From Words to Event Operations
Events are sets of outcomes, so ordinary words about whether conditions hold can be translated into set operations. “Not” points to the outcomes outside an event, “or” to a union, and “and” to an intersection. Making those translations precisely matters: in probability, a small change in wording can describe a different event.
Let \(\Omega\) be a sample space and let \(A\) and \(B\) be events, meaning subsets of \(\Omega\). An outcome belongs to an event when it satisfies the condition that event represents. The operations below are defined by asking, for each outcome in \(\Omega\), whether it belongs to the resulting set.
The word “or” in \(A\cup B\) is inclusive: an outcome satisfying both conditions is included. If a statement intends that exactly one of two conditions holds, it must say so; that event excludes the overlap \(A\cap B\). In contrast, “and” requires both conditions at once.
Worked Examples With Finite Sample Spaces
Worked Example: One Roll of a Die
Roll a standard six-sided die once, with \(\Omega=\{1,2,3,4,5,6\}\). Let \(A\) be the event “the result is even” and \(B\) the event “the result is greater than 3.” List the complement, union, and intersection.
First write each event as a set: \(A=\{2,4,6\}\) and \(B=\{4,5,6\}\).
Not \(A\). The outcomes in \(\Omega\) that are not in \(A\) are \(1,3,5\), so \(A^c=\{1,3,5\}\).
\(A\) or \(B\). Include every outcome appearing in either set, listing an outcome only once: \(A\cup B=\{2,4,5,6\}\). Outcomes \(4\) and \(6\) satisfy both conditions, but inclusive “or” still includes them.
\(A\) and \(B\). Keep only outcomes appearing in both sets: \(A\cap B=\{4,6\}\).
The phrases “even or greater than 3” and “even and greater than 3” therefore describe different events: the first has four outcomes and the second has two.
Worked Example: Two Coin Tosses
Toss a coin twice and record the ordered outcomes, so \(\Omega=\{HH,HT,TH,TT\}\). Let \(A\) be “the first toss is heads” and \(B\) be “the second toss is heads.”
The event sets are \(A=\{HH,HT\}\) and \(B=\{HH,TH\}\). “Not \(A\)” means the first toss is not heads, giving \(A^c=\{TH,TT\}\).
For “\(A\) or \(B\),” include outcomes with heads on the first toss, the second toss, or both. Thus \(A\cup B=\{HH,HT,TH\}\). The outcome \(HH\) belongs to both events and is included because “or” is inclusive.
For “\(A\) and \(B\),” both tosses must be heads. Therefore \(A\cap B=\{HH\}\). If a question instead asks for exactly one head, the event is \(\{HT,TH\}\): it includes neither \(HH\) nor \(TT\). This is not the inclusive union \(A\cup B\).
Worked Example: Patients Meeting One or Both Criteria
A finite screening set contains five patients, represented by \(\Omega=\{p_1,p_2,p_3,p_4,p_5\}\). Let \(A\) be the event that a patient has symptom A, and \(B\) the event that a patient has symptom B. Suppose the recorded sets are \(A=\{p_1,p_2,p_4\}\) and \(B=\{p_2,p_3\}\).
The complement \(B^c\) contains the patients in the screening set who do not have symptom B: \(B^c=\{p_1,p_4,p_5\}\).
The event “symptom A or symptom B” is \(A\cup B=\{p_1,p_2,p_3,p_4\}\). Patient \(p_2\) has both symptoms and is counted once as an outcome in the union.
The event “symptom A and symptom B” is \(A\cap B=\{p_2\}\). Only \(p_2\) meets both criteria. The event “symptom A but not symptom B” is \(A\cap B^c=\{p_1,p_4\}\), because it requires membership in \(A\) and exclusion from \(B\).
Two Basic Facts and Their Proofs
The definitions immediately give useful checks on a translation. The complement reverses membership in an event: every possible outcome is either in \(A\) or outside \(A\), and none can be both. The following result states this precisely.
Proof. Take any outcome \(\omega\in\Omega\). If \(\omega\in A\), then it belongs to \(A\cup A^c\). If \(\omega\notin A\), the definition of complement gives \(\omega\in A^c\), so it again belongs to the union. Thus every outcome in \(\Omega\) belongs to \(A\cup A^c\), and both sets are subsets of \(\Omega\); hence \(A\cup A^c=\Omega\). No outcome can be both in \(A\) and not in \(A\), so \(A\cap A^c=\varnothing\). \(\square\)
A second useful fact concerns counting a union in a finite sample space. Adding the sizes of \(A\) and \(B\) counts outcomes in the overlap twice, so that overlap must be subtracted once.
Proof. Separate \(A\) into the outcomes in \(A\cap B\) and those in \(A\) but not \(B\). These two sets are disjoint and together make up \(A\), so \(|A|=|A\cap B|+|A\setminus B|\). Likewise, \(|B|=|A\cap B|+|B\setminus A|\). The union consists of the three disjoint sets \(A\cap B\), \(A\setminus B\), and \(B\setminus A\). Therefore \[ |A\cup B|=|A\cap B|+|A\setminus B|+|B\setminus A|. \] Adding the two earlier expressions for \(|A|\) and \(|B|\) counts \(|A\cap B|\) twice, while the union expression counts it once. Subtracting one copy gives \(|A\cup B|=|A|+|B|-|A\cap B|\). \(\square\)
Worked Example: Checking a Union Count
For the patient sets in the previous example, \(A=\{p_1,p_2,p_4\}\) and \(B=\{p_2,p_3\}\). We have \(|A|=3\), \(|B|=2\), and \(|A\cap B|=1\). The finite-set formula gives \[ |A\cup B|=3+2-1=4. \] Directly listing the union gives \(\{p_1,p_2,p_3,p_4\}\), which also has four outcomes. If we had simply added 3 and 2, patient \(p_2\) would have been counted twice.
Translate the Whole Condition
For a longer phrase, identify the condition attached to each event first, then combine the events in the same order as the words. For example, “A or B, but not C” means \((A\cup B)\cap C^c\): the outcome must satisfy at least one of the first two conditions and must fail the third. “A and not B” means \(A\cap B^c\). Parentheses help make the intended grouping visible.
When working with finite sample spaces, a reliable method is to list \(\Omega\), write each named event as a subset, and then apply the membership rule for the requested operation. For a complement, compare against all of \(\Omega\), not just the outcomes already listed in another event. For a union, gather outcomes from either set without duplicating them. For an intersection, retain only those appearing in both.
Common Mistakes
- Treating “or” as exclusive by default. In event notation, \(A\cup B\) includes outcomes in both events. Use an exactly-one condition only when the wording calls for it.
- Using the wrong universe for a complement. The event \(A^c\) contains outcomes of \(\Omega\) outside \(A\). If \(\Omega\) changes, the complement may change too.
- Confusing “and” with “or.” An intersection keeps outcomes satisfying both conditions; a union keeps outcomes satisfying at least one.
- Counting overlap twice. When listing a union, write each outcome once. When using counts, subtract the size of the intersection once.
Check Your Understanding
For each question, use the stated sample space and distinguish inclusive “or” from an exactly-one condition when needed.
- Let \(\Omega=\{1,2,3,4,5,6\}\), \(A=\{1,2,3\}\), and \(B=\{3,4,5\}\). Find \(A^c\), \(A\cup B\), and \(A\cap B\).
- In two coin tosses, let \(A\) mean “the first toss is tails” and \(B\) mean “the second toss is tails.” List the outcomes in \(A\cup B\) and in \(A\cap B\).
- Explain why the event “\(A\) or \(B\)” includes the overlap unless the problem explicitly asks for exactly one condition.
- If finite events \(C\) and \(D\) have \(|C|=8\), \(|D|=6\), and \(|C\cap D|=3\), how many outcomes are in \(C\cup D\)?
- Write “event \(A\) occurs, but event \(B\) does not” using complement, union, or intersection notation.