8.2 Mutually Exclusive Events and the Addition Rule

Learning Objectives

By the end of this section, you will be able to:

In this section, you will learn to:

Describe compound events using union, intersection, and complement.
Identify mutually exclusive events.
Use the Addition Rule to compute probabilities of unions of events.

8.2.1 Set Operations on Events

In the previous chapter we saw how to take the union, intersection, and complement of sets. Events in a probability experiment are themselves subsets of the sample space, so the same operations apply — and they let us describe compound questions like "$A$ or $B$," "$A$ and $B$," and "not $A$."

Definition 8.4: Union, Intersection, Complement of Events

Let $E$ and $F$ be events in a sample space $S$.

The union $E \cup F$ is the set of outcomes that lie in $E$, in $F$, or in both.
The intersection $E \cap F$ is the set of outcomes that lie in both $E$ and $F$.
The complement $\overline{E}$ is the set of outcomes in $S$ that are not in $E$.

A useful consequence of the complement: if $n(S) = n$ and $n(E) = k$, then $n(\overline{E}) = n - k$, so

\[P(\overline{E}) \;=\; \frac{n - k}{n} \;=\; 1 - \frac{k}{n} \;=\; 1 - P(E).\]

Context Pause: Why bother re-using set notation for events?

Probability is built on counting, and counting works the same whether the objects are numbers, students, or experimental outcomes. Reusing the language of sets — union, intersection, complement — means every fact you already know about sets carries straight into probability. Drawing a Venn diagram for an event problem isn't a clever trick; it's the natural picture.

Insight Note: The complement rule is your "easy mode"

Whenever an event sounds like at least one, not all, or fewer than two, ask whether its complement is easier to describe. "At least one head in five tosses" has 31 outcomes; its complement, "no heads at all," has just 1. Computing $P(\overline{E}) = 1 - P(E)$ often saves a lot of casework.

Example 8.2.1

A card is drawn from a standard deck. Determine whether the events $E$ and $F$ below are mutually exclusive (i.e., whether $E \cap F = \varnothing$).

$E = \{\text{the card is an ace}\}$, $F = \{\text{the card is a heart}\}$.

Solution

The Ace of Hearts belongs to both events, so

\[E \cap F = \{\text{Ace of Hearts}\} \neq \varnothing.\]

Answer: $E$ and $F$ are not mutually exclusive.

Example 8.2.2

Two dice are rolled. Determine whether the events below are mutually exclusive.

$G = \{\text{the sum of the faces is 6}\}$, $H = \{\text{one die shows a 4}\}$.

Solution

List both events as sets of ordered pairs:

\[G = \{(1,5),(2,4),(3,3),(4,2),(5,1)\}, \qquad H = \{(2,4),(4,2)\, \text{and any other pair with a 4}\}.\]

Restricting $H$ to pairs that intersect $G$:

\[G \cap H = \{(2,4),(4,2)\} \neq \varnothing.\]

Answer: $G$ and $H$ are not mutually exclusive.

Example 8.2.3

A family has three children. Determine whether the following events are mutually exclusive.

$M = \{\text{the family has at least one boy}\}$, $N = \{\text{the family has all girls}\}$.

Solution

Listing the events:

\[M = \{BBB, BBG, BGB, BGG, GBB, GBG, GGB\}, \qquad N = \{GGG\}.\]

These are disjoint sets:

\[M \cap N = \varnothing.\]

Answer: $M$ and $N$ are mutually exclusive.

Try It Now 8.2.1

A single die is rolled. Are the events $E = \{\text{a multiple of 3 shows}\}$ and $F = \{\text{a 2 shows}\}$ mutually exclusive?

Solution

$E = \{3, 6\}$ and $F = \{2\}$. Their intersection is empty, so

\[E \cap F = \varnothing.\]

Answer: Yes, $E$ and $F$ are mutually exclusive.

8.2.2 Mutually Exclusive Events

The events that interest us most often are those whose outcomes do not overlap.

Definition 8.5: Mutually Exclusive Events

Two events $E$ and $F$ are mutually exclusive (also called disjoint) if they share no outcomes:

\[E \cap F = \varnothing, \quad \text{equivalently} \quad P(E \cap F) = 0.\]

Context Pause: "Mutually exclusive" is a yes/no property of events

"Mutually exclusive" is the language for events that cannot both happen in the same trial. Drawing a single card cannot give you both a king and a 7 — those two events are mutually exclusive. But it can give you both a king and a heart (the King of Hearts), so those two events are not mutually exclusive. The probability $P(E \cap F)$ measures the overlap; mutually exclusive means that overlap is zero.

8.2.3 The Addition Rule for Probability

When we want $P(E \cup F)$ — the probability that $E$ happens, or $F$ happens, or both — a quick attempt is to add $P(E) + P(F)$. That works only when the events do not overlap. Otherwise we double-count the overlap and have to subtract it back out.

Example 8.2.4

A die is rolled. Find the probability that the result is an even number or a number greater than four.

Solution

Step 1 — Identify the sample space and events.

\[S = \{1,2,3,4,5,6\}, \quad E = \{2, 4, 6\}, \quad F = \{5, 6\}.\]

Step 2 — Naive addition is wrong.

$P(E) = 3/6$ and $P(F) = 2/6$. Adding gives $5/6$, but that double-counts the outcome $6$, which lies in both events.

Step 3 — Use $E \cup F$ directly.

\[E \cup F = \{2, 4, 5, 6\}, \qquad n(E \cup F) = 4.\]

\[P(E \cup F) \;=\; \frac{4}{6} \;=\; \frac{2}{3}.\]

Step 4 — Equivalent form using the overlap.

Since $E \cap F = \{6\}$ and $P(E \cap F) = 1/6$,

\[P(E \cup F) \;=\; P(E) + P(F) - P(E \cap F) \;=\; \frac{3}{6} + \frac{2}{6} - \frac{1}{6} \;=\; \frac{4}{6}.\]

Answer: $P(E \cup F) = 2/3$.

The pattern in Example 8.2.4 is the Addition Rule — it always adjusts for the overlap.

Addition Rule for Probability

For any events $E$ and $F$ in the same sample space,

\[P(E \cup F) \;=\; P(E) + P(F) - P(E \cap F).\]

When $E$ and $F$ are mutually exclusive, $P(E \cap F) = 0$, so the rule simplifies to

\[P(E \cup F) \;=\; P(E) + P(F).\]

Insight Note: The Addition Rule is just inclusion–exclusion

If you draw a Venn diagram for two events, you will see that $n(E) + n(F)$ counts the lens-shaped overlap twice. Subtracting $n(E \cap F)$ once corrects the count. Dividing through by $n(S)$ turns the count into a probability — and that's the Addition Rule.

Example 8.2.5

In a club, 20% of the students take Finite Mathematics, 30% take Statistics, and 10% take both. What percent take Finite Mathematics or Statistics?

Solution

Let $F = \{\text{takes Finite Math}\}$ and $S = \{\text{takes Statistics}\}$. Then $P(F) = 0.20$, $P(S) = 0.30$, $P(F \cap S) = 0.10$.

\[P(F \cup S) \;=\; 0.20 + 0.30 - 0.10 \;=\; 0.40.\]

Answer: $40\%$ of students take at least one of the two courses.

Try It Now 8.2.2

If $P(E) = 0.4$, $P(F) = 0.5$, and $P(E \cup F) = 0.7$, find $P(E \cap F)$.

Solution

Solve the Addition Rule for the overlap:

\[P(E \cap F) = P(E) + P(F) - P(E \cup F) = 0.4 + 0.5 - 0.7 = 0.2.\]

Answer: $P(E \cap F) = 0.2$.

8.2.4 Two-Way Tables and the Addition Rule

A two-way (cross-classification) table neatly displays a sample space split by two attributes. Probabilities of unions and intersections are read off the table directly.

Example 8.2.6

The table below is the distribution of the 100 U.S. senators of the 114th Congress (January 2015) by political party and gender.

| | Male (M) | Female (F) | Total | |--------------|---------:|-----------:|------:| | Democrats (D)| 30 | 14 | 44 | | Republicans (R)| 48 | 6 | 54 | | Other (T) | 2 | 0 | 2 | | Total | 80 | 20 | 100 |

A senator is selected at random. Find:

a) $P(M \cap D)$ — male and Democrat.

b) $P(M \cup D)$ — male or Democrat.

c) $P(F \cap R)$.

d) $P(F \cup R)$.

Solution

Part a — Intersection. The "male Democrat" cell shows 30:

\[P(M \cap D) \;=\; \frac{30}{100} \;=\; 0.30.\]

Part b — Union via Addition Rule. $P(M) = 80/100$, $P(D) = 44/100$, and we just computed $P(M \cap D) = 30/100$. So

\[P(M \cup D) = \frac{80}{100} + \frac{44}{100} - \frac{30}{100} = \frac{94}{100} = 0.94.\]

Part c — Intersection. The "female Republican" cell shows 6:

\[P(F \cap R) \;=\; \frac{6}{100} \;=\; 0.06.\]

Part d — Union. $P(F) = 20/100$, $P(R) = 54/100$, $P(F \cap R) = 6/100$:

\[P(F \cup R) = \frac{20}{100} + \frac{54}{100} - \frac{6}{100} = \frac{68}{100} = 0.68.\]

Try It Now 8.2.3

Using the Senate table above, find $P(F \cup T)$ — female or Other party.

Solution

$P(F) = 20/100$, $P(T) = 2/100$, and the "female Other" cell is 0, so $P(F \cap T) = 0$. The events are mutually exclusive in the data:

\[P(F \cup T) = \frac{20}{100} + \frac{2}{100} - 0 = \frac{22}{100} = 0.22.\]

Answer: $0.22$.

Problem Set 8.2

Determine whether the following pairs of events are mutually exclusive.

8.2.1 $A = \{\text{a person earns more than } \MATH020{,}000\}$.

8.2.2 A card is drawn from a deck. $C = \{\text{it is a King}\}$, $D = \{\text{it is a heart}\}$.

8.2.3 A die is rolled. $E = \{\text{an even number shows}\}$, $F = \{\text{a number greater than 3 shows}\}$.

8.2.4 Two dice are rolled. $G = \{\text{the sum of the dice is 8}\}$, $H = \{\text{one die shows a 6}\}$.

8.2.5 Three coins are tossed. $I = \{\text{two heads come up}\}$, $J = \{\text{at least one tail comes up}\}$.

8.2.6 A family has three children. $K = \{\text{first born is a boy}\}$, $L = \{\text{the family has children of both sexes}\}$.

Use the Addition Rule to find the following probabilities.

8.2.7 A card is drawn from a deck. Events $C$ and $D$ are: $C = \{\text{it is a king}\}$, $D = \{\text{it is a heart}\}$. Find $P(C \cup D)$.

8.2.8 A die is rolled. Events $E$ and $F$ are: $E = \{\text{an even number shows}\}$, $F = \{\text{a number greater than 3 shows}\}$. Find $P(E \cup F)$.

8.2.9 Two dice are rolled. Events $G$ and $H$ are: $G = \{\text{the sum of the dice is 8}\}$, $H = \{\text{exactly one die shows a 6}\}$. Find $P(G \cup H)$.

8.2.10 Three coins are tossed. Events $I$ and $J$ are: $I = \{\text{two heads come up}\}$, $J = \{\text{at least one tail comes up}\}$. Find $P(I \cup J)$.

Use the Addition Rule to find the following probabilities.

8.2.11 At a college, 20% of the students take Finite Mathematics, 30% take Statistics, and 10% take both. What percent of students take Finite Mathematics or Statistics?

8.2.12 This quarter there is a 50% chance that Jason will pass Accounting, a 60% chance that he will pass English, and an 80% chance that he will pass at least one of these two courses. What is the probability that he will pass both Accounting and English?

Questions 8.2.13–8.2.20 refer to the following table, the distribution of Democratic and Republican U.S. senators by gender in the 114th Congress as of January 2015.

Use this table to determine the following probabilities.

8.2.13 $P(M \cap D)$

8.2.14 $P(F \cap R)$

8.2.15 $P(M \cup D)$

8.2.16 $P(F \cup R)$

8.2.17 $P(\overline{M} \cup R)$

8.2.18 $P(M \cup F)$

8.2.19 Are the events $F$ and $R$ mutually exclusive? Use probabilities to support your conclusion.

8.2.20 Are the events $F$ and $T$ mutually exclusive? Use probabilities to support your conclusion.

Use the Addition Rule to find the following probabilities.

8.2.21 If $P(E) = 0.5$, $P(F) = 0.4$, and $E$ and $F$ are mutually exclusive, find $P(E \cap F)$.

8.2.22 If $P(E) = 0.4$, $P(F) = 0.2$, and $E$ and $F$ are mutually exclusive, find $P(E \cup F)$.

8.2.23 If $P(E) = 0.3$, $P(E \cup F) = 0.6$, and $P(E \cap F) = 0.2$, find $P(F)$.

8.2.24 If $P(E) = 0.4$, $P(F) = 0.5$, and $P(E \cup F) = 0.7$, find $P(E \cap F)$.

8.2.25 In a box of assorted cookies, 36% of cookies contain chocolate and 12% of cookies contain nuts. 8% of cookies have both chocolate and nuts. Sean is allergic to chocolate and nuts. Find the probability that a cookie has chocolate or nuts (he can't eat it).

8.2.26 At a college, 72% of courses have final exams and 46% of courses require research papers. 32% of courses have both a research paper and a final exam. Let $F$ be the event that a course has a final exam and $R$ be the event that a course requires a research paper. Find the probability that a course requires a final exam or a research paper.

Problem 8.2.1 Solution

Step 1 — Compare the events: A person cannot earn more than \$25,000 and less than \$20,000 simultaneously.

Step 2 — Conclude:

\[A \cap B = \varnothing.\]

Answer: $A$ and $B$ are mutually exclusive.

Problem 8.2.2 Solution

Step 1 — Identify the overlap: The King of Hearts is both a King and a heart, so it lies in both events.

Step 2 — Conclude:

\[C \cap D = \{\text{King of Hearts}\} \neq \varnothing.\]

Answer: $C$ and $D$ are not mutually exclusive.

Problem 8.2.3 Solution

Step 1 — List the events: $E = \{2, 4, 6\}$ (even) and $F = \{4, 5, 6\}$ (greater than 3).

Step 2 — Compute the intersection:

\[E \cap F = \{4, 6\} \neq \varnothing.\]

Answer: $E$ and $F$ are not mutually exclusive.

Problem 8.2.4 Solution

Step 1 — Sum-equals-8 outcomes:

\[G = \{(2,6),(3,5),(4,4),(5,3),(6,2)\}.\]

Step 2 — Outcomes where one die shows a 6: All ordered pairs containing a 6 — many — but those that also lie in $G$ are $(2,6)$ and $(6,2)$.

Step 3 — Conclude:

\[G \cap H = \{(2,6),(6,2)\} \neq \varnothing.\]

Answer: $G$ and $H$ are not mutually exclusive.

Problem 8.2.5 Solution

Step 1 — List the events: $I = \{HHT, HTH, THH\}$ (exactly two heads) and $J = $ all outcomes with at least one tail $= S \setminus \{HHH\}$.

Step 2 — Intersection: Every outcome in $I$ has at least one tail, so

\[I \cap J = I = \{HHT, HTH, THH\} \neq \varnothing.\]

Answer: $I$ and $J$ are not mutually exclusive.

Problem 8.2.6 Solution

Step 1 — List the events: $K = \{BBB, BBG, BGB, BGG\}$ (first born is a boy). $L$ excludes the all-same-gender outcomes $BBB$ and $GGG$, so $L = \{BBG, BGB, BGG, GBB, GBG, GGB\}$.

Step 2 — Intersection:

\[K \cap L = \{BBG, BGB, BGG\} \neq \varnothing.\]

Answer: $K$ and $L$ are not mutually exclusive.

Problem 8.2.7 Solution

Step 1 — Component probabilities: $P(C) = 4/52$ (kings), $P(D) = 13/52$ (hearts), $P(C \cap D) = 1/52$ (King of Hearts).

Step 2 — Apply the Addition Rule:

\[P(C \cup D) = \frac{4}{52} + \frac{13}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13}.\]

Answer: $4/13$.

Problem 8.2.8 Solution

Step 1 — Identify events on a single die: $E = \{2, 4, 6\}$, $F = \{4, 5, 6\}$, $E \cap F = \{4, 6\}$.

Step 2 — Apply the Addition Rule:

\[P(E \cup F) = \frac{3}{6} + \frac{3}{6} - \frac{2}{6} = \frac{4}{6} = \frac{2}{3}.\]

Answer: $2/3$.

Problem 8.2.9 Solution

Step 1 — Count $G$: sum equals 8 → $(2,6),(3,5),(4,4),(5,3),(6,2)$, so $n(G) = 5$.

Step 2 — Count $H$: exactly one die shows a 6 → 10 ordered pairs containing exactly one 6 (excluding $(6,6)$).

Step 3 — Count $G \cap H$: $(2,6)$ and $(6,2)$, so $n(G \cap H) = 2$.

Step 4 — Apply the Addition Rule:

\[P(G \cup H) = \frac{5}{36} + \frac{10}{36} - \frac{2}{36} = \frac{13}{36}.\]

Answer: $13/36$.

Problem 8.2.10 Solution

Step 1 — Identify events for three coins: $I = \{HHT, HTH, THH\}$, so $P(I) = 3/8$. $J$ excludes only $HHH$, so $P(J) = 7/8$. Every outcome in $I$ has at least one tail, so $I \cap J = I$ and $P(I \cap J) = 3/8$.

Step 2 — Apply the Addition Rule:

\[P(I \cup J) = \frac{3}{8} + \frac{7}{8} - \frac{3}{8} = \frac{7}{8}.\]

Answer: $7/8$.

Problem 8.2.11 Solution

Step 1 — Set up: $P(F) = 0.20$, $P(S) = 0.30$, $P(F \cap S) = 0.10$.

Step 2 — Apply the Addition Rule:

\[P(F \cup S) = 0.20 + 0.30 - 0.10 = 0.40.\]

Answer: $40\%$.

Problem 8.2.12 Solution

Step 1 — Solve the Addition Rule for the intersection: $P(A \cup E) = P(A) + P(E) - P(A \cap E)$, so

\[P(A \cap E) = P(A) + P(E) - P(A \cup E) = 0.50 + 0.60 - 0.80 = 0.30.\]

Answer: $30\%$.

Problem 8.2.13 Solution

Step 1 — Read the table cell: The "male Democrat" cell is 30 out of 100.

Step 2 — Compute:

\[P(M \cap D) = \frac{30}{100} = \frac{3}{10}.\]

Answer: $0.30$.

Problem 8.2.14 Solution

Step 1 — Read the table cell: The "female Republican" cell is 6 out of 100.

Step 2 — Compute:

\[P(F \cap R) = \frac{6}{100} = \frac{3}{50}.\]

Answer: $0.06$.

Problem 8.2.15 Solution

Step 1 — Component totals: $P(M) = 80/100$, $P(D) = 44/100$, $P(M \cap D) = 30/100$.

Step 2 — Apply the Addition Rule:

\[P(M \cup D) = \frac{80 + 44 - 30}{100} = \frac{94}{100} = \frac{47}{50}.\]

Answer: $0.94$.

Problem 8.2.16 Solution

Step 1 — Component totals: $P(F) = 20/100$, $P(R) = 54/100$, $P(F \cap R) = 6/100$.

Step 2 — Apply the Addition Rule:

\[P(F \cup R) = \frac{20 + 54 - 6}{100} = \frac{68}{100} = \frac{17}{25}.\]

Answer: $0.68$.

Problem 8.2.17 Solution

Step 1 — $\overline{M}$ is "not male" = female: $P(\overline{M}) = 20/100$. $P(R) = 54/100$. $\overline{M} \cap R$ is "female and Republican" = 6/100.

Step 2 — Apply the Addition Rule:

\[P(\overline{M} \cup R) = \frac{20 + 54 - 6}{100} = \frac{68}{100} = \frac{17}{25}.\]

Answer: $0.68$.

Problem 8.2.18 Solution

Step 1 — $M$ and $F$ are mutually exclusive: A senator is either male or female (no overlap), so $P(M \cap F) = 0$.

Step 2 — Apply the Addition Rule:

\[P(M \cup F) = \frac{80}{100} + \frac{20}{100} - 0 = 1.\]

Answer: $1$ (every senator is in $M$ or $F$).

Problem 8.2.19 Solution

Step 1 — Look at $P(F \cap R)$: The female Republican cell is 6, so $P(F \cap R) = 6/100 = 0.06 > 0$.

Step 2 — Conclude: Because the intersection probability is not zero, the events overlap.

Answer: $F$ and $R$ are not mutually exclusive.

Problem 8.2.20 Solution

Step 1 — Look at $P(F \cap T)$: The female "Other" cell is 0, so $P(F \cap T) = 0$.

Step 2 — Conclude: With zero overlap probability the events cannot occur together.

Answer: $F$ and $T$ are mutually exclusive (in this dataset).

Problem 8.2.21 Solution

Step 1 — Apply the definition of mutually exclusive:

\[E \cap F = \varnothing \;\Longrightarrow\; P(E \cap F) = 0.\]

Answer: $P(E \cap F) = 0$.

Problem 8.2.22 Solution

Step 1 — Mutually exclusive simplifies the rule:

\[P(E \cup F) = P(E) + P(F) = 0.4 + 0.2 = 0.6.\]

Answer: $0.6$.

Problem 8.2.23 Solution

Step 1 — Solve the Addition Rule for $P(F)$:

\[P(E \cup F) = P(E) + P(F) - P(E \cap F) \;\Longrightarrow\; P(F) = P(E \cup F) - P(E) + P(E \cap F).\]

Step 2 — Plug in:

\[P(F) = 0.6 - 0.3 + 0.2 = 0.5.\]

Answer: $0.5$.

Problem 8.2.24 Solution

Step 1 — Solve the Addition Rule for the intersection:

\[P(E \cap F) = P(E) + P(F) - P(E \cup F) = 0.4 + 0.5 - 0.7 = 0.2.\]

Answer: $0.2$.

Problem 8.2.25 Solution

Step 1 — Set up the events: $C = \{\text{contains chocolate}\}$, $N = \{\text{contains nuts}\}$. $P(C) = 0.36$, $P(N) = 0.12$, $P(C \cap N) = 0.08$.

Step 2 — Apply the Addition Rule:

\[P(C \cup N) = 0.36 + 0.12 - 0.08 = 0.40.\]

Answer: $40\%$ of cookies contain chocolate or nuts (so Sean cannot eat them).

Problem 8.2.26 Solution

Step 1 — Set up: $P(F) = 0.72$, $P(R) = 0.46$, $P(F \cap R) = 0.32$.

Step 2 — Apply the Addition Rule:

\[P(F \cup R) = 0.72 + 0.46 - 0.32 = 0.86.\]

Answer: $86\%$.