The Region Outside Both Circles
In Drawing a Venn Diagram of Two Events, you placed the overlap \(A\cap B\) in the shared part of the circles and used the circles together to represent the union \(A\cup B\). There is one more region in the Venn diagram: the part of the sample space outside both circles. That region represents outcomes for which neither event occurs.
Be careful to distinguish “not \(A\)” from “neither \(A\) nor \(B\).” The complement \(A^c\) includes every outcome outside circle \(A\), including outcomes that are in \(B\) only. “Neither” is more specific: the outcome must be outside circle \(A\) and outside circle \(B\).
The complement rule from The Complement Rule makes the calculation direct. The union contains all outcomes in at least one of the two events. Everything not in that union is in neither event. Therefore, subtract the probability of the union from 1.
If you know the event probabilities and their overlap, the union formula from the previous tutorial gives another way to calculate the answer. First find the probability inside the circles, then subtract that union probability from 1.
This expression also explains why the overlap matters. Adding \(P(A)\) and \(P(B)\) counts outcomes in both events twice, so subtract the overlap once to get the probability in at least one circle. Then subtract that union from 1 to get the outside region.
Fill the Other Regions, Then Find the Remainder
A Venn diagram gives a visual way to organize the calculation. If the individual event probabilities and overlap are known, place the overlap first. Then subtract it from each event probability to find A only and B only, as in the earlier tutorial. Add the three regions inside the circles to find the union. The probability left over is neither.
All four regions—A only, both, B only, and neither—partition the entire sample space. Each possible outcome belongs to exactly one of them. Consequently, their probabilities add to 1. This gives a useful check on your work as well as a way to find a missing region.
| Venn diagram region | Meaning | How to find it |
|---|---|---|
| A only | \(A\) occurs but \(B\) does not | \(P(A)-P(A\cap B)\) |
| Both | Both \(A\) and \(B\) occur | \(P(A\cap B)\) |
| B only | \(B\) occurs but \(A\) does not | \(P(B)-P(A\cap B)\) |
| Neither | Neither \(A\) nor \(B\) occurs | \(1-P(A\cup B)\) |
The outside-region calculation does not require you to know whether \(A\) and \(B\) are independent. Use the probabilities that the problem gives, along with the overlap or union if available. Do not assume a relationship between the events unless the question provides a reason to do so.
Worked Example: Use Event Probabilities and the Overlap
Worked Example: Use Event Probabilities and the Overlap
Suppose a community program tracks whether a randomly selected participant attended a garden workshop and whether the participant attended a food-waste workshop. Let \(A\) mean the participant attended the garden workshop, and let \(B\) mean the participant attended the food-waste workshop. Suppose \(P(A)=0.41\), \(P(B)=0.35\), and \(P(A\cap B)=0.16\). Find the probability that the participant attended neither workshop.
Plan: Find the union probability, which is the probability the participant attended at least one workshop. Then use the complement rule to find the probability of attending neither.
Find the union: Add the two event probabilities and subtract their overlap, which would otherwise be counted twice.
Find neither: The event “neither workshop” is the complement of attending at least one workshop.
Check with the four regions: A only has probability \(0.41-0.16=0.25\), the overlap has probability \(0.16\), and B only has probability \(0.35-0.16=0.19\). The inside regions total \(0.25+0.16+0.19=0.60\), leaving \(1-0.60=0.40\) outside both circles.
Conclude: The probability that a randomly selected participant attended neither workshop is \(0.40\), or 40%. This is the model probability for the outside-both region, not the probability of attending neither workshop among only those who attended one or more.
Use Counts to Find the Outside Region
When a problem gives counts for a group, the same region logic applies. You can find the number in neither by subtracting the number in the union from the total. Then divide by the total to express the result as a probability for a randomly selected person from that group.
A useful order is to find the number in the overlap, then the A-only and B-only counts, and then the number in the union. Alternatively, add the A and B counts and subtract the overlap to get the union count directly. Either method must count each person in the union exactly once.
Worked Example: Find Neither from Survey Counts
In an invented survey of 240 residents, 86 said they compost food scraps, 102 said they grow at least some of their own vegetables, and 38 did both. Let \(A\) be the event that a randomly selected resident composts food scraps, and let \(B\) be the event that the resident grows vegetables. Find the probability that a randomly selected resident did neither.
Find the union count: Add the two counts and subtract the 38 residents counted in both groups. This gives the number who compost, grow vegetables, or do both.
Find the neither count: There are 240 residents altogether, so the number outside both circles is the total minus the union count.
Convert to a probability: Each of the 240 residents is represented once in the full group, so divide the neither count by 240.
Check with the regions: The A-only count is \(86-38=48\), and the B-only count is \(102-38=64\). Adding A only, both, and B only gives \(48+38+64=150\), matching the union count. The four counts total \(48+38+64+90=240\).
Conclude: For a resident selected at random from the 240 surveyed residents, the probability of composting neither food scraps nor growing vegetables is \(0.375\), or 37.5%.
When the Union Is Already Known
Sometimes the problem gives the probability of \(A\cup B\) directly. In that case, there is no need to reconstruct the regions inside the circles. Subtract the given union probability from 1. If you want to show a complete diagram, you can still use the event probabilities and overlap to check the inside regions.
Worked Example: Complement a Given Union
A randomly selected device in a fictional quality-control batch may have a loose connector, event \(A\), or a scratched case, event \(B\). Suppose \(P(A)=0.44\), \(P(B)=0.39\), and \(P(A\cap B)=0.18\). Find the probability that a device has neither issue.
Find the union: The probability of at least one issue is the probability of a loose connector or a scratched case, including devices with both issues. Use the overlap formula.
Find the complement: The probability of neither issue is the probability outside both circles.
Check the diagram: A only is \(0.44-0.18=0.26\); both is \(0.18\); and B only is \(0.39-0.18=0.21\). The inside regions total \(0.26+0.18+0.21=0.65\), so the outside region is \(0.35\). All four regions total \(0.26+0.18+0.21+0.35=1\).
Conclude: The probability that a randomly selected device has neither a loose connector nor a scratched case is \(0.35\), or 35%.
Check the Meaning and the Arithmetic
The phrase “neither \(A\) nor \(B\)” identifies the outcomes outside both circles. It does not mean only that \(A\) fails to occur: some outcomes in \(A^c\) may still belong to \(B\). For example, the B-only region is part of \(A^c\), but it is not part of the neither region.
Before accepting a result, check that the probabilities refer to the same sample space and that the four regions fit together. A negative probability for neither, or a probability greater than 1, signals inconsistent inputs or an arithmetic mistake. For valid region probabilities, the sum of the three inside regions cannot exceed 1.
Common Mistakes and AP Exam Tips
- Using \(1-P(A)\) to find neither. That gives \(P(A^c)\), which includes B-only outcomes. For neither, use \(1-P(A\cup B)\).
- Subtracting only one event probability from 1. Neither means outside both circles, so first find the probability in at least one circle—the union.
- Adding \(P(A)\) and \(P(B)\) without subtracting the overlap. Outcomes in both events would be counted twice. Use \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\), or add the three distinct inside regions.
- Mixing counts and probabilities. If using counts, keep the calculations in counts until finding the neither count, then divide by the total. If using probabilities, use probabilities throughout.
- Giving a number without saying what it represents. State that the result is the probability that a randomly selected individual or outcome belongs to neither event, in the context given.
For a clear AP response, identify the union as “at least one of the two events,” show the complement calculation, and interpret the answer in context. If the problem provides the event probabilities and overlap, showing how you found the union makes the reasoning easy to follow. A labelled Venn diagram can also make clear that the answer belongs outside both circles.
Check Your Understanding
Use the complement of the union or the outside region of a Venn diagram to answer each question.
- If \(P(A)=0.62\), \(P(B)=0.47\), and \(P(A\cap B)=0.25\), find \(P(\text{neither }A\text{ nor }B)\).
- In a group of 180 people, 72 are in event \(A\), 65 are in event \(B\), and 27 are in both. How many are in neither, and what is the probability of neither for a person selected at random from the group?
- If \(P(A\cup B)=0.71\), find the probability that neither event occurs.
- Explain why \(1-P(A)\) does not generally equal the probability that neither \(A\) nor \(B\) occurs.
- If A only has probability \(0.19\), both has probability \(0.12\), and B only has probability \(0.31\), find the probability of neither and check that all four regions add to 1.