Deriving the Addition Rule by Counting Overlap

Why an Overlap Must Be Counted Once

When either of two events can satisfy a question, the event of interest is their union. If the events overlap, adding their probabilities directly counts the shared outcomes twice. The addition rule corrects that double count by subtracting the probability of the overlap once. Here we derive that correction by separating the union into regions that do not overlap.

For example, a numbered outcome might belong to event $A$, event $B$, both events, or neither. The union $A\cup B$ includes outcomes in $A$, in $B$, or in both. The intersection $A\cap B$ consists precisely of the outcomes that belong to both. These set operations let us express the counting argument exactly.

Definition: The union $A\cup B$ is the event that $A$ occurs, $B$ occurs, or both occur. The intersection $A\cap B$ is the event that both $A$ and $B$ occur. The events are disjoint (or mutually exclusive) when $A\cap B$ is empty.

The derivation will first use a finite sample space in which all outcomes are equally likely. Let $\Omega$ be that sample space, let $N$ be its number of outcomes, and let $|E|$ denote the number of outcomes in an event $E$. In this model, $P(E)=|E|/N$. The key is to count the union by splitting it into disjoint pieces.

Deriving the Rule by Partitioning

Imagine the usual two-circle Venn diagram. The part of circle $A$ outside circle $B$ contains outcomes in $A$ only. The shared lens contains outcomes in $A\cap B$. The part of $B$ outside $A$ contains outcomes in $B$ only. The union includes all three of these regions, but not the area outside both circles.

A partition breaks an event into pieces that are disjoint and together make up the entire event. We can partition $A\cup B$ into $A$ and the part of $B$ that is not already in $A$, written $B\setminus A$. These pieces have no outcomes in common. We can also partition $B$ into $B\setminus A$ and $A\cap B$. The first partition counts the union without duplication; the second tells us how much of $B$ was left out of the first count.

Partition the union.
$A\cup B$ is the disjoint union of $A$ and $B\setminus A$, so its number of outcomes is $|A|+|B\setminus A|$.

Partition event $B$.
$B$ consists of $B\setminus A$ and $A\cap B$, so $|B|=|B\setminus A|+|A\cap B|$.

Substitute the second count into the first.
Rearranging gives $|B\setminus A|=|B|-|A\cap B|$, which can be substituted into the union count.

Putting those equalities together gives a count identity that holds for any two subsets of a finite sample space:

|A\cup B|=|A|+|B\setminus A|=|A|+|B|-|A\cap B|.

If the $N$ sample-space outcomes are equally likely, divide every count in this identity by $N$. Since the probability of an event is its count divided by $N$, the result is the two-event addition rule:

Theorem: For two events $A$ and $B$ in a finite sample space with equally likely outcomes, $$ P(A\cup B)=P(A)+P(B)-P(A\cap B). $$ The intersection is subtracted once because adding $P(A)$ and $P(B)$ counts every outcome in $A\cap B$ twice, while the union should count each outcome only once.

The same partition explains the rule in a probability model that is not based on equally likely outcomes. Probability is additive for disjoint events: if $C$ and $D$ are disjoint, then $P(C\cup D)=P(C)+P(D)$. Apply this fact first to the disjoint parts of $A\cup B$, then to the disjoint parts of $B$:

\begin{aligned} P(A\cup B)&=P(A)+P(B\setminus A),\\ P(B)&=P(B\setminus A)+P(A\cap B),\\ P(B\setminus A)&=P(B)-P(A\cap B),\\ P(A\cup B)&=P(A)+P(B)-P(A\cap B). \end{aligned}

This justification needs no assumption that $A$ and $B$ are independent. It only uses the way the events overlap and the additivity of probability for disjoint events. Independence is a separate relationship between events; it is not a condition for the addition rule.

Worked Examples

Worked Example: Even and Large Results on a Die

Hypothetical model. Roll a fair six-sided die once. Its sample space is $\Omega=\{1,2,3,4,5,6\}$, and the six outcomes are equally likely. Let $A$ be the event that the result is even, and $B$ the event that the result is greater than 3. Find the probability that at least one event occurs.

List the events: $A=\{2,4,6\}$ and $B=\{4,5,6\}$. Their intersection is $A\cap B=\{4,6\}$, while their union is $A\cup B=\{2,4,5,6\}$. Thus $|A|=3$, $|B|=3$, $|A\cap B|=2$, and $|A\cup B|=4$.

The direct count gives $P(A\cup B)=4/6=2/3$. The addition rule gives the same result:

P(A\cup B)=\frac{3}{6}+\frac{3}{6}-\frac{2}{6} =\frac{4}{6}=\frac{2}{3}\approx 0.667.

The two separate probabilities add to $6/6$, which counts the outcomes 4 and 6 twice. Subtracting the intersection's $2/6$ removes exactly those duplicate counts. Under this fair-die model, the probability of rolling an even number or a number greater than 3 is $2/3$.

Worked Example: Students in Two Activities

Hypothetical data. In a group of 40 students, 18 are enrolled in biology, 15 are taking statistics, and 7 are doing both. Suppose one of the 40 students is selected at random, so each student is equally likely to be selected. What is the probability that the selected student is enrolled in biology or taking statistics?

Let $A$ be the event that the selected student is enrolled in biology, and $B$ the event that the student is taking statistics. The sample space contains 40 equally likely students. There are 18 outcomes in $A$, 15 in $B$, and 7 in their intersection.

To count the union, start with $18+15=33$, then remove the 7 students counted in both groups. The number in at least one group is $33-7=26$. In probability terms:

P(A\cup B)=\frac{18}{40}+\frac{15}{40}-\frac{7}{40} =\frac{26}{40}=0.650.

For this random selection from the stated group, the probability of choosing a student who is in biology, statistics, or both is $0.650$. This calculation describes the 40-student group; by itself, it does not establish the same proportion for a larger population.

Worked Example: Multiples of 2 or 3

Hypothetical model. Choose one tile uniformly at random from 12 tiles labelled 1 through 12. Let $A$ be the event that the label is a multiple of 2, and $B$ the event that it is a multiple of 3. Find the probability that the chosen label is a multiple of 2 or 3.

The outcomes in $A$ are $\{2,4,6,8,10,12\}$, so $|A|=6$. The outcomes in $B$ are $\{3,6,9,12\}$, so $|B|=4$. The shared outcomes are $\{6,12\}$, so $|A\cap B|=2$. Each of the 12 labels is equally likely.

The count identity gives $|A\cup B|=6+4-2=8$. Checking by listing, $A\cup B=\{2,3,4,6,8,9,10,12\}$, which also has 8 outcomes. Therefore:

P(A\cup B)=\frac{6}{12}+\frac{4}{12}-\frac{2}{12} =\frac{8}{12}=\frac{2}{3}\approx 0.667.

The probability that the selected label is divisible by 2 or by 3 is $2/3$. The labels 6 and 12 satisfy both conditions, so they appear in both of the first two counts and must be subtracted once.

Reading the Rule in a Venn Diagram

A Venn diagram provides a visual check on the algebra. Call the three regions inside $A\cup B$ “$A$ only,” “both,” and “$B$ only.” If their probabilities are $a$, $c$, and $b$, respectively, then $P(A)=a+c$, $P(B)=b+c$, and $P(A\cap B)=c$. Adding the first two probabilities gives $a+b+2c$; subtracting $c$ leaves $a+b+c$, the probability of the union.

The region outside both circles does not appear in these sums because it is not part of $A\cup B$. This is why the rule concerns the union of the two events, rather than every outcome in the sample space.

Key takeaway: In the sum $P(A)+P(B)$, every outcome in exactly one event is counted once and every outcome in both events is counted twice. Subtracting $P(A\cap B)$ once leaves every outcome in the union counted exactly once.

Common Mistakes and Useful Checks

Adding without correcting for overlap. The reasoning “either event can happen, so add their probabilities” overlooks outcomes that satisfy both. The correction is to subtract $P(A\cap B)$ once.
Subtracting the overlap twice. The intersection is counted twice in the initial sum, but should be counted once in the union. Removing one copy is sufficient.
Assuming independence is required. It is not. The rule follows from partitioning the events. If two events are independent, that may help calculate $P(A\cap B)$, but it is not needed for the addition rule itself.
Using the disjoint-events shortcut when there is overlap. The simplified equation $P(A\cup B)=P(A)+P(B)$ applies only when $P(A\cap B)=0$, as it does for disjoint events.

One quick check is that a calculated probability for a union must be at least as large as either individual probability, because the union contains each event. It must also be no greater than 1. If a result fails either check, review the event counts, the intersection, and the subtraction.

Check Your Understanding

Use a partition into disjoint regions to reason through these questions.

In a finite equally likely sample space, why does counting $|A|+|B|$ count the outcomes in $A\cap B$ twice?
A group has 25 people in event $A$, 19 in event $B$, and 8 in both. How many are in at least one event?
If $A$ and $B$ are disjoint, what does the addition rule reduce to, and why?
Does the addition rule require $A$ and $B$ to be independent? Explain using the partition argument.
In a Venn diagram, what is represented by the part of $B$ outside $A$, and how does it help form the union?