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Probability Models and Laws · Tutorial 4 of 994

Events as Sets of Outcomes

You will learn to describe an event precisely by listing the outcomes that satisfy its defining condition.

Beginner 9 min read

What You'll Learn

  • Define an event as a subset of a sample space.
  • Distinguish an outcome from an event containing one or more outcomes.
  • Translate “exactly two successes” into a set of ordered outcomes.
  • Represent a drawing condition when draw order matters.
  • Write a sum condition for two dice as a set of outcome pairs.
  • Check whether an event description matches the chosen sample space.

From a Verbal Condition to a Set

A probability model starts with a sample space: the collection of outcomes the experiment could produce. A question about an experiment usually singles out some of those outcomes. For example, when tossing a coin three times, “exactly two heads” describes the outcomes HHT, HTH, and THH. The description refers to a collection, not to a single result. Representing that collection as a set makes the question precise.

The sample space matters because it determines what counts as one outcome. If the outcomes of three coin tosses are recorded in order, HHT and HTH are different outcomes. If the experiment recorded only the total number of heads, they would not be distinguished. A verbal event must therefore be translated using the sample space actually chosen for the experiment.

Definition: An event is a subset of the sample space \(\Omega\). An outcome \(\omega\) belongs to an event \(A\) exactly when \(\omega\in A\). Thus an event consists of all outcomes in \(\Omega\) that meet its defining condition. In a finite sample space, an event can be specified by listing its outcomes or by giving a condition that determines membership.

An outcome and an event are different kinds of objects, even when the event contains only one outcome. For instance, if \(\Omega=\{H,T\}\), then \(H\) is an outcome and \(\{H\}\) is an event. The event that a toss results in heads is the set \(\{H\}\), not the outcome \(H\) itself. Events can also contain several outcomes, or—in the finite setting—no outcomes or every outcome.

Writing Events with Conditions

A useful general form for an event is \(\{\omega\in\Omega:\text{condition on }\omega\}\). Read this as “the set of outcomes \(\omega\) in \(\Omega\) for which the stated condition holds.” The condition does not create new outcomes; it selects some of the outcomes already in the sample space.

For example, let \(\Omega=\{1,2,3,4,5,6\}\) describe one roll of a fair six-sided die. The event “the result is even” can be written \(\{2,4,6\}\), or in condition form as \(\{x\in\Omega:x\text{ is even}\}\). Both notations specify the same subset of \(\Omega\). The notation with a condition is especially useful when listing every outcome would be inconvenient.

Worked Example: Exactly Two Successes in Three Trials

Suppose three trials are recorded in order, and each trial is classified as a success, \(S\), or a failure, \(F\). The sample space is

$$ \Omega=\{SSS,SSF,SFS,SFF,FSS,FSF,FFS,FFF\}. $$

Let \(A\) be the event “exactly two successes.” Check each outcome by counting its \(S\) symbols. \(SSF\), \(SFS\), and \(FSS\) each contain two successes. The outcomes \(SSS\) and \(FFF\) contain three and zero successes, respectively; the other three outcomes contain one success each. Therefore,

$$ A=\{SSF,SFS,FSS\}. $$

The order of the trials is why there are three outcomes in \(A\): the failure can occur in the first, second, or third position. “Exactly two successes” is the event, while \(SSF\) is one outcome in that event.

Events Depend on the Chosen Outcomes

The same physical experiment can be described by different sample spaces. The event must be translated into the language of the chosen space. If a sample space records only a summary, then outcomes that differ in unrecorded details are represented by the same summary. That may be adequate for one question but not another.

Worked Example: One Red and One Blue Ball

A bag contains two distinct red balls, \(R_1\) and \(R_2\), and one blue ball, \(B_1\). Two balls are drawn without replacement, and the order of drawing is recorded. The outcomes are ordered pairs of distinct balls. For example, \((R_1,B_1)\) means that \(R_1\) was drawn first and \(B_1\) second.

Let \(A\) be the event “one red ball and one blue ball are drawn.” The first ball can be red and the second blue, giving \((R_1,B_1)\) and \((R_2,B_1)\). Or the first ball can be blue and the second red, giving \((B_1,R_1)\) and \((B_1,R_2)\). Hence

$$ A=\{(R_1,B_1),(R_2,B_1),(B_1,R_1),(B_1,R_2)\}. $$

Each listed pair has distinct balls, as required by drawing without replacement, and each contains one red and one blue ball. By contrast, “two red balls” describes a different event; it is not the event \(A\). This check against the verbal condition is an important way to catch a mismatched event description.

If the experiment instead recorded only the colors in order, its sample space could be \(\{(R,R),(R,B),(B,R)\}\), since two blue balls cannot be drawn. In that space, the same event is \(\{(R,B),(B,R)\}\). The two representations use different outcomes because the sample spaces record different information.

Translating Numerical Conditions

For experiments with numerical outcomes, an event condition often refers to a function of the recorded outcome. With two dice, for instance, the outcome is an ordered pair of face values, and the sum is calculated from that pair. The condition “sum above seven” selects the pairs whose coordinate values add to more than seven.

Worked Example: A Sum Above Seven on Two Dice

Roll two six-sided dice and record the first die followed by the second. The sample space is

$$ \Omega=\{(i,j):i,j\in\{1,2,3,4,5,6\}\}. $$

Let \(B\) be the event that the sum is greater than seven. Its condition form is

$$ B=\{(i,j)\in\Omega:i+j>7\}. $$

To list its outcomes, group them by the first coordinate. If \(i=1\), even the largest possible sum is \(1+6=7\), so there are no qualifying pairs. If \(i=2\), only \(j=6\) works. If \(i=3\), the possible second coordinates are \(5,6\); if \(i=4\), they are \(4,5,6\); if \(i=5\), they are \(3,4,5,6\); and if \(i=6\), they are \(2,3,4,5,6\). Thus

$$ B=\{(2,6),(3,5),(3,6),(4,4),(4,5),(4,6), (5,3),(5,4),(5,5),(5,6),(6,2),(6,3),(6,4),(6,5),(6,6)\}. $$

The list contains \(1+2+3+4+5=15\) ordered pairs. Each listed pair sums to at least eight, and every pair with sum greater than seven appears in the appropriate first-coordinate group. The check also shows why \((1,6)\) is excluded: its sum is seven, not greater than seven.

Two Useful Facts About Event Descriptions

Conditions on outcomes are a convenient way to define events. More generally, suppose \(T\) assigns a value \(T(\omega)\) to every outcome \(\omega\in\Omega\), and \(C\) is a set of possible values of \(T\). The outcomes for which \(T\) falls in \(C\) form the event

$$ A=\{\omega\in\Omega:T(\omega)\in C\}. $$

For a pair of dice, take \(T(i,j)=i+j\) and \(C=\{8,9,10,11,12\}\). Then this definition gives the event of a sum above seven.

Result: If \(T\) is a function defined on every outcome in \(\Omega\), and \(C\) is a subset of its value space, then \(\{\omega\in\Omega:T(\omega)\in C\}\) is an event.

Proof. By its definition, every element of \(\{\omega\in\Omega:T(\omega)\in C\}\) is an element of \(\Omega\). Therefore this set is a subset of \(\Omega\), which is exactly the definition of an event. The condition is well-defined because \(T\) assigns a value to every \(\omega\in\Omega\), so membership of that outcome can be checked. \(\square\)

A verbal description can sometimes be expressed in different ways. Two descriptions specify the same event precisely when they select the same outcomes. This provides a simple, rigorous way to verify that alternative lists or formulas agree.

Result: Let \(A\) and \(D\) be subsets of \(\Omega\). They are equal if and only if, for every \(\omega\in\Omega\), \(\omega\in A\) exactly when \(\omega\in D\).

Proof. If \(A=D\), then they contain exactly the same elements, so for every \(\omega\in\Omega\), membership in \(A\) is equivalent to membership in \(D\). Conversely, suppose membership is equivalent for every \(\omega\in\Omega\). Every element of \(A\) then belongs to \(D\), and every element of \(D\) belongs to \(A\). The sets contain precisely the same elements, so \(A=D\). \(\square\)

For example, \(\{(i,j)\in\Omega:i+j>7\}\) and the explicit list for \(B\) in the dice example describe the same event because the list includes exactly the pairs satisfying the condition. The proof rule is also useful when an event is described in words: check each candidate outcome against the condition rather than relying on how the description sounds.

Common Mistake: Treating a Description as an Outcome

A common error is to write a verbal condition as though it were itself one outcome. For example, in the three-trial experiment, “exactly two successes” is not one of the eight ordered strings in \(\Omega\). It describes three outcomes, so the event is \(\{SSF,SFS,FSS\}\). Similarly, in the ball-drawing example, “one red and one blue” is not an ordered pair of balls; it specifies which ordered pairs belong to the event.

The correction is to ask two questions in order: What is one outcome in this sample space? Then, which of those outcomes satisfy the condition? Writing the event as a subset answers the second question without confusing a property with an outcome.

A second common mistake is to ignore order after the sample space has recorded it. In the dice example, \((2,6)\) and \((6,2)\) are distinct outcomes, even though both have the same sum. A condition about the sum can include both, but it does not make them the same outcome. In the ball example, \((R_1,B_1)\) and \((B_1,R_1)\) are also distinct because the draws occurred in a different order.

Key takeaway: An event is a set of outcomes, not the verbal condition used to describe it. State the sample space first, then list or characterize exactly those outcomes that satisfy the condition.

Check Your Understanding

For each question, identify the relevant outcomes or explain the distinction being tested.

  1. A coin is tossed twice, with order recorded. Write the event “exactly one head” as a subset of the sample space.
  2. For two ordered die rolls, write the condition-form event that the first die shows a larger number than the second.
  3. In a three-trial success/failure experiment, explain why \(SFS\) is an outcome but “exactly two successes” is an event.
  4. A bag contains \(R_1,R_2,B_1\), and two balls are drawn in order without replacement. List the outcomes in the event “the first ball is blue.”
  5. State the membership check that would show two different descriptions represent the same event.