Common Errors with the Addition Rule

Why Addition-Rule Errors Happen

A union probability answers a question about whether event $A$, event $B$, or both occur. The word “or” includes the overlap. As discussed in The General Addition Rule and Why We Subtract the Overlap, adding $P(A)$ and $P(B)$ counts outcomes in $A\cap B$ twice, so the general addition rule subtracts that intersection once.

The rule is short, but mistakes often come from not matching each number to the event it describes. A probability for “neither,” for example, is not the probability of “both.” And a probability for all of $A$ is not the probability of the overlap. Before calculating, say what each term represents in words.

Formula: For any two events $A$ and $B$, add their probabilities and subtract the probability that both occur: $$P(A\cup B)=P(A)+P(B)-P(A\cap B).$$ The quantity subtracted is specifically $P(A\cap B)$, the overlap.

If the events are mutually exclusive, there is no overlap, so $P(A\cap B)=0$. In that special case, adding the two event probabilities is correct. The earlier tutorials What Mutually Exclusive Events Mean and The Addition Rule for Mutually Exclusive Events explain how to recognize and use that case. Do not assume events are mutually exclusive just because they have different names or describe different categories.

A Quick Audit Before You Add

A useful habit is to label the events and the overlap before substituting numbers. Ask: “Can one outcome satisfy both descriptions?” If yes, the events overlap, and the simple sum $P(A)+P(B)$ counts those outcomes twice. Then check what each supplied probability actually describes.

Translate the event.
Write what $A$, $B$, and $A\cup B$ mean in context. The union is “A, B, or both.”

Identify the overlap.
Determine whether an outcome can be in both events. If so, identify $P(A\cap B)$ or find it from the information given.

Subtract only the overlap.
Use the general addition rule. Do not subtract a complement, an entire event, or “neither” unless you are using a separate, correctly identified method.

Check the result.
The union must be at least as large as either individual event probability and cannot exceed 1. If you know the overlap, the uncorrected sum should exceed the union by exactly that overlap.

These checks can catch errors even when the arithmetic itself is correct. For example, if an answer for $P(A\cup B)$ is smaller than $P(A)$, something is wrong: every outcome in $A$ is also in the union. Likewise, a result above 1 cannot be a probability. A plausible-looking answer still needs the event meanings and calculation checked.

Worked Example: Adding Two Overlapping Travel Options

In an invented survey, 54% of residents say they use a bike-share service, 41% say they use a bus pass, and 18% use both. Let $A$ mean “uses bike-share” and $B$ mean “uses a bus pass.” A student calculates $P(A\text{ or }B)=0.54+0.41=0.95$. Find and explain the error, then calculate the correct probability.

State: The requested event is using bike-share, using a bus pass, or using both. Since 18% use both, $A$ and $B$ overlap.

Plan: The student’s sum counts the residents who use both once as members of $A$ and again as members of $B$. Use the general addition rule and subtract the overlap once.

Do: Substitute the three probabilities:

P(A\cup B)=0.54+0.41-0.18=0.77

The incorrect sum is $0.95$, which is $0.18$ greater than the correct union: $0.95-0.77=0.18$. That difference matches the overlap that was counted twice. As a second check, the probability $0.77$ is at least $0.54$ and $0.41$, and it is no greater than 1.

Conclude: The probability that a randomly selected resident in this survey uses bike-share, a bus pass, or both is 0.77. The student’s answer overcounts by 0.18 because it adds the overlap without correcting for the double count.

Subtract the Probability of “Both,” Not “Neither”

A frequent subtraction error is choosing a number that sounds relevant but describes a different region of the sample space. The general addition rule subtracts the intersection, the outcomes in both events. “Neither” describes outcomes outside the union. Those probabilities refer to opposite parts of the Venn diagram and cannot be swapped.

If the probability of neither event is given, you can find the union by using the complement rule: $P(A\cup B)=1-P(\text{neither }A\text{ nor }B)$. This is a valid alternative because “neither” is the complement of the union, as covered in Probability of “Not” and “Neither” Events. It is not valid to subtract the probability of neither from $P(A)+P(B)$.

Worked Example: Subtracting the Wrong Region

An invented survey of households reports that $P(A)=0.62$ for having a meal-kit subscription, $P(B)=0.47$ for using a grocery-delivery service, and $P(\text{neither }A\text{ nor }B)=0.21$. A student calculates $0.62+0.47-0.21=0.88$ as the probability of having at least one service. Find the error and the correct union probability.

State: The requested event is having a meal-kit subscription, using grocery delivery, or having both. The reported probability $0.21$ describes households with neither service, not households with both.

Plan: Since neither is the complement of the union, find the union by subtracting the neither probability from 1. To identify why the student’s calculation fails, also find the actual overlap using the general addition rule.

Do: First, use the complement:

P(A\cup B)=1-0.21=0.79

Now find the overlap from the general addition rule rearranged to solve for $P(A\cap B)$:

P(A\cap B)=P(A)+P(B)-P(A\cup B) $$ $$ P(A\cap B)=0.62+0.47-0.79=0.30

Thus, the correct subtraction in the addition rule is $0.30$, not $0.21$. Subtracting the neither probability from the sum gives $0.62+0.47-0.21=0.88$, which is too large by $0.09$. The correct result also passes a region check: $0.32$ is in $A$ only, $0.17$ is in $B$ only, and $0.30$ is in both, for a union of $0.32+0.17+0.30=0.79$.

Conclude: The probability that a randomly selected household in this survey has at least one of the two services is 0.79. The student subtracted the probability of neither, which is not the intersection required by the general addition rule.

Do Not Subtract an Entire Event

Another mistake is subtracting $P(A)$ or $P(B)$ in place of $P(A\cap B)$. An event probability includes both its exclusive region and its overlap. Subtracting the whole event removes too much; the correction should remove only the part counted twice. This distinction is clear when you compare the named regions in a Venn diagram.

Worked Example: Finding the Intersection Without Replacing It

In an invented survey about a community website, $P(A)=0.56$ of residents use its mobile app, $P(B)=0.39$ use its web version, and $P(A\cup B)=0.71$ use at least one version. One student says the overlap is $0.56+0.39-0.39=0.56$, subtracting all of $P(B)$. Find the actual overlap and explain why the student’s subtraction is wrong.

State: Here, $A\cap B$ means residents who use both the mobile app and the web version. The value $0.39$ is the probability of using the web version, whether or not the resident also uses the app; it is not given as the probability of both.

Plan: Rearrange the general addition rule to find the intersection from the two individual event probabilities and the union. Then check that putting the intersection back into the rule reproduces the stated union.

Do: Calculate the intersection:

P(A\cap B)=P(A)+P(B)-P(A\cup B) $$ $$ P(A\cap B)=0.56+0.39-0.71=0.24

Check the union:

$$ 0.56+0.39-0.24=0.71 $$

The student subtracts $P(B)=0.39$, the entire web-version group, rather than the overlap $0.24$. That calculation gives $0.56$, not the intersection. In this example, it incorrectly treats every web-version user as if that person also used the app. The actual probability of web-only use is $0.39-0.24=0.15$, so the web-version group contains both web-only users and users in the overlap.

Conclude: The probability that a randomly selected resident uses both versions is 0.24. The subtraction must use the intersection, because only that region is counted in both $P(A)$ and $P(B)$.

Common Mistakes and AP Exam Tips

Adding overlapping event probabilities without subtracting. This counts the intersection twice. A full-credit correction identifies the overlap and subtracts $P(A\cap B)$ once.
Assuming different labels mean disjoint events. Ask whether one outcome can satisfy both definitions. A person can use two services, attend two activities, or belong to two groups at once.
Subtracting “neither” in the general addition rule. “Neither” is outside the union; the intersection is inside both events. If neither is given, use $1-P(\text{neither})$ to find the union instead.
Subtracting $P(A)$ or $P(B)$ instead of the overlap. An entire event includes its exclusive region as well as the intersection. Subtract only the probability of both.
Subtracting the overlap twice. The sum counts the overlap twice, but the union should count it once, so one subtraction corrects the total. Subtracting twice removes it completely.
Accepting a result without checking its size. A union cannot be smaller than either event, and no probability can exceed 1. These checks do not replace the rule, but they can expose an arithmetic or setup error.

On an AP response, do more than write a corrected number. Name what the probabilities mean, state that the events overlap (if they do), show the rule with the intersection in the subtraction term, and give the answer in context. If you use the complement, identify “neither” as the complement of the union. That explanation shows that you chose the subtraction for a reason rather than by guesswork.

Key takeaway: In the general addition rule, subtract only $P(A\cap B)$, the probability of both events. If you are given “neither,” use it as the complement of the union; do not insert it as the overlap.

Check Your Understanding

For each situation, identify the error or calculate the requested probability. Be clear about which event each probability describes.

In a survey, $P(A)=0.48$, $P(B)=0.37$, and $P(A\cap B)=0.12$. A student says $P(A\cup B)=0.85$. What did the student overlook, and what is the correct union probability?
For two events, the probability of neither is 0.26. A student subtracts 0.26 from $P(A)+P(B)$ as the overlap. Explain why that subtraction is not appropriate. What probability does 0.26 help you find directly?
If $P(A)=0.64$, $P(B)=0.42$, and $P(A\cup B)=0.78$, find $P(A\cap B)$. Show a check using the general addition rule.
A student calculates $P(A)+P(B)-P(A)=P(A\cup B)$. Explain what region the student has subtracted and why it is not generally the correct quantity.
Suppose a calculated union probability is 0.43 while $P(A)=0.51$. What does this tell you about the calculation? State a general size check that applies to every union.