Inclusion-Exclusion for Two Events

Why Overlap Changes a Probability

Suppose an outcome can satisfy event $A$, event $B$, or both. The event that at least one occurs is the union $A\cup B$. Adding $P(A)$ and $P(B)$ seems like a natural way to find its probability, but that sum counts outcomes in both events twice. To get the union probability, the shared part must be counted only once.

This adjustment is the two-event inclusion-exclusion formula. It applies to any two events in a probability model; the events do not need to be equally likely or disjoint. The topic is aligned with the SOA VEE Mathematical Statistics syllabus.

Definition: For events $A$ and $B$, the union $A\cup B$ is the event that $A$ occurs, $B$ occurs, or both occur. Their intersection $A\cap B$ is the event that both $A$ and $B$ occur. The events are disjoint if $A\cap B=\varnothing$, so they cannot occur together.

The words “or” and “at least one” generally refer to the union and include the possibility that both events occur. This is sometimes called inclusive “or.” By contrast, “exactly one” excludes the intersection. Keeping those meanings distinct is essential when translating a problem into events.

Deriving the Two-Event Formula

The formula follows from splitting the union into nonoverlapping regions. The union consists of the outcomes in $A$ but not $B$, the outcomes in $B$ but not $A$, and the outcomes in both. These three regions are disjoint, so finite additivity lets us add their probabilities.

P(A\cup B) = P(A\setminus B)+P(B\setminus A)+P(A\cap B).

The event $A$ itself consists of the part in $A$ but not $B$, together with the intersection. Thus $P(A\setminus B)=P(A)-P(A\cap B)$. Likewise, $P(B\setminus A)=P(B)-P(A\cap B)$. Substituting both expressions into the disjoint-region sum gives

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

Theorem: For any two events $A$ and $B$, $$ P(A\cup B)=P(A)+P(B)-P(A\cap B). $$ The intersection is subtracted once because adding $P(A)$ and $P(B)$ includes it twice, while the union should include it once.

When $A$ and $B$ are disjoint, $P(A\cap B)=0$, and the formula reduces to $P(A\cup B)=P(A)+P(B)$. That is the finite additivity rule for disjoint events. For overlapping events, omitting the intersection correction gives a sum that is too large by exactly $P(A\cap B)$.

Worked Applications

Worked Example: Two Training Modules in a Hypothetical Staff Survey

A hypothetical organization has 200 staff members. One staff member is selected uniformly at random from the roster. Of the 200, 118 completed Module A, 91 completed Module B, and 54 completed both. Find the probability that the selected staff member completed at least one module, neither module, or exactly one module.

Model and method. The sample space is the roster of 200 staff members, with every member having probability $1/200$. Let $A$ be the event that the selected staff member completed Module A, and $B$ the event that they completed Module B. The supplied counts give $P(A)=118/200=0.590$, $P(B)=91/200=0.455$, and $P(A\cap B)=54/200=0.270$.

At least one module. Apply inclusion-exclusion:

P(A\cup B)=0.590+0.455-0.270=0.775.

Equivalently, the number who completed at least one is $118+91-54=155$, and $155/200=0.775$.

Neither module. “Neither” is the complement of the union, so its probability is $1-0.775=0.225$. This corresponds to $200-155=45$ staff members, and $45/200=0.225$.

Exactly one module. The number who completed A but not B is $118-54=64$. The number who completed B but not A is $91-54=37$. These regions are disjoint, so the count completing exactly one is $64+37=101$, giving $101/200=0.505$.

Check and common mistake. The four disjoint categories have counts 54 (both), 64 (A only), 37 (B only), and 45 (neither). They sum to $54+64+37+45=200$. Simply adding the two completion counts would give $118+91=209$, incorrectly counting the 54 staff who completed both twice.

Worked Example: Combining Two Probabilities With a Known Overlap

For a hypothetical randomly selected service request, let $A$ mean that it is flagged for a document review and $B$ mean that it is flagged for a timing review. Suppose the probability model assigns $P(A)=0.42$, $P(B)=0.35$, and $P(A\cap B)=0.18$. Find the probability of at least one flag, exactly one flag, and no flags.

Model and method. The values are probabilities for the same randomly selected request. The events can occur together, and their overlap is specified as $0.18$. Apply inclusion-exclusion for the union:

P(A\cup B)=0.42+0.35-0.18=0.59.

Thus the probability of at least one flag is $0.59$. The probability of no flags is the complement, $1-0.59=0.41$.

To find the probability of exactly one flag, separate the two nonoverlapping regions. The probability of $A$ but not $B$ is $0.42-0.18=0.24$. The probability of $B$ but not $A$ is $0.35-0.18=0.17$. Therefore,

P(\text{exactly one})=0.24+0.17=0.41.

Check and common mistake. The four categories have probabilities $0.18$ (both), $0.24$ (A only), $0.17$ (B only), and $0.41$ (neither), which sum to $1.00$. Adding $0.42+0.35=0.77$ would be the wrong union probability: it counts the $0.18$ overlap twice. Subtracting it once gives $0.59$.

Worked Example: Recovering an Unknown Intersection

In a hypothetical risk-monitoring model, event $A$ means that a randomly selected day has elevated demand, and event $B$ means that it has a staffing alert. The model gives $P(A)=0.51$, $P(B)=0.44$, and $P(A\cup B)=0.76$. Find the probability of both events, exactly one event, and neither event.

Model and method. These are probabilities for the same day under the stated model. Rearrange inclusion-exclusion to isolate the unknown intersection:

P(A\cap B)=P(A)+P(B)-P(A\cup B) =0.51+0.44-0.76=0.19.

The intersection probability is $0.19$. For exactly one event, subtract the intersection from each individual event and add the two remaining disjoint parts:

\begin{aligned} P(\text{A only})&=0.51-0.19=0.32,\\ P(\text{B only})&=0.44-0.19=0.25,\\ P(\text{exactly one})&=0.32+0.25=0.57. \end{aligned}

The probability of neither event is $1-P(A\cup B)=1-0.76=0.24$. As a check, $0.19+0.32+0.25+0.24=1.00$.

Common mistake avoided. The union probability cannot be subtracted from either individual probability alone to obtain the intersection. Instead, rearrange the full formula, keeping both $P(A)$ and $P(B)$: their sum minus the union leaves precisely the overlap.

Bounds and Exam Checks

The formula also gives useful checks on supplied probabilities. Since an intersection is contained in each event, its probability cannot exceed either event probability. The union contains each event, so it cannot have a probability smaller than either one. A valid set of values must satisfy

0\leq P(A\cap B)\leq \min\{P(A),P(B)\}, \qquad \max\{P(A),P(B)\}\leq P(A\cup B)\leq 1.

There is also a lower bound on the intersection. From inclusion-exclusion and $P(A\cup B)\leq 1$, $P(A\cap B)=P(A)+P(B)-P(A\cup B)\geq P(A)+P(B)-1$. Combining this with the fact that probabilities are nonnegative gives

\max\{0,\ P(A)+P(B)-1\} \leq P(A\cap B) \leq \min\{P(A),P(B)\}.

These bounds are not a replacement for the formula when the overlap is known. They are a quick way to notice inconsistent data or arithmetic. For instance, a claimed intersection larger than $P(A)$ cannot be correct because every outcome in $A\cap B$ is already in $A$.

A reliable solution method is to name the two events, identify whether the question asks for at least one, exactly one, both, or neither, and then select the relevant relationship. “At least one” uses the union formula; “both” uses the intersection; “neither” is the complement of the union; and “exactly one” is the sum of the A-only and B-only regions. Check that the resulting probabilities are between zero and one and that any disjoint categories used in the calculation account for the full sample space.

At least one: add the event probabilities, then subtract the overlap once.
Exactly one: remove the intersection from each event, then add the two remaining parts.
Neither: subtract the union probability from one.
Disjoint events: the intersection is zero, so no overlap correction is needed.

Check Your Understanding

For each problem, identify the relevant overlap before calculating. Final answers are included for checking.

A hypothetical probability model gives $P(A)=0.31$, $P(B)=0.46$, and $P(A\cap B)=0.12$. Find the probability of at least one event, neither event, and exactly one event. (Answer: union $0.65$; neither $0.35$; exactly one $0.53$.)
Among 160 staff members in a hypothetical roster, 96 have completed a safety briefing, 72 have completed a data-protection briefing, and 38 have completed both. One person is selected uniformly at random. Find the probabilities of at least one briefing, neither briefing, and exactly one briefing. (Answer: $0.8125$, $0.1875$, and $0.5750$, respectively.)
A hypothetical model gives $P(A)=0.48$, $P(B)=0.39$, and $P(A\cup B)=0.73$. Find the intersection and the probability of exactly one event. (Answer: intersection $0.14$; exactly one $0.59$.)
For two events in a hypothetical model, $P(A)=0.62$, $P(B)=0.53$, and the overlap is $0.40P(B)$. Find the union, neither, and exactly-one probabilities. (Answer: intersection $0.212$; union $0.938$; neither $0.062$; exactly one $0.726$.)
A hypothetical model gives $P(A)=0.68$, $P(B)=0.47$, and $P(A\cup B)=0.92$. Find the probability of both events, exactly one event, and neither event; check that the values are feasible. (Answer: intersection $0.23$; exactly one $0.69$; neither $0.08$; feasible.)

What You'll Learn