5.1 Sample Spaces, Outcomes, Events, and Set Operations

A sample space is a list of all possible outcomes of an experiment and is denoted \(\Omega\). Examples of sample spaces are:

  • Rolling a single die: \(\Omega = \{1,2,3,4,5,6\}\)
  • Tossing a coin: \(\Omega = \{H,T\}\)
  • Grades: \(\Omega = \{A+,A,A-, \dots, F\}\)
  • Number of calls to a fire station in a 24-hour period: \(\Omega = \{0,1,2 \dots \}\)

An event is a subset of the sample space. Examples are

  • Rolling an even number: \(E = \{2,4,6\}\)
  • Rolling a number less or equal to 4: \(S = \{1,2,3,4\}\)
  • More than 5 calls to the fire station: \(F = \{6,7,\dots \}\)

There are several set operations and we can use Venn diagrams to illustrate them:

  • Intersection: The intersection \(W\) of two sets \(X\) and \(Y\) is the set of elements that are in both \(X\) and \(Y\). We write \(W=X\cap Y\).
  • Empty or Null Sets: The empty set or the null set (\(\emptyset\)) is the set with no elements. For example, if the sets \(A\) and \(B\) contain no common elements then these two sets are said to be disjoint, e.g., odd and even numbers: \(A\cap B=\emptyset\).
  • Unions: The union of two sets \(A\) and \(B\) is the set of all elements in one or the other of the sets. We write \(C=A \cup B\).
  • Complements: The complement of a set \(X\) is the set of elements of the universal set \(U\) that are not elements of \(X\), and is written \(X^{c}\).

For a discrete sample space \(\Omega\), the probability of an event is a non-negative number, i.e., \(Pr(A) \geq 0\), for any subset \(A\) of \(\Omega\). In addition, we have \(Pr(\Omega) = 1\), i.e., all the probabilities of the outcomes in the sample space sum up to 1.

For example, if we flip a coin, the sample space is \(\Omega = \{H,T\}\). The corresponding probabilities are \(Pr(H)=0.5\) and \(Pr(T)=0.5\). The sum of both is equal to 1. Keep in mind that \(Pr(A^c)=1-Pr(A)\). It is sometimes easier to calculate the complement of an event than the event itself. If \(A,B, C,\dots\) is a finite or infinite sequence of mutually exclusive events — that is events which cannot happen at the same time — from the sample space \(\Omega\), then \[P\left(A \cup B \cup C \cup \cdots \right)=P(A)+P(B)+P(C)+\cdots\] To illustrate this, suppose you are flipping a coin three times. The eight events in \(\Omega\) are \[\Omega = \{HHH,HHT,HTH,HTT,THH,THT,TTH,TTT\}\] The probability of each event is equally likely, i.e., \[Pr\left(E_i \right)=1/8\] for \(i=1,2,3,\dots,8\). If your event of interest \(A\) is that exactly two of the three tosses results are heads, then \(A=\{HHT,HTH,THH\}\). By summing up the probabilities associated with those events, we find that \(Pr(A)\): \[Pr(A)=Pr(HHT)+Pr(HTH)+Pr(THH)=\frac{1}{8} +\frac{1}{8} +\frac{1}{8} =\frac{3}{8}\] Note that the same result can be obtained by using a probability tree. A probability tree has the advantage that it represents the possible outcome and probabilities in a graphical manner.