The Overlap Is Counted Twice
In The General Addition Rule, you learned to subtract \(P(A\cap B)\) when finding \(P(A\cup B)\). The reason is a counting issue: adding \(P(A)\) and \(P(B)\) includes outcomes in both events twice, even though each outcome in the union should contribute only once. A Venn diagram makes that extra count visible.
Picture two circles inside a rectangle representing the sample space. One circle represents event A, and the other represents event B. Where the circles overlap represents \(A\cap B\), the event that both A and B occur. The entire area inside either circle represents \(A\cup B\). The overlap is part of the union; it is not a separate area outside the events.
You can track the count by splitting the circles into three non-overlapping regions: A only, the overlap, and B only. The probability \(P(A)\) includes A only and the overlap. The probability \(P(B)\) includes B only and the overlap. Adding those two probabilities therefore includes the overlap two times. But \(P(A\cup B)\) includes A only, the overlap, and B only, each once.
The extra copy is the difference between these expressions. Subtracting the overlap once from the sum leaves the union. This is the reasoning behind the general addition rule from the previous tutorial.
Reading the Venn Diagram by Regions
A useful way to check the logic is to label each part of the diagram and verify which parts belong to each probability. The table below describes the regions in a two-circle Venn diagram.
| Venn-diagram region | In A? | In B? | In \(A\cup B\)? |
|---|---|---|---|
| A only | Yes | No | Yes |
| Overlap, \(A\cap B\) | Yes | Yes | Yes |
| B only | No | Yes | Yes |
| Outside both circles | No | No | No |
The table highlights the central point: the overlap appears in both individual event totals, but only once in the union. The area outside both circles does not appear in either event probability or in the union. It is not part of the correction.
The same logic works with counts as well as probabilities. If every person in a group is equally likely to be selected, convert each count to a probability by dividing by the total. Since all the counts have the same denominator, correcting the double count can be done with the counts first. The result should agree with using probabilities directly.
Worked Example: Trail Users and Garden Volunteers
Imagine a community group of 40 people. In this invented example, 22 people use a local walking trail, 18 volunteer in a community garden, and 8 do both. One person is selected at random. What is the probability that the person uses the trail or volunteers in the garden?
Worked Example: Correcting the Double Count in a Venn Diagram
Define the events: Let \(T\) be the event that the selected person uses the trail, and let \(G\) be the event that the person volunteers in the garden. We want \(P(T\cup G)\). Since 8 people do both, the circles overlap.
Map the Venn regions: The 8 people who do both go in the shared part of the circles. The trail-only region has \(22-8=14\) people, and the garden-only region has \(18-8=10\). These three regions make up the union.
| Venn region | Number of people |
|---|---|
| Trail only | 14 |
| Both trail and garden | 8 |
| Garden only | 10 |
| Neither | 8 |
See the double count: Adding the trail and garden counts gives \(22+18=40\). But there are only \(14+8+10=32\) people in the union. The difference is 8 because each of the 8 people in both groups was included once in the trail count and once in the garden count. To correct the total, subtract one copy of those 8 people:
Calculate and check: Divide the union count by the 40 people in the group. Equivalently, use the general addition rule with all three probabilities written over 40.
A check using the Venn regions gives \((14+8+10)/40=32/40=0.80\). Both methods count the overlap once.
Conclude: The probability that a randomly selected person in this community group uses the trail or volunteers in the garden is \(0.80\), or 80%.
Notice that the 8 people who do both remain among the 32 people in the union. The subtraction only removes their extra appearance in the sum \(22+18\). It does not remove them from the group of people who meet at least one event condition.
Worked Example: Two Features on a Phone
Suppose 100 people in an invented technology survey are asked which phone-unlocking features they use. Of these people, 65 use face unlocking, 55 use fingerprint unlocking, and 40 use both features. A person is selected at random from the 100. Find the probability that the person uses face unlocking or fingerprint unlocking.
Worked Example: Face or Fingerprint Unlocking
State: Let \(F\) represent using face unlocking and \(R\) represent using fingerprint unlocking. The desired event is \(F\cup R\). Because 40 people use both, adding the two feature counts without a correction would count those people twice.
Plan: Use the general addition rule, then check it by organizing the counts into the regions of a Venn diagram. All counts come from the same group of 100 people.
Do: The face-only count is \(65-40=25\), and the fingerprint-only count is \(55-40=15\). The regions inside the circles total \(25+40+15=80\). The same total follows by subtracting one copy of the overlap from the sum of the feature counts:
Now express the counts as probabilities and apply the rule:
The region check agrees: \(25/100+40/100+15/100=0.80\). The 40 people using both features count once in the union, though they contributed to both of the original feature counts.
Conclude: The probability that a randomly selected person in this survey group uses face unlocking or fingerprint unlocking is \(0.80\), or 80%.
Worked Example: Observations with Two Features
Consider 60 nature observations recorded for an invented field activity. In 21 observations, a bird nest is present; in 26, a flowering plant is present; and in 9, both are present. If one observation is selected at random, what is the probability that it includes a bird nest or a flowering plant?
Worked Example: Bird Nest or Flowering Plant
Define the events: Let \(N\) be the event that an observation includes a bird nest, and let \(P\) be the event that it includes a flowering plant. We are finding \(P(N\cup P)\). An observation can have both features, so the overlap matters.
Arrange the regions: The overlap is 9 observations. The nest-only region contains \(21-9=12\) observations, and the plant-only region contains \(26-9=17\). Together, the regions inside the circles contain \(12+9+17=38\) observations.
Use the rule and verify: Adding the two original counts gives \(21+26=47\), which counts each of the 9 observations in both twice. Subtracting 9 leaves 38 observations in the union.
The region check gives \((12+9+17)/60=38/60\), the same probability. The remaining \(60-38=22\) observations are outside both circles and do not belong to the event “nest or plant.”
Conclude: The probability that a randomly selected observation includes a bird nest or a flowering plant is \(38/60\), approximately \(0.6333\), or about 63.33%.
Common Mistakes and AP Exam Tips
- Adding the two event probabilities and stopping. If events overlap, the sum includes the shared outcomes twice. A full-credit calculation identifies the overlap and subtracts it once.
- Subtracting the overlap twice. The sum contains two copies of the overlap, but the union should contain one. Removing one copy is enough; removing both would incorrectly exclude outcomes that satisfy both events.
- Thinking “subtract” means the overlap is not in the union. “A or B” includes outcomes in A, in B, or in both. The subtraction corrects the arithmetic, not the definition of the event.
- Subtracting the wrong region. Use \(P(A\cap B)\), the probability of both events. Do not subtract the probability of A only, B only, or neither.
- Mixing counts and probabilities carelessly. You can subtract counts when using counts, or subtract probabilities when using probabilities. Do not subtract a count from a probability, and keep the denominator consistent when converting counts.
- Forgetting a check. When the region counts are available, add A only, overlap, and B only. That total should match the union found with the rule.
For a clear AP-style explanation, name the events, state that they overlap, show the sum and the overlap subtraction, and interpret the result in context. For example: “The 8 people who use both the trail and garden are counted in each event total. Subtracting one copy gives \(22+18-8=32\) people in the union. Therefore, the probability that a randomly selected person uses the trail or volunteers in the garden is \(32/40=0.80\).”
Check Your Understanding
Use Venn-diagram regions to explain the overlap correction and calculate each requested probability.
- In a group of 50 residents, 28 use a bus route, 19 use a bike route, and 7 use both. How many residents are in the union? What is the probability that a randomly selected resident uses the bus route or bike route?
- A survey of 80 people finds that 46 use a reusable bottle, 38 bring a lunch from home, and 20 do both. How many people are in the overlap, and how many are in the union?
- Explain why an outcome in \(A\cap B\) appears twice in \(P(A)+P(B)\) but only once in \(P(A\cup B)\).
- In a group of 60 people, 25 belong to event A and 30 belong to event B. If 10 belong to both, find the number in A only, B only, and the union. Check the union by both methods.
- True or false: Subtracting \(P(A\cap B)\) from the sum means that outcomes in both events are excluded from \(A\cup B\). Explain your answer.