Adding Probabilities When Events Overlap

Why “Or” Needs Care

Suppose a student is asked whether they participate in a science club, a music club, or both. If we want the probability that the student participates in the science club or the music club, adding the two participation probabilities directly can count students in both clubs twice. The calculation must account for the overlap.

In probability, “A or B” ordinarily means that event \(A\) occurs, event \(B\) occurs, or both occur. This is inclusive “or.” It differs from “exactly one of A and B,” which excludes outcomes where both occur.

Definition: The union \(A \cup B\), read “A or B,” is the event that at least one of \(A\) and \(B\) occurs. The intersection \(A \cap B\), read “A and B,” is the event that both occur. Events overlap when their intersection contains at least one outcome.

The addition rule applies to any two events in the same probability model. It requires the probabilities of \(A\), \(B\), and their intersection. It does not require the events to be equally likely or independent. Independence is a different property and is not needed for this calculation.

Theorem: Addition Rule for Two Events. For any events \(A\) and \(B\),

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

The term \(P(A \cap B)\) is subtracted because the outcomes in the intersection are included in \(P(A)\) and again in \(P(B)\). Subtracting that shared probability once leaves those outcomes counted once in the probability of the union.

Why the Rule Works

A short justification comes from dividing the union into three disjoint parts: outcomes in \(A\) but not \(B\), outcomes in \(B\) but not \(A\), and outcomes in both. Because these parts are disjoint, their probabilities add.

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

Here \(A\setminus B\) means outcomes that are in \(A\) and not in \(B\). The three parts on the right do not share outcomes. By additivity for disjoint events,

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

Also, \(A\) consists of the disjoint parts \(A\setminus B\) and \(A\cap B\), while \(B\) consists of \(B\setminus A\) and \(A\cap B\). Therefore,

P(A)+P(B) =P(A\setminus B)+P(B\setminus A)+2P(A\cap B).

The sum \(P(A)+P(B)\) has the intersection twice, whereas \(P(A\cup B)\) has it once. Subtracting one copy gives the addition rule. This reasoning also shows why the intersection probability is subtracted in every case: if \(P(A\cap B)=0\), subtracting it changes nothing.

Worked Example: Participation in Two Clubs

In a hypothetical survey of 120 students, 54 report participating in a science club, 42 report participating in a music club, and 24 report participating in both. If one student is selected at random from these 120, find the probability that the student participates in at least one of the clubs.

Model and check. Treat each of the 120 students as equally likely to be selected, so each event probability is the corresponding fraction of students. The “both” group is part of each club group, so it is the intersection that must be subtracted.

Let \(A\) be the event that the selected student participates in the science club and \(B\) the event that the student participates in the music club. Then

P(A)=\frac{54}{120}=0.45,\qquad P(B)=\frac{42}{120}=0.35,\qquad P(A\cap B)=\frac{24}{120}=0.20.

Apply the addition rule and substitute the probabilities:

P(A\cup B)=0.45+0.35-0.20=0.60.

The probability is \(0.60\), or 60%. In this hypothetical group, 60% of the students participate in at least one of the two clubs. Adding 45% and 35% without subtracting the 20% in both would give 80%, counting those students twice.

Applying the Rule to a Discrete Experiment

When outcomes are equally likely, probabilities can be obtained from the number of outcomes in an event divided by the total number of outcomes. Once the event probabilities and intersection probability are known, the addition rule works exactly as before.

Worked Example: Rolling an Even Number or a Number Greater Than Three

Imagine rolling one fair six-sided die. Find the probability of rolling an even number or a number greater than 3. Assume the six faces \(1,2,3,4,5,6\) are equally likely.

Let \(A\) be rolling an even number, so \(A=\{2,4,6\}\). Let \(B\) be rolling a number greater than 3, so \(B=\{4,5,6\}\). Each event has three of the six equally likely outcomes, and their intersection is \(\{4,6\}\), with two outcomes. Thus,

P(A)=\frac{3}{6},\qquad P(B)=\frac{3}{6},\qquad P(A\cap B)=\frac{2}{6}.

Substitute into the addition rule:

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

The event \(A\cup B\) has outcomes \(\{2,4,5,6\}\), which confirms the result: four of the six equally likely faces satisfy at least one condition. The probability of rolling an even number or a number greater than 3 is \(2/3\).

Overlap in Outcomes Does Not Always Mean Positive Probability

An important distinction is between events sharing an outcome and their intersection having positive probability. Events overlap as sets if \(A\cap B\) contains an outcome. But in some probability models, an event containing an outcome can still have probability zero. The addition rule remains valid: use the actual value of \(P(A\cap B)\), whether it is positive or zero.

Worked Example: Events Sharing a Single Point

Let \(X\) be uniformly distributed on the interval from 0 to 1. For this model, the probability that \(X\) falls in an interval is the interval’s length, and any single point has probability zero. Define \(A\) as \(X\leq 0.5\) and \(B\) as \(X\geq 0.5\). Find \(P(A\cup B)\).

The events share the outcome \(X=0.5\), so \(A\cap B\) is not empty. However, their intersection contains only that one point, which has probability zero in this continuous model. The interval lengths give

P(A)=0.5,\qquad P(B)=0.5,\qquad P(A\cap B)=0.

The addition rule gives

P(A\cup B)=0.5+0.5-0=1.

Indeed, every value from 0 to 1 satisfies at least one of \(X\leq 0.5\) or \(X\geq 0.5\). The events share an outcome, but the overlap contributes zero probability. This does not change the rule or require a different formula.

A Reliable Calculation Routine

Before substituting numbers, identify what “or” means in the question and name the intersection explicitly. The following routine keeps the calculation tied to the events rather than to a memorized pattern.

Name the events.
State what \(A\) and \(B\) represent in the experiment or population.

Identify the union and intersection.
Translate “A or B” as at least one event, and determine what it means for both events to occur.

Find the three probabilities.
Obtain \(P(A)\), \(P(B)\), and \(P(A\cap B)\) from the stated model or data.

Substitute, calculate, and interpret.
Use \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\), then state the result as a probability in context.

As a basic check, the answer must be between 0 and 1. It must also be at least as large as either individual event probability, because every outcome in \(A\) or \(B\) is included in their union. If a calculation violates either check, revisit the probabilities or the subtraction.

Common Mistakes and Why the Rule Matters

Adding without correcting for shared outcomes. The reasoning “A or B means add the probabilities” gives \(P(A)+P(B)\), which counts the intersection twice. Subtract \(P(A\cap B)\) once.
Subtracting the wrong quantity. It is the probability of “A and B,” not the probability of “A or B,” that is subtracted. Identify the intersection before calculating.
Treating “or” as “exactly one.” The standard event \(A\cup B\) includes outcomes where both events occur. If a question explicitly asks for exactly one, that is a different event.
Assuming overlap must have positive probability. A nonempty intersection can have probability zero, as when two intervals share only an endpoint in a continuous model. Always subtract \(P(A\cap B)\); if it is zero, the subtraction has no numerical effect.
Assuming independence is required. It is not. The addition rule applies to dependent as well as independent events, provided the intersection probability is correctly specified.

This rule matters whenever a question asks for the chance of at least one of two conditions, such as a person having either of two characteristics or an outcome meeting either of two criteria. Correctly accounting for overlap ensures that the union probability neither double-counts shared outcomes nor mistakenly ignores them.

Key takeaway: For any two events, add their probabilities and subtract the probability that both occur. The intersection may have positive probability or probability zero; the rule is the same.

Check Your Understanding

For each question, identify the relevant events and explain how the addition rule applies.

A hypothetical survey finds \(P(A)=0.38\), \(P(B)=0.27\), and \(P(A\cap B)=0.11\). Calculate \(P(A\cup B)\) and interpret it as “at least one.”
In a fair six-sided die roll, let \(A\) be rolling 1 or 2 and \(B\) be rolling 2 or 3. Find \(P(A)\), \(P(B)\), \(P(A\cap B)\), and \(P(A\cup B)\).
In your own words, why does the addition rule subtract \(P(A\cap B)\)?
Can two events share an outcome while their intersection has probability zero? Explain what the addition rule does in that case.
Explain why independence is not a requirement for the addition rule.