mutually exclusive.

*$\{A_1, B_2\}$ is a collection of
mutually exclusive and collectively exhaustive events, and is thus a partition.
A: By definition mutually exclusive events are those satisfying $A\cap B = \varnothing$. This means these events can't occur together, that is, probability of both of them happening at the same time is 0, so $P(A \cap B) = 0$.
The sum of their probabilities can be anything less than or equal to 1. If there are no other elements in the sample set $S$ then $A \cup B = S$ and $$P(S) = P(A) + P(B) - P(A \cap B) = P(A) + P(B) = 1$$
