From Outcomes to Events
An experiment can have several possible outcomes, but a question about it often groups those outcomes by a condition of interest. For example, after rolling a die, we might ask whether the result is even. The possible results are outcomes; the collection of results that satisfy the condition is an event.
Writing events as sets makes verbal conditions precise. It also gives a consistent way to combine conditions: “A or B,” “A and B,” and “not A” each correspond to a standard operation on sets. In this tutorial, the emphasis is on identifying which outcomes belong to each event, not on calculating their probabilities.
An event can contain one outcome, several outcomes, every outcome, or no outcomes. The event \(S\) contains all possible outcomes, while the empty set \(\varnothing\) contains none. If an experiment has a finite sample space, it is often practical to write an event by listing its outcomes. Another option is set-builder notation: for instance, \(\{x\in S: x\text{ is even}\}\) means the set of outcomes \(x\) in \(S\) that are even.
Combining Events With Set Operations
Suppose \(A\) and \(B\) are events in the same sample space \(S\). The operations below turn two events into new events, or describe the outcomes that satisfy a negated condition.
Each definition can be checked outcome by outcome. An outcome is in \(A\cup B\) if it is in at least one of the two sets. It is in \(A\cap B\) only if it is in both sets. It is in \(A^c\) if it is a possible outcome in \(S\) but is not in \(A\).
In ordinary language, “or” can sometimes mean “one or the other, but not both.” In the definition of a union, “or” is inclusive: an outcome that belongs to both events is included. If the intention is to exclude outcomes in both, that is a different condition and must be stated explicitly.
Worked Example: Two Coin Tosses
Suppose a coin is tossed twice and each outcome records the result of the first toss followed by the result of the second. The sample space is \(S=\{HH,HT,TH,TT\}\), where \(H\) means heads and \(T\) means tails. Let \(A\) be the event that the first toss is heads, and let \(B\) be the event that exactly one toss is heads. Find \(A\), \(B\), \(A\cup B\), \(A\cap B\), and \(A^c\).
To identify \(A\), select outcomes whose first letter is \(H\): \(A=\{HH,HT\}\). To identify \(B\), select outcomes with exactly one \(H\): \(B=\{HT,TH\}\).
For the union, include every outcome that appears in \(A\), in \(B\), or in both. Thus \(A\cup B=\{HH,HT,TH\}\). The outcome \(HT\) is in both events, but it is listed only once in the set.
For the intersection, keep only outcomes in both sets. The only shared outcome is \(HT\), so \(A\cap B=\{HT\}\).
For the complement, start with all of \(S\) and remove the outcomes in \(A\). Therefore \(A^c=\{TH,TT\}\). These are exactly the outcomes in which the first toss is not heads.
Translating Conditions Into Sets
A reliable translation begins by writing down the sample space and defining each event in words. Then apply the set operation that matches the logical condition. Union means at least one condition is true; intersection means both are true; complement means the condition is false for an outcome in the sample space.
Write the sample space \(S\), using outcomes detailed enough to determine whether each condition is true.
Translate each named condition into the outcomes that satisfy it, or write it as a condition on an outcome.
Use \(\cup\) for “or,” \(\cap\) for “and,” and a complement for “not,” then verify the included outcomes against the original wording.
Worked Example: Rolling a Die
A fair six-sided die is rolled once. For this example, the sample space is \(S=\{1,2,3,4,5,6\}\). Let \(A\) be the event that the result is even and \(B\) the event that the result is greater than 4. Find the sets corresponding to “\(A\) or \(B\),” “\(A\) and \(B\),” and “not \(B\).”
The even results are \(A=\{2,4,6\}\). The results greater than 4 are \(B=\{5,6\}\).
“\(A\) or \(B\)” is \(A\cup B\). Collecting the outcomes in either set gives \(A\cup B=\{2,4,5,6\}\). In particular, 6 is included because it belongs to both \(A\) and \(B\); union uses inclusive “or.”
“\(A\) and \(B\)” is \(A\cap B\). The only result that is both even and greater than 4 is 6, so \(A\cap B=\{6\}\).
“Not \(B\)” is \(B^c=S\setminus B\). Removing 5 and 6 from the sample space leaves \(B^c=\{1,2,3,4\}\). The sample space matters: the complement includes all possible die results that fail the condition, not numbers outside the die's possible outcomes.
Events Can Describe Several Features at Once
An outcome does not have to be a single number or label. It can record several features together. When outcomes have multiple parts, an event is still a subset of the sample space; the condition simply checks one or more parts of each outcome. Choosing the outcome format carefully prevents ambiguity.
Worked Example: Two Recorded Features
A clinic uses a hypothetical screening exercise in which each recorded outcome consists of a symptom status and a test result. Write an outcome as an ordered pair \((s,t)\), where \(s\) is \(Y\) or \(N\) for symptom present or absent, and \(t\) is \(+\) or \(-\) for a positive or negative test. The sample space is \(S=\{(Y,+),(Y,-),(N,+),(N,-)\}\). Let \(A\) mean symptoms are present and \(B\) mean the test result is positive. Find each event and the set for “symptoms are absent or the test is positive.”
The event \(A\) includes the outcomes with \(Y\) as their first component: \(A=\{(Y,+),(Y,-)\}\). The event \(B\) includes the outcomes with \(+\) as their second component: \(B=\{(Y,+),(N,+)\}\).
The phrase “symptoms are absent or the test is positive” is \(A^c\cup B\). Since the sample space contains only outcomes with either \(Y\) or \(N\) for symptom status, \(A^c=\{(N,+),(N,-)\}\). Combining these outcomes with those in \(B\) gives \(A^c\cup B=\{(Y,+),(N,+),(N,-)\}\).
The outcome \((Y,+)\) is included because the test is positive, even though symptoms are present. The outcome \((N,-)\) is included because symptoms are absent, even though the test is negative. Only \((Y,-)\) satisfies neither part of the “or” condition, so it is excluded.
This example concerns only how outcomes and conditions are represented. It does not assert that any test is accurate or that one outcome is more likely than another.
A Short Justification: Why the Translation Works
The set definitions are not arbitrary symbols for familiar words. They work because they classify every possible outcome according to whether each condition is true. For a union, an outcome qualifies when at least one membership test succeeds. For an intersection, both membership tests must succeed. For a complement, membership in the sample space is required, and membership in the original event is excluded.
For example, to verify that \(A\cup B\) represents “\(A\) or \(B\),” take any possible outcome \(\omega\). By the definition of union, \(\omega\in A\cup B\) exactly when \(\omega\in A\) or \(\omega\in B\). Those are precisely the outcomes for which at least one of the two conditions is true. The same outcome-by-outcome check establishes the meaning of intersection and complement:
The phrase “\(\omega\in S\)” in the complement statement is essential. Complements are always relative to a specified universe of possible outcomes. If the experiment changes, then the sample space may change, and so may the complement.
Common Mistakes and Why This Matters
- Treating “or” as “one but not both.” This reasoning would wrongly remove outcomes shared by \(A\) and \(B\). In a union, an outcome in both events is included. Use a more specific phrase if the situation requires exactly one condition to hold.
- Using “and” for an outcome satisfying only one condition. An outcome belongs to \(A\cap B\) only when it belongs to both sets. Check every part of the combined condition.
- Taking a complement without naming the sample space. “Not \(A\)” means the outcomes in \(S\) that are not in \(A\). It does not mean every imaginable outcome outside \(A\).
- Leaving outcomes too vague. If an outcome records only “positive” but the condition concerns both symptoms and a test, the sample space has not preserved enough information. Represent outcomes with all features needed to evaluate the conditions.
Representing events correctly matters because probability statements are about sets of possible outcomes. Before assigning or calculating a probability, one must know which outcomes the event contains. Writing the event as a subset makes the intended condition visible and provides a direct way to check whether an outcome belongs.
Check Your Understanding
For each question, identify the outcomes that satisfy the stated condition and express the event using set notation where appropriate.
- Two coin tosses have sample space \(S=\{HH,HT,TH,TT\}\). Let \(A\) be the event that the second toss is heads. List \(A\) and \(A^c\).
- For a die roll with \(S=\{1,2,3,4,5,6\}\), let \(A=\{2,4,6\}\) and \(B=\{4,5,6\}\). List \(A\cup B\) and \(A\cap B\).
- Explain why an outcome belonging to both \(A\) and \(B\) must be included in \(A\cup B\).
- A sample space records pairs \((\text{day},\text{weather})\). Describe what additional detail may be needed in \(S\) to decide whether an outcome belongs to the event “it rained on Tuesday.”
- In a sample space \(S=\{a,b,c,d\}\), an event is \(A=\{a,c\}\). Find \(A^c\), and state which outcomes in \(S\) it contains.