Recall that given a random experiment, then the outcome space (or sample space) \(\mathbf{S}\) is the collection of all possible outcomes of the random experiment.

- Event
- denoted with capital letters \(A, B, C\), ... — is just a subset of the sample space \(\mathbf{S}\). That is, for example \(A\subset \mathbf{S}\), where "\(\subset\)" denotes "is a subset of."

##
Example 2-3
Section* *

Suppose we randomly select a student, and ask them "how many pairs of jeans do you own?". In this case our sample space \(\mathbf{S}\) is:

\(\mathbf{S} = \{0, 1, 2, 3, ...\}\)

We could theoretically put some realistic upper limit on that sample space, but who knows what it would be? So, let's leave it as accurate as possible. Now let's define some events.

If \(A\) is the event that a randomly selected student owns no jeans:

\(A\) = student owns none = \(\{0\}\)

If \(B\) is the event that a randomly selected student owns some jeans:

\(B\) = student owns some = \(\{1, 2, 3, ...\}\)

If \(C\) is the event that a randomly selected student owns no more than five pairs of jeans:

\(C\) = student owns no more than five pairs = \(\{0, 1, 2, 3, 4, 5\}\)

And, if \(D\) is the event that a randomly selected student owns an odd number of pairs of jeans:

\(D\) = student owns an odd number = \(\{1, 3, 5, ...\}\)

##
Review
Section* *

Since events and sample spaces are just sets, let's review the algebra of sets:

- \(\emptyset\) is the "
**null set**" (or "**empty set**") - \(C\cup D\) = "
**union**" = the elements in \(C\) or \(D\) or both - \(A\cap B\) = "
**intersection**" = the elements in \(A\) and \(B\). If (A\cap B=\emptyset\), then \(A\) and \(B\) are called "**mutually exclusive events**" (or "**disjoint events**"). - \(D^\prime=D^c\)= "
**complement"**= the elements not in \(D\) - If \(E\cup F\cup G...=\mathbf{S}\), then \(E, F, G\), and so on are called "
**exhaustive events**."

##
Example 2-3 Continued
Section* *

Let's revisit the previous "how many pairs of jeans do you own?" example. That is, suppose we randomly select a student, and ask them "how many pairs of jeans do you own?". In this case our sample space *S* is:

\(\mathbf{S} = \{0, 1, 2, 3, ...\}\)

Now, let's define some composite events.

The union of events \(C\) and \(D\) is the event that a randomly selected student either owns no more than five pairs or owns an odd number. That is:

\(C\cup D=\{0, 1, 2, 3, 4, 5, 7, 9, ...\}\)

The intersection of events \(A\) and \(B\) is the event that a randomly selected student owes no pairs and owes some pairs of jeans. That is:

\(A\cap B = \{0\} \cap \{1, 2, 3, ...\}\) = the empty set \(\emptyset\)

The complement of event \(D\) is the event that a randomly selected student owes an even number of pairs of jeans. That is:

\(D^\prime= \{0, 2, 4, 6, ...\}\)

If \(E = \{0, 1\}\), \(F = \{2, 3\}\), \(G = \{4, 5\}\) and so on, so that:

\(E\cup F\cup G\cup ...=\mathbf{S}\)

then \(E, F, G\), and so on are exhaustive events.