Addition Rule and Mutually Exclusive Events

How to find the probability that at least one of two things happens.

At the 2023 US Open tennis tournament, a statistician tracked every first serve Rafael Nadal hit. On any given point, Nadal either hit the serve into the service box, or he double-faulted, or something in between. The statistician wanted to know: what is the probability that a randomly chosen point ended with a fault or with an ace? Those two outcomes don't overlap — you can't fault and ace on the same serve. But not every pair of outcomes is that clean. Sometimes two events can happen at the same time, and counting them both risks counting some outcomes twice. The addition rule exists to handle that problem precisely.

Start with two events, $A$ and $B$ , drawn from the same sample space. The probability that $A$ or $B$ occurs — meaning at least one of them happens — is given by the addition rule:

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

The symbol $A \cup B$ means the union of $A$ and $B$ : everything that belongs to $A$ , or $B$ , or both. The symbol $A \cap B$ means the intersection: outcomes that belong to both at the same time. You subtract $P(A \cap B)$ because when you added $P(A)$ and $P(B)$ , any outcome in both events got counted twice. Subtracting the overlap corrects that.

Here is a concrete picture. A standard deck has 52 cards. Pick one at random. Let $A$ be the event "the card is a heart" and $B$ be the event "the card is a face card." There are 13 hearts, 12 face cards, and 3 cards that are both (jack, queen, and king of hearts). Without the correction, adding KaTeX can only parse string typed expression counts those 3 cards twice. Subtracting $\frac$ fixes it.

Drawing a heart or a face card from a standard deck

P(A) = \frac{13}{52}, \quad P(B) = \frac{12}{52}, \quad P(A \cap B) = \frac{3}{52}

Count the hearts, the face cards, and the cards that are both.

P(A \cup B) = \frac{13}{52} + \frac{12}{52} - \frac{3}{52}

Plug into the addition rule. The three cards that are both a heart and a face card were counted twice, so subtract them once.

P(A \cup B) = \frac{22}{52} = \frac{11}{26} \checkmark

Simplify by dividing numerator and denominator by 2.

Now for the special case that comes up constantly on the Regents. Two events are mutually exclusive when they cannot both happen at the same time — their intersection is empty. Formally, $P(A \cap B) = 0$ . Rolling a 2 and rolling a 5 on one die roll. Being born in July and being born in October. Drawing a club and drawing a heart from one card draw. When events are mutually exclusive, the addition rule simplifies:

P(A \cup B) = P(A) + P(B)

The subtraction term disappears because there is nothing to subtract. This simpler form only applies when you have confirmed the events cannot overlap. Using it when they can overlap is one of the most common errors on this exam.

Rolling a 4 or an odd number on a standard die

A = \{4\}, \quad B = \{1, 3, 5\}

List out which outcomes belong to each event.

P(A) = \frac{1}{6}, \quad P(B) = \frac{3}{6}

Count the outcomes. 4 is even, so it cannot be odd — these events don't overlap.

P(A \cap B) = 0

4 is not odd. There is no outcome that satisfies both events at once.

P(A \cup B) = \frac{1}{6} + \frac{3}{6} = \frac{4}{6} = \frac{2}{3} \checkmark

Since the events are mutually exclusive, just add. No subtraction needed.

There is one more rule worth knowing here. Every event $A$ has a complement, written $A^c$ or $A'$ , which contains every outcome in the sample space that is not in $A$ . An event and its complement are always mutually exclusive and together cover the entire sample space, so their probabilities must add to 1:

P(A) + P(A^c) = 1

Rearranged: $P(A^c) = 1 - P(A)$ . This is the complement rule, and it saves real time. If a problem asks for the probability of something not happening, subtract from 1.

Finding the probability of not drawing a spade

P(\text{spade}) = \frac{13}{52} = \frac{1}{4}

There are 13 spades in a 52-card deck.

P(\text{not spade}) = 1 - \frac{1}{4}

The complement of drawing a spade is drawing anything else. Subtract from 1.

P(\text{not spade}) = \frac{3}{4} \checkmark

Three out of every four cards is not a spade.

The Venn diagram below shows how two overlapping events sit inside a sample space. The overlap in the middle is $A \cap B$ — the part you subtract. When those circles don't touch at all, the events are mutually exclusive.

Interactive graph — scroll to zoom, drag to pan

Practice Questions

P(A) = 0.45, \quad P(B) = 0.30, \quad P(A \cap B) = 0.15. \text{ Find } P(A \cup B).

P(A) = \frac{2}{5}, \quad P(B) = \frac{1}{4}, \quad A \text{ and } B \text{ are mutually exclusive. Find } P(A \cup B).

P(A \cup B) = 0.78, \quad P(A) = 0.55, \quad P(B) = 0.40. \text{ Find } P(A \cap B).

Regents Corner

On Part II and Part III of the Algebra II Regents, addition rule problems often appear as multi-step questions where you are given a two-way frequency table and asked to find a union probability. Read the table carefully to identify the overlap before applying the formula. Show the formula, show the substitution, and show the arithmetic as separate lines — partial credit is available if your setup is correct even if you arithmetic-error the final answer.

A student sees that two events are mutually exclusive and writes P(A or B) = P(A) · P(B) instead of P(A) + P(B), confusing the rule for mutually exclusive events with the multiplication rule for independent events. These are different situations with different rules. Mutually exclusive means they cannot happen together, so you add their probabilities. Independent means knowing one gives no information about the other, so you multiply. Never multiply when the question is asking for a union.

Conditional Probability and Independence