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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.

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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.

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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 $$P(S) = P(A) + P(B) - P(A \cap B) = P(A) + P(B) = 1$$

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    $\begingroup$ The first sentence of this answer misses (or obscures) a crucial point, which is that the definition of mutual exclusivity entails $A \cap B = \varnothing$ which is a stronger statement than $\mathbb P (A \cap B ) = 0$. A minor tweak to the wording would resolve this. Cheers. $\endgroup$ – cardinal Apr 16 '13 at 12:44
  • $\begingroup$ Fair point, edited. $\endgroup$ – sashkello Apr 16 '13 at 12:50
  • $\begingroup$ How did you come across the formula P(S)=P(A)+P(B)−P(A∩B)=P(A)+P(B)=1? Thank you $\endgroup$ – user30894 Sep 30 '13 at 23:32
  • $\begingroup$ @SamiHaddad 1. See pics here en.wikipedia.org/wiki/Intersection_(set_theory). 2. Second equality is from the first paragraph of my answer. 3. Third equality is definition of S. $\endgroup$ – sashkello Oct 1 '13 at 0:14

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