Exactly One Is Not the Same as “A or B”
In the previous tutorial, Venn Diagrams for Disjoint and Overlapping Events, you used the regions A only, both, B only, and neither to organize probabilities. This tutorial focuses on a particular combination of those regions: the probability that exactly one of \(A\) and \(B\) occurs.
The phrase exactly one of \(A\) or \(B\) means that \(A\) occurs without \(B\), or \(B\) occurs without \(A\). It does not include outcomes in the overlap. By contrast, the ordinary event “\(A\) or \(B\)” includes A only, B only, and both. Reading the wording carefully determines whether the overlap belongs in the answer.
The Venn-diagram formula for the two exclusive regions is already familiar from the previous tutorial. A only has probability \(P(A)-P(A\cap B)\), and B only has probability \(P(B)-P(A\cap B)\). These regions cannot occur together on one outcome, so to find the probability of A only or B only, add their probabilities.
Building the Exactly-One Formula
Adding the two exclusive-region probabilities gives a useful formula. Notice that the intersection is subtracted from each event probability: once to remove the overlap from \(A\), and again to remove it from \(B\). That is why the intersection is multiplied by 2.
This formula applies whether \(A\) and \(B\) are disjoint or overlapping. If they are disjoint, \(P(A\cap B)=0\), so the probability of exactly one is simply \(P(A)+P(B)\). When they overlap, the shared outcomes must be excluded because those outcomes have both events, not exactly one.
There is another way to see the same calculation. The general addition rule, from The General Addition Rule, finds the probability of \(A\) or \(B\), including both: \(P(A)+P(B)-P(A\cap B)\). To change that inclusive-or probability into an exactly-one probability, remove the overlap that was included once. The result subtracts the intersection twice overall: once to correct the double count in the addition, and once more to exclude the both region.
Worked Example: Exactly One of Two App Features
Consider an invented model for users of a reading app. Let \(A\) mean that a randomly selected user turns on a night-reading setting, and let \(B\) mean that the user turns on a text-to-speech setting. Suppose \(P(A)=0.46\), \(P(B)=0.35\), and \(P(A\cap B)=0.12\). Find the probability that a user turns on exactly one of the two settings.
State: We want A only or B only. Users who turn on both settings must not be counted as having exactly one.
Plan: Use the given intersection to find each exclusive region. Add those two probabilities. As a check, use \(P(A)+P(B)-2P(A\cap B)\).
Do: The probability of A only is:
The probability of B only is:
The two exclusive regions do not overlap, so add them:
Check with the combined formula:
The region probabilities are all possible: A only is 0.34, both is 0.12, and B only is 0.23. Together, these regions give \(P(A\cup B)=0.69\), which is at most 1. The neither region is \(1-0.69=0.31\), so the four regions sum to 1.
Conclude: In this invented app-user model, the probability that a randomly selected user turns on exactly one of the two settings is 0.57, or 57%.
When the Overlap Is Not Given Directly
Sometimes a problem gives \(P(A)\), \(P(B)\), and the probability of at least one event, \(P(A\cup B)\), but leaves out the overlap. The earlier tutorial Finding \(P(A\text{ and }B)\) from the Addition Rule showed how to recover the intersection from those three values. Once you know the intersection, the exactly-one formula can be used.
An alternative is to notice that “exactly one” consists of the union without the both region. Thus, subtract the intersection from the union. Either approach works, as long as the intersection is removed so outcomes in both events are not part of the answer.
Worked Example: Use the Union to Find Exactly One
In an invented model of participants at a community workshop, let \(A\) mean that a randomly selected participant attends the repair session, and let \(B\) mean that the participant attends the reuse session. Suppose \(P(A)=0.64\), \(P(B)=0.52\), and \(P(A\cup B)=0.88\). Find the probability that a participant attends exactly one session.
State: The union includes participants who attend either session or both. We need to exclude those who attend both.
Plan: First use the general addition rule rearranged to find the intersection. Then substitute the event probabilities and the intersection into the exactly-one formula.
Do: Recover the probability of attending both sessions:
Now use the exactly-one formula:
Check by finding each exclusive region. A only is \(0.64-0.28=0.36\), and B only is \(0.52-0.28=0.24\). Their sum is \(0.36+0.24=0.60\). Another check is to subtract the both region from the union: \(0.88-0.28=0.60\).
Conclude: In this invented workshop model, 60% of participants attend exactly one of the repair and reuse sessions.
Using Counts in a Two-Way Table
When information is presented as counts, the same reasoning still applies. First identify the counts in the A-only and B-only categories. Their sum is the count of outcomes with exactly one event. To turn that count into a probability for a randomly selected individual from the full group, divide by the grand total, as in Using Two-Way Tables to Find Probabilities.
A table can make the distinction between “exactly one” and “at least one” especially clear. The exactly-one count is the sum of A only and B only. The at-least-one count also includes both.
Worked Example: Exactly One Activity from a Two-Way Table
Suppose an invented questionnaire is completed by 120 members of a recreation center. Let \(A\) mean that a member uses the indoor climbing wall and \(B\) mean that the member uses the fitness studio. The recorded counts are shown below.
| B: Uses studio | Not B: Does not use studio | Total | |
|---|---|---|---|
| A: Uses climbing wall | 18 | 27 | 45 |
| Not A: Does not use climbing wall | 21 | 54 | 75 |
| Total | 39 | 81 | 120 |
Find the probability that a randomly selected member uses exactly one of the two facilities.
State: Exactly one means climbing wall only or studio only. Members who use both facilities do not count toward this event.
Plan: Add the A-only count and B-only count, then divide by the grand total of 120. As a second method, find \(P(A)\), \(P(B)\), and \(P(A\cap B)\), and apply the formula.
Do: The two exclusive counts are 27 and 21. Therefore:
Check using probabilities from the table. The climbing-wall probability is \(45/120=0.375\), the studio probability is \(39/120=0.325\), and the both probability is \(18/120=0.150\). Substituting gives:
The approaches agree. The union includes \(27+18+21=66\) members, so the “at least one” probability is \(66/120=0.55\). Removing the both group gives \((66-18)/120=48/120=0.40\), another check.
Conclude: For a random selection from these 120 invented responses, the probability that the member uses exactly one of the two facilities is 0.40, or 40%.
Common Mistakes and AP Exam Tips
- Including the overlap in an exactly-one answer. The union \(A\cup B\) includes outcomes in both events, but exactly one does not. Subtract the both region from the union, or subtract the intersection from each event before adding.
- Subtracting the intersection only once in the formula. \(P(A)+P(B)-P(A\cap B)\) is the probability of the inclusive “A or B.” For exactly one, the intersection must be subtracted twice: \(P(A)+P(B)-2P(A\cap B)\).
- Adding event probabilities when the events overlap. The sum \(P(A)+P(B)\) counts outcomes in both twice. Use the exclusive-region calculations or the formula, and check whether the events are actually disjoint.
- Assuming exactly one means either event, including both. In ordinary probability wording, “A or B” usually means at least one, so it includes both. Look for the word exactly and explicitly identify which regions qualify.
- Forgetting to use a common probability model or denominator. The probabilities must describe the same chance process and the same definitions of \(A\) and \(B\). With counts, divide the exclusive total by the appropriate grand total for the stated random selection.
- Giving only a number without interpreting it. For a full-credit explanation, identify A only and B only, show how their probabilities are combined, and state what the result means in context. If useful, verify that the exclusive-region probabilities are nonnegative and their sum does not exceed 1.
One final check is to compare exactly one with at least one. Exactly one cannot have a larger probability than at least one, because every exactly-one outcome is included among the at-least-one outcomes. The difference between those probabilities is precisely \(P(A\cap B)\).
Check Your Understanding
For each question, identify the relevant regions, show the calculation, and interpret the result when a context is provided.
- Suppose \(P(A)=0.58\), \(P(B)=0.43\), and \(P(A\cap B)=0.16\). Find the probability that exactly one event occurs.
- Events \(C\) and \(D\) are disjoint, with \(P(C)=0.19\) and \(P(D)=0.34\). Find the probability that exactly one occurs. Explain why the calculation is simpler here.
- Given \(P(A)=0.70\), \(P(B)=0.48\), and \(P(A\cup B)=0.90\), find \(P(A\cap B)\), then find the probability that exactly one event occurs.
- A two-way table has 35 outcomes in A only, 12 in both, 28 in B only, and 45 in neither. For a random selection from all the outcomes, find the probability of exactly one event.
- Explain the difference between \(P(A\cup B)\) and the probability that exactly one of \(A\) and \(B\) occurs. Which Venn-diagram region is treated differently?