In $P(E_1, E_2)$, the comma is read as an and.

However, consider a random variable with $\Omega=$ {1,2,3,4,5,6} and $E_1$={1,2,3}, $E_2=${3,4,5,6}

Am I right in assuming that a comma in the event and sample space definitions means xor?

Therefore, can you conclude that $P(E_1)+P(E_2)=1$?

EDIT: Would $P(E_1)=P(1,2,3)$ would mean 'probability of 1 xor 2 xor 3' or 'probability of 1 and 2 and 3'. The latter would obviously be 0 since mutually exclusive events can't be realized simultaneously.

  • $\begingroup$ If the events are pairwise disjoint, then xor (i.e. symmetric difference of sets) is the same as or (union). $\endgroup$
    – Berci
    Oct 18, 2012 at 10:04
  • $\begingroup$ Anyway, as $3\notin E_1\mathrm{xor} E_2$, $P(E_1\mathrm{xor} E_2)=5/6$. $\endgroup$
    – Berci
    Oct 18, 2012 at 10:08

2 Answers 2


In usual mathematical notation, commas do not have any specific fixed meaning — they just serve to separate things in a list of things.

Events are, formally, subsets of the sample space $\Omega$. While there are several ways of defining events, in your example the events $E_1$ and $E_2$ are literally written as sets by enumerating their members (i.e. the primitive outcomes that are included in the events) inside a pair of braces.

In this enumerative set notation, the commas simply separate the members of the set, so that you can e.g. tell the difference between the sets $\{1,2\}$ and $\{12\}$.

Now, since a set is the union of the singleton sets (i.e. the elementary events) containing each of its members, you could certainly interpret the set $\{1,2\}$ as $\{1\} \cup \{2\}$. And since the members of a sets are always distinct, and their singleton sets thus always disjoint, you could also replace the union $\{1\} \cup \{2\}$ with the symmetric set difference $\{1\} \mathbin\triangle \{2\}$ (which is just the set-theory version of XOR, just like $\cup$ is the set-theory version of OR). But that's really overthinking it, in my opinion — fundamentally, $\{1,2\}$ is just a set containing a list of members, and that's how you should think about it.

As for notation like $P(E_1, E_2)$, what's inside the parentheses is just a list of events, separated by commas to make it clear that we're not talking about some single event named $E_1E_2$ or whatever. With events represented by single-letter variables, that's arguably not really necessary, but it becomes useful when we have more complicated events like $X = Y$ or $A \setminus B$.

Now, by common convention, a list of events is interpreted as the intersection (AND) of those events, such that $P(E_1,E_2) = P(E_1 \cap E_2)$ or, using logical connectives instead of set-theory ones, $P(E_1 \land E_2)$. However, that convention is by no means universal, so if you want to be sure to avoid ambiguity, you should explicitly use $\cap$ (or $\land$) to denote the intersection of events.

However, I do personally confess to often being lazy and just using comma-separated lists of events, despite the risk of confusing people unfamiliar with the convention, simply because a comma is quicker to type and takes less space than $\cap$. (Besides, not everyone is familiar with set operators or logical connective symbols either, so when I do explicitly include the connectives while writing to a general audience, I often spell them out as in $P(E_1 \text{ and } E_2)$.)


No. As you said, comma is read as an and, hence it corresponds to the intersection of sets. In your example $E_1\cap E_2=\{3\}$.

Let's say you're throwing a dice at a world time (state) $t_0$. Then you can view the probability space $\Omega$ as the space of possible continuations of $t_0$, so we can call elements $\omega\in\Omega$ as possible worlds or possible continuation.

Let's assume, that a probability measure $P$ is given on this (whatever) space $\Omega$. That is, $P$ assigns values in $[0,1]$ to some subsets of $\Omega$, called events (these are the measurable subsets, wrt. $P$), and $P$ acts like area ('measure'), i.e. $\sigma$-additive.

In the given example of dice, we can introduce a random variable $X$, that for a given continuation $\omega$ assigns the number $\in\{1,2,3,4,5,6\}$ on the top face.

If we assume that each value has the same probability and the probability of those possible continuations which cannot assign any value (e.g. the dice is eaten by an eagle:) is Zero, then $P(X=n)=\displaystyle\frac16$ for all $n\in\{1,2,3,4,5,6\}$.

In your notation, $\Omega$ is identified with the range of the random variable, which can always be done, and then you can write it also as $P(\{3\})=1/6$.

For the other question, $P(E_1)=3/6$ and $P(E_2)=4/6$ so $P(E_1)+P(E_2)=7/6$.

Anyway, it is always true that $$P(E_1)+P(E_2) = P(E_1\cap E_2) + P(E_1\cup E_2).$$

  • $\begingroup$ But then, wouldn't $P(E_1)$=0, since the outcomes are mutually exclusive and it asks for all of them happening? $\endgroup$ Oct 18, 2012 at 9:52
  • $\begingroup$ $P(E_1)=3/6$ but it can depend on $P$. $\endgroup$
    – Berci
    Oct 18, 2012 at 9:53
  • $\begingroup$ Assuming the die is biased, can you still conclude P(E1)+P(E2)=1? $\endgroup$ Oct 18, 2012 at 9:57
  • 1
    $\begingroup$ No. $7/6$. Continued the answer in a possible way.. $\endgroup$
    – Berci
    Oct 18, 2012 at 10:05
  • $\begingroup$ Upvoted your post, but I selected the one below. Thanks for all your information. $\endgroup$ Oct 19, 2012 at 4:49

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