---
title: "Probability of Union — AP Stats Definition & Formula"
description: "Probability of union, P(A ∪ B), is the chance that A or B (or both) occurs, found with the addition rule. A core Unit 4 formula on the AP Stats exam."
canonical: "https://fiveable.me/ap-stats/key-terms/probability-of-union"
type: "key-term"
subject: "AP Statistics"
unit: "Unit 2"
---

# Probability of Union — AP Stats Definition & Formula

## Definition

The probability of the union of events A and B, written P(A ∪ B), is the probability that A or B (or both) will occur. On AP Stats, you calculate it with the addition rule: P(A ∪ B) = P(A) + P(B) − P(A ∩ B), subtracting the overlap so it isn't counted twice.

## What It Is

The probability of a union answers the question "what's the [chance](/ap-stats/unit-3 "fv-autolink") that at least one of these events happens?" In symbols, P(A ∪ B) is the probability that [event A](/ap-stats/key-terms/event-a "fv-autolink") occurs, event B occurs, or both occur. The CED defines it directly in VAR-4.E.3, and it lives in Topic 4.6.

The formula comes from a counting fix. If you just add P(A) and P(B), any outcome sitting in both events gets counted twice. So the addition rule subtracts the overlap once: P(A ∪ B) = P(A) + P(B) − P(A ∩ B). Picture a Venn diagram where the two circles overlap. Adding the circles double-counts the middle, and subtracting P(A ∩ B) fixes it. If A and B are [mutually exclusive](/ap-stats/unit-2/mutually-exclusive-events/study-guide/iBljImMDLJ8bSWeWXbi6 "fv-autolink") (they can't happen together), the overlap is zero and the formula collapses to P(A) + P(B).

## Why It Matters

This is the heart of Topic 4.6 in [Unit 4](/ap-stats/unit-4 "fv-autolink") (Probability, Random Variables, and Probability Distributions) and directly supports learning objective 4.6.A, which asks you to calculate probabilities for independent events and for unions of two events. The union pairs with [independence](/ap-stats/key-terms/independence "fv-autolink") in the same topic for a reason. When A and B are independent, you can compute the overlap as P(A ∩ B) = P(A) · P(B) (VAR-4.E.2) and then plug it into the addition rule. That two-step combo, multiply for the intersection, then add and subtract for the union, is one of the most common probability moves on the exam. Union probabilities also feed forward into everything built on probability rules later in Unit 4, like probability distributions and "at least one" calculations.

## Connections

### [Addition rule (Unit 4)](/ap-stats/key-terms/addition-rule)

The [addition rule](/ap-stats/key-terms/addition-rule "fv-autolink") is the formula; the probability of union is the thing it computes. VAR-4.E.4 states the rule, and every union problem on the exam runs through it. Know them as one package.

### Independent events (Unit 4)

Independence gives you a shortcut for the overlap term. If A and B are independent, P(A ∩ B) = P(A) · P(B), so you can find P(A ∪ B) even when the [intersection](/ap-stats/key-terms/intersection "fv-autolink") isn't handed to you. Topic 4.6 tests both ideas together on purpose.

### [Law of total probability (Unit 4)](/ap-stats/key-terms/law-of-total-probability)

Both ideas are about piecing together a probability from parts. The union combines two events into "either one happens," while the [law of total probability](/ap-stats/key-terms/law-of-total-probability "fv-autolink") splits an event across mutually exclusive cases and adds them up. Same Venn-diagram thinking, opposite directions.

### Mutually exclusive events (Unit 4)

Mutually exclusive is the special case where the overlap is empty. Then P(A ∪ B) is just P(A) + P(B), no subtraction needed. Spotting whether events can happen together tells you which version of the formula to use.

## On the AP Exam

Union probabilities show up almost exclusively in multiple-choice and in the calculation steps of probability FRQs. Typical stems give you two-way table counts, a Venn diagram, or stated probabilities and ask for P(A or B). Practice questions on this topic ask exactly what the addition rule calculates and when to apply it, so the tested skill is recognition plus execution. Watch for two traps. First, forgetting to subtract P(A ∩ B), which double-counts the overlap and usually matches a wrong answer choice. Second, assuming events are mutually exclusive or independent without justification. If a problem says independent, multiply to get the intersection first, then run the addition rule. Always show the formula with values plugged in when this appears in an FRQ, since communication earns partial credit.

## Probability of union vs Probability of intersection

The union, P(A ∪ B), is the probability that A OR B (or both) happens, found with the addition rule. The intersection, P(A ∩ B), is the probability that A AND B both happen, found by multiplying when events are independent. A quick gut check helps. "Or" makes the event bigger, so the union is always at least as large as either individual probability. "And" makes the event smaller. If your "or" answer came out tiny or your "and" answer came out huge, you mixed them up.

## Key Takeaways

- P(A ∪ B) is the probability that event A or event B or both will occur, which the CED calls the probability of the union (VAR-4.E.3).
- The addition rule says P(A ∪ B) = P(A) + P(B) − P(A ∩ B), and the subtraction exists to undo the double-counting of outcomes in both events.
- If A and B are mutually exclusive, P(A ∩ B) = 0, so the union is just P(A) + P(B).
- If A and B are independent, find the overlap first using P(A ∩ B) = P(A) · P(B), then plug it into the addition rule.
- "Or" means union and uses addition; "and" means intersection and uses multiplication. Translating the words correctly is half the problem.
- A union probability can never be greater than 1 or smaller than either P(A) or P(B), which is a fast way to sanity-check your answer.

## FAQs

### What is the probability of union in AP Stats?

It's the probability that event A or event B (or both) occurs, written P(A ∪ B). You calculate it with the addition rule: P(A ∪ B) = P(A) + P(B) − P(A ∩ B). It's covered in Topic 4.6 of Unit 4.

### Why do you subtract P(A ∩ B) in the addition rule?

Because outcomes that are in both A and B get counted once in P(A) and again in P(B). Subtracting the intersection removes the double-count. In a Venn diagram, it's the overlapping middle region you'd otherwise count twice.

### Is P(A ∪ B) the same as P(A and B)?

No. P(A ∪ B) is "A or B or both" and uses the addition rule, while P(A ∩ B) is "A and B together" and is found by multiplication when events are independent. The union is always at least as big as the intersection.

### Does 'or' in probability mean A or B but not both?

No. In statistics, "or" is inclusive, so P(A ∪ B) includes the case where both A and B happen. That's exactly why the formula subtracts P(A ∩ B) once instead of twice.

### How do I find P(A ∪ B) if the problem only says the events are independent?

Use independence to get the intersection first: P(A ∩ B) = P(A) · P(B) (this works if and only if the events are independent, per VAR-4.E.2). Then plug that into P(A ∪ B) = P(A) + P(B) − P(A ∩ B). This two-step combo is the classic Topic 4.6 question.

## Related Study Guides

- [2.7 Independent Events and Unions of Events](/ap-stats/unit-2/independent-events-unions-events/study-guide/aMOuOhtDQIJAtgZx3JNz)

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