From “A or B” to “Neither A nor B”
In Three Mutually Exclusive Events in a Row, you added probabilities for separate categories. This tutorial combines that addition rule with the complement rule. The combination is useful when a survey question asks for the probability that a respondent chose neither of two specified options.
Let \(A\) be the event that a randomly selected respondent chooses option A, and let \(B\) be the event that the respondent chooses option B. If each respondent can choose only one option, then \(A\) and \(B\) are mutually exclusive, or disjoint. The event “A or B” is their union, \(A\cup B\). Its complement is the event that neither A nor B occurs.
The addition rule for disjoint events gives \(P(A\cup B)=P(A)+P(B)\). The complement rule then says that the probability of neither event is 1 minus the probability of their union. Together, the rules give a direct calculation.
The survey setting matters. If respondents must select exactly one answer from a list, then selecting A and selecting B are disjoint events. A respondent who selects some other answer belongs to neither event. If instead respondents may select several answers, one person might select both A and B. The events would overlap, and you would need the general addition rule from The General Addition Rule before taking the complement.
A useful way to picture the calculation is to divide the full group into three regions: A only, B only, and neither. When A and B are disjoint, there is no overlap region. The union probability accounts for A and B; subtracting it from 1 leaves the neither region.
A Reliable Calculation
Start by stating what \(A\) and \(B\) mean for one randomly selected respondent. Then check that the survey’s response rules make the events disjoint. Add \(P(A)\) and \(P(B)\) to find the probability of at least one of them, and subtract that total from 1. Finally, translate the result into the survey context.
Specify exactly what it means for the selected respondent to be in \(A\) and in \(B\).
Confirm that one respondent cannot satisfy both event definitions under the survey’s response rules.
Add \(P(A)+P(B)\) to find \(P(A\cup B)\), then calculate \(1-P(A\cup B)\).
Describe the probability of neither response in context. If counts are available, check that the number outside A and B gives the same result.
Worked Examples
Worked Example: Neither of Two Library Choices
Imagine an invented survey of 400 library visitors. Each visitor names one preferred service. Of the visitors surveyed, 132 name digital borrowing and 96 name in-person programs. Find the probability that a randomly selected visitor named neither of these services.
State: Let \(A\) be the event that the selected visitor names digital borrowing, and let \(B\) be the event that the visitor names in-person programs. We want \(P((A\cup B)^c)\), the probability of neither response.
Plan: Each visitor names one preferred service, so a visitor cannot name both as the single preferred service. Thus, \(A\) and \(B\) are disjoint. Find their probabilities from the total of 400 visitors, add them, and subtract the union probability from 1.
Do:
Check using counts. The number who named neither option is \(400-132-96=172\). The corresponding proportion is \(172/400=0.43\), agreeing with the complement calculation.
Conclude: In this invented survey, the probability that a randomly selected visitor named neither digital borrowing nor in-person programs as their preferred service is \(0.43\), or 43%.
The subtraction from 1 is not a special rule for surveys; it is the complement rule applied to the union. The addition is justified separately, by the fact that these single-choice responses cannot occur together for one visitor.
Worked Example: Neither of Two Community Priorities
In an invented community survey, each respondent selects one priority for a neighborhood improvement project. A total of 500 people respond. Of these, 170 select safer street crossings and 115 select more public seating. What is the probability that a randomly selected respondent selected neither priority?
State: Let \(A\) be the event that a respondent selects safer street crossings, and let \(B\) be the event that the respondent selects more public seating. We seek the probability of neither \(A\) nor \(B\).
Plan: Each respondent selects only one priority, so \(A\) and \(B\) are mutually exclusive. Convert the counts to probabilities using 500 respondents, add those probabilities, and take the complement.
Do:
As a count check, \(500-170-115=215\) respondents selected another priority. The proportion is \(215/500=0.43\), which matches the probability calculation.
Conclude: The probability that a randomly selected respondent chose neither safer street crossings nor more public seating is \(0.43\). In this survey, 43% selected another priority.
Worked Example: Interpreting “Not A or B” Carefully
An invented school survey asks each student to choose one preferred time for a workshop: morning, afternoon, evening, or another time. Suppose 120 of 400 students choose morning and 100 choose evening. Let \(A\) be choosing morning and \(B\) be choosing evening. Compare “neither morning nor evening” with the literal event “not morning, or evening.”
State: The intended “neither morning nor evening” event is \((A\cup B)^c\). The phrase “not morning, or evening,” without parentheses, could instead mean \(A^c\cup B\): the student does not choose morning, or chooses evening.
Plan: Because each student chooses exactly one time, \(A\) and \(B\) are disjoint. For the “neither” event, add the probabilities of morning and evening, then subtract from 1. For \(A^c\cup B\), notice that choosing evening already means not choosing morning, so \(B\) is contained in \(A^c\).
Do: First calculate the neither probability:
For the other interpretation, \(A^c\cup B=A^c\) because every student choosing evening is already in the event “not morning.” Therefore:
Conclude: The probability of choosing neither morning nor evening is \(0.45\). The probability of “not morning, or evening,” interpreted as \(A^c\cup B\), is \(0.70\). To remove ambiguity in a written response, use “neither A nor B” or write \((A\cup B)^c\) when you mean the complement of the union.
Common Mistakes and AP Exam Tips
- Taking the complement of only one event. “Neither A nor B” is the complement of \(A\cup B\), not just \(A^c\) or \(B^c\). First find the probability of A or B, then subtract that union from 1.
- Forgetting to check whether the events overlap. You can use \(P(A\cup B)=P(A)+P(B)\) only when \(A\) and \(B\) are disjoint. If respondents may choose both options, use the general addition rule and subtract the overlap once.
- Subtracting the two probabilities separately from 1 and adding the results. That would count some outcomes more than once. For disjoint A and B, the correct calculation is \(1-[P(A)+P(B)]\).
- Assuming all survey categories are automatically disjoint. A “choose one” survey question creates mutually exclusive response events. A “select all that apply” question does not necessarily do so. Use the survey rules, not just the category names, to decide.
- Leaving “or” unclear. In ordinary language, “not A or B” can be read in more than one way. A full-credit response makes the event precise. If the question means neither event, write \((A\cup B)^c\) or say “neither A nor B.”
- Reporting only a number. State what the probability represents and identify why addition is valid. For example: “Because each respondent chooses one priority, the events are disjoint; the probability of choosing neither is \(1-[P(A)+P(B)]\).”
When a survey gives counts, a count check is often a simple way to catch an error: subtract the counts in A and B from the total, then divide the remaining count by the total. For disjoint categories, this should agree with the complement calculation. If the two answers disagree, recheck the event definitions, the total, and whether any respondents were counted in both categories.
Check Your Understanding
For each question, define the events, explain whether they are disjoint, and show the probability calculation.
- In a survey where each person chooses one favorite activity, 28% choose reading and 17% choose gardening. Find the probability of choosing neither.
- In an invented survey of 250 residents, 70 select a new playground and 55 select a community garden as their single top priority. Find the probability that a randomly selected resident selected neither.
- A survey allows respondents to select all transportation methods they use. Explain why adding the proportions who use buses and bicycles may not give the probability that a respondent uses a bus or a bicycle.
- For disjoint events \(A\) and \(B\), write a formula for the probability of neither event and explain why the complement rule applies.
- Explain the difference between \((A\cup B)^c\) and \(A^c\cup B\). Which one represents “neither A nor B”?