Complement, Addition, and Multiplication Rules | Malone Math

The Central Idea

The language of events determines the probability rule to use.

\[ \boxed{ \begin{aligned} \text{not }A &\longrightarrow \text{complement rule},\\ A\text{ or }B &\longrightarrow \text{addition rule},\\ A\text{ and }B &\longrightarrow \text{multiplication rule}. \end{aligned} } \]

The main task is not memorizing isolated formulas. It is identifying the event structure correctly before applying arithmetic.

The Complement Rule

An event \(A\) and its complement \(A^c\) partition the sample space:

\[ A\cup A^c=S \]

and

\[ A\cap A^c=\varnothing. \]

Therefore,

\[ P(A)+P(A^c)=1. \]

Solving for either probability gives

\[ \boxed{P(A^c)=1-P(A)} \]

and

\[ \boxed{P(A)=1-P(A^c)}. \]

When the Complement Rule Is Useful

The complement rule is especially useful when the opposite event is easier to describe or count.

Common phrases that suggest a complement include:

not;
none;
at least one;
does not occur;
fails to happen.

Many “at least one” problems are easier to solve by computing the probability of “none” and subtracting from \(1\).

Example: A Direct Complement

Suppose

\[ P(A)=0.37. \]

Then

\[ P(A^c)=1-0.37=\boxed{0.63}. \]

Example: At Least One Head

Three fair coins are tossed.

Let \(A\) be the event “at least one head.”

The complement is

\[ A^c=\{\text{no heads}\}. \]

No heads means

\[ TTT. \]

Thus,

\[ P(A^c)=\frac18, \]

\[ P(A)=1-\frac18=\boxed{\frac78}. \]

The Addition Rule for Disjoint Events

If \(A\) and \(B\) are mutually exclusive, then

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

Their probabilities add directly:

\[ \boxed{P(A\cup B)=P(A)+P(B)}. \]

This is the special addition rule for disjoint events.

Example: Disjoint Die Events

Roll one fair die.

Let

\[ A=\{1\} \]

and

\[ B=\{6\}. \]

These events are disjoint, so

\[ P(A\cup B) = \frac16+\frac16 = \boxed{\frac13}. \]

Why Direct Addition Fails for Overlapping Events

If \(A\) and \(B\) overlap, then

\[ P(A)+P(B) \]

counts the overlap

\[ A\cap B \]

twice.

The general addition rule corrects that double counting.

The General Addition Rule

For any two events \(A\) and \(B\),

\[ \boxed{ P(A\cup B) = P(A)+P(B)-P(A\cap B) }. \]

The intersection is subtracted once because it was included once in \(P(A)\) and once again in \(P(B)\).

Example: Overlapping Events

Suppose

\[ P(A)=0.60, \qquad P(B)=0.45, \qquad P(A\cap B)=0.25. \]

Then

\[ P(A\cup B) = 0.60+0.45-0.25 = \boxed{0.80}. \]

Example: Heart or King

One card is drawn from a standard \(52\)-card deck.

Let \(A\) be “heart” and \(B\) be “king.”

Then

\[ P(A)=\frac{13}{52}, \] \[ P(B)=\frac4{52}, \]

and the king of hearts lies in the overlap:

\[ P(A\cap B)=\frac1{52}. \]

Therefore,

\[ P(A\cup B) = \frac{13}{52}+\frac4{52}-\frac1{52} = \boxed{\frac{16}{52}=\frac4{13}}. \]

The Phrase “Or” Usually Signals a Union

In probability,

\[ A\text{ or }B \]

normally means

\[ A\cup B. \]

This is inclusive “or,” so outcomes in both events are included unless the problem explicitly says “but not both” or “exactly one.”

Exactly One of Two Events

“Exactly one of \(A\) and \(B\)” excludes the overlap.

The event is

\[ (A\setminus B)\cup(B\setminus A). \]

In terms of probabilities,

\[ \boxed{ P(\text{exactly one of }A,B) = P(A)+P(B)-2P(A\cap B) }. \]

Neither Event

“Neither \(A\) nor \(B\)” means

\[ (A\cup B)^c. \]

Therefore,

\[ \boxed{ P(\text{neither }A\text{ nor }B) = 1-P(A\cup B) }. \]

Using the addition rule,

\[ P(\text{neither}) = 1-P(A)-P(B)+P(A\cap B). \]

The Multiplication Rule Begins with Conditional Probability

The probability that both \(A\) and \(B\) occur depends on how likely \(B\) is when \(A\) has already occurred.

This conditional probability is written

\[ P(B\mid A). \]

Read this as

“the probability of \(B\) given \(A\).”

The General Multiplication Rule

The joint probability of \(A\) and \(B\) is

\[ \boxed{ P(A\cap B)=P(A)P(B\mid A) }. \]

Equivalently,

\[ \boxed{ P(A\cap B)=P(B)P(A\mid B) }. \]

These formulas apply whether or not the events are independent.

Why the Multiplication Rule Works

The conditional probability

\[ P(B\mid A) \]

measures the fraction of the probability of \(A\) that also lies in \(B\).

Multiplying

\[ P(A) \]

by that conditional fraction produces the probability of the overlap:

\[ P(A\cap B). \]

Example: Two Draws Without Replacement

A standard deck contains \(52\) cards, including \(4\) aces. Two cards are drawn without replacement.

The probability that the first card is an ace is

\[ \frac4{52}. \]

Given that the first card was an ace, \(3\) aces remain among \(51\) cards:

\[ P(\text{second ace}\mid\text{first ace})=\frac3{51}. \]

Therefore,

\[ P(\text{two aces}) = \frac4{52}\cdot\frac3{51} = \boxed{\frac1{221}}. \]

Sequential Probability Trees

A probability tree organizes multistage experiments.

Along a path, probabilities multiply.

Across mutually exclusive completed paths, probabilities add.

This gives a useful operational rule:

\[ \boxed{\text{multiply along branches, add across disjoint paths}}. \]

Example: Two Coin Tosses

For two fair independent coin tosses,

\[ P(HH) = P(H)\,P(H\mid H). \]

Independence gives

\[ P(H\mid H)=P(H)=\frac12. \]

Therefore,

\[ P(HH) = \frac12\cdot\frac12 = \boxed{\frac14}. \]

Independent Events

Events \(A\) and \(B\) are independent when the occurrence of one does not change the probability of the other.

One equivalent condition is

\[ P(B\mid A)=P(B), \]

when \(P(A)>0\).

Under independence, the general multiplication rule simplifies.

The Multiplication Rule for Independent Events

If \(A\) and \(B\) are independent, then

\[ P(B\mid A)=P(B). \]

Therefore,

\[ \boxed{ P(A\cap B)=P(A)P(B) }. \]

This simplified product rule should be used only when independence is justified.

Example: Independent Dice

Two fair dice are rolled independently.

Let \(A\) be “the first die shows \(6\)” and \(B\) be “the second die shows \(6\).”

Then

\[ P(A)=\frac16, \qquad P(B)=\frac16. \]

Therefore,

\[ P(A\cap B) = \frac16\cdot\frac16 = \boxed{\frac1{36}}. \]

Dependence Changes the Second Factor

When events are dependent,

\[ P(B\mid A)\neq P(B) \]

in general.

The multiplication rule remains

\[ \boxed{ P(A\cap B)=P(A)P(B\mid A) }. \]

The conditional probability must reflect the changed situation.

With Replacement Often Preserves the Original Probabilities

Suppose one object is selected from a collection and then returned before the next selection.

If the selection mechanism is otherwise unchanged, the second draw begins from the same composition as the first.

This often makes repeated draws independent.

Without Replacement Usually Creates Dependence

If a selected object is not returned, the composition of the remaining collection changes.

Therefore, later probabilities generally depend on earlier outcomes.

The correct multiplication rule must use conditional probabilities.

Example: Two Red Cards Without Replacement

A standard deck has \(26\) red cards.

The probability that the first card is red is

\[ \frac{26}{52}. \]

Given that the first card was red, \(25\) red cards remain among \(51\) cards.

Therefore,

\[ P(\text{two red cards}) = \frac{26}{52}\cdot\frac{25}{51} = \boxed{\frac{25}{102}}. \]

Example: Two Red Cards With Replacement

If the first card is returned and the deck is reshuffled before the second draw, then each draw has probability

\[ \frac{26}{52}=\frac12 \]

of being red.

Thus,

\[ P(\text{two red cards}) = \frac12\cdot\frac12 = \boxed{\frac14}. \]

At Least One in Repeated Independent Trials

Suppose an event has probability \(p\) on each of \(n\) independent trials.

The probability that it fails on one trial is

\[ 1-p. \]

The probability that it fails on all \(n\) trials is

\[ (1-p)^n. \]

Therefore,

\[ \boxed{ P(\text{at least one occurrence}) = 1-(1-p)^n }. \]

Example: At Least One Six in Four Rolls

A fair die is rolled four times independently.

The probability of not rolling a \(6\) on one roll is

\[ \frac56. \]

The probability of no sixes in all four rolls is

\[ \left(\frac56\right)^4. \]

Therefore,

\[ \boxed{ P(\text{at least one }6) = 1-\left(\frac56\right)^4 }. \]

Three-Event Multiplication Rule

For three events,

\[ \boxed{ P(A\cap B\cap C) = P(A)\, P(B\mid A)\, P(C\mid A\cap B) }. \]

Each stage is conditioned on all relevant information from the earlier stages.

General Chain Rule

For events \(A_1,A_2,\ldots,A_n\),

\[ \boxed{ \begin{aligned} P(A_1\cap\cdots\cap A_n) &= P(A_1) P(A_2\mid A_1)\\ &\quad\cdot P(A_3\mid A_1\cap A_2) \cdots P(A_n\mid A_1\cap\cdots\cap A_{n-1}). \end{aligned} } \]

This is the multiplication or chain rule for a sequence of events.

Example: Three Aces Without Replacement

Three cards are drawn from a standard deck without replacement.

The probability that all three are aces is

\[ \frac4{52} \cdot \frac3{51} \cdot \frac2{50}. \]

Thus,

\[ \boxed{ P(\text{three aces}) = \frac4{52}\cdot\frac3{51}\cdot\frac2{50} }. \]

Addition and Multiplication Work Together

Many probability problems contain several disjoint ways for an event to occur.

Within each path, probabilities multiply.

Across mutually exclusive paths, probabilities add.

This combination is one of the most important recurring structures in elementary probability.

Example: Exactly One Head in Two Tosses

Exactly one head can occur through two disjoint paths:

\[ HT \]

\[ TH. \]

Each path has probability

\[ \frac12\cdot\frac12=\frac14. \]

Therefore,

\[ P(\text{exactly one head}) = \frac14+\frac14 = \boxed{\frac12}. \]

Example: Exactly One Six in Two Rolls

Exactly one six can occur as

\[ (6,\text{not }6) \]

\[ (\text{not }6,6). \]

These paths are disjoint.

Therefore,

\[ P(\text{exactly one }6) = \frac16\cdot\frac56 + \frac56\cdot\frac16 = \boxed{\frac{10}{36}=\frac5{18}}. \]

Mutually Exclusive and Independent Are Different

Mutually exclusive events satisfy

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

Independent events satisfy

\[ P(A\cap B)=P(A)P(B). \]

These are different relationships.

Nontrivial Mutually Exclusive Events Are Not Independent

Suppose

\[ P(A)>0 \]

and

\[ P(B)>0. \]

If \(A\) and \(B\) are mutually exclusive, then

\[ P(A\cap B)=0. \]

But

\[ P(A)P(B)>0. \]

Therefore,

\[ P(A\cap B)\neq P(A)P(B), \]

so the events are not independent.

Testing Independence with Probabilities

If the relevant probabilities are known, one way to test independence is to compare

\[ P(A\cap B) \]

with

\[ P(A)P(B). \]

\[ \boxed{ P(A\cap B)=P(A)P(B) }, \]

then \(A\) and \(B\) are independent.

Example: Testing Independence

Suppose

\[ P(A)=0.4, \qquad P(B)=0.5, \qquad P(A\cap B)=0.2. \]

Since

\[ P(A)P(B)=0.4(0.5)=0.2, \]

we have

\[ P(A\cap B)=P(A)P(B). \]

Therefore, \(A\) and \(B\) are independent.

Example: Detecting Dependence

Suppose

\[ P(A)=0.4, \qquad P(B)=0.5, \qquad P(A\cap B)=0.30. \]

But

\[ P(A)P(B)=0.20. \]

Since

\[ 0.30\neq0.20, \]

the events are dependent.

The Addition Rule for Three Events

For three events,

\[ \boxed{ \begin{aligned} P(A\cup B\cup C) &= P(A)+P(B)+P(C)\\ &\quad -P(A\cap B) -P(A\cap C) -P(B\cap C)\\ &\quad +P(A\cap B\cap C). \end{aligned} } \]

Pairwise overlaps are subtracted, and the triple overlap is added back because it was then removed too many times.

Connection to Inclusion–Exclusion

The general addition rule is the probability form of the inclusion–exclusion principle.

Counting and probability therefore share the same structural correction for overlapping sets.

This connection becomes especially useful in counting-based probability.

Example: Three Overlapping Events

Suppose

\[ P(A)=0.5,\quad P(B)=0.4,\quad P(C)=0.3, \] \[ P(A\cap B)=0.2,\quad P(A\cap C)=0.1,\quad P(B\cap C)=0.15, \]

and

\[ P(A\cap B\cap C)=0.05. \]

Then

\[ \begin{aligned} P(A\cup B\cup C) &= 0.5+0.4+0.3\\ &\quad-0.2-0.1-0.15+0.05\\ &=\boxed{0.80}. \end{aligned} \]

“At Most One” Can Be Handled by a Complement

For two events \(A\) and \(B\), “at most one occurs” means “not both.”

Therefore,

\[ P(\text{at most one}) = 1-P(A\cap B). \]

Thus,

\[ \boxed{ P(\text{at most one of }A,B) = 1-P(A\cap B) }. \]

“At Least One” Across Several Events

If \(A_1,\ldots,A_n\) represent occurrence on separate opportunities, then the event “at least one occurs” is

\[ A_1\cup\cdots\cup A_n. \]

Its complement is

\[ A_1^c\cap\cdots\cap A_n^c. \]

Therefore,

\[ \boxed{ P(\text{at least one}) = 1-P(\text{none}) }. \]

Example: Defective Items

Suppose each independently produced item has probability

\[ 0.02 \]

of being defective.

The probability that one item is not defective is

\[ 0.98. \]

For five independent items,

\[ P(\text{none defective})=0.98^5. \]

Therefore,

\[ \boxed{ P(\text{at least one defective}) = 1-0.98^5 }. \]

Example: Success Followed by Failure

Suppose independent trials have probability

\[ P(\text{success})=0.7. \]

Then

\[ P(\text{failure})=0.3. \]

The probability of success followed by failure is

\[ 0.7(0.3)=\boxed{0.21}. \]

Example: A Conditional Branch

Suppose

\[ P(A)=0.6 \]

and

\[ P(B\mid A)=0.4. \]

Then

\[ P(A\cap B) = 0.6(0.4) = \boxed{0.24}. \]

No independence assumption is needed because the conditional probability is given directly.

Recovering a Conditional Probability

From

\[ P(A\cap B)=P(A)P(B\mid A), \]

if \(P(A)>0\), then

\[ \boxed{ P(B\mid A) = \frac{P(A\cap B)}{P(A)} }. \]

This formula will be studied directly in the later lesson on conditional probability.

Example: Recovering a Conditional Probability

Suppose

\[ P(A)=0.5 \]

and

\[ P(A\cap B)=0.2. \]

Then

\[ P(B\mid A) = \frac{0.2}{0.5} = \boxed{0.4}. \]

Common Error: Adding When You Should Multiply

“\(A\) and \(B\)” refers to

\[ A\cap B, \]

not \(A\cup B\).

Sequential or joint occurrence generally requires a multiplication rule, not an addition rule.

Common Error: Multiplying When You Should Add

If an event can occur through several mutually exclusive alternatives, their probabilities add.

For example, “roll a \(1\) or a \(6\)” on one fair die is

\[ \frac16+\frac16=\frac13, \]

not

\[ \frac16\cdot\frac16. \]

Common Error: Forgetting the Overlap in the Addition Rule

For overlapping events,

\[ P(A\cup B) \neq P(A)+P(B) \]

in general.

The correct formula is

\[ \boxed{ P(A\cup B) = P(A)+P(B)-P(A\cap B) }. \]

Common Error: Assuming Independence Without Justification

The formula

\[ P(A\cap B)=P(A)P(B) \]

requires independence.

Without replacement, changing populations, conditional selection rules, and many real-world processes create dependence.

When independence is not justified, use

\[ P(A\cap B)=P(A)P(B\mid A). \]

Common Error: Confusing Disjointness with Independence

Disjoint events cannot happen together.

Independent events do not change one another's probabilities.

Nontrivial disjoint events with positive probability are not independent.

Common Error: Forgetting That Probabilities Change Without Replacement

If an object is removed and not replaced, both the numerator and denominator can change before the next draw.

Reusing the original fraction at every stage can therefore be incorrect.

Common Error: Using the Complement of the Wrong Event

“At least one success” is complemented by

“no successes.”

It is not complemented by “exactly one failure” or by “at least one failure.”

Translate the event carefully before subtracting from \(1\).

A Reliable Procedure for “Not” Problems

Define the event \(A\).
Describe its exact complement \(A^c\).
Compute whichever of \(P(A)\) or \(P(A^c)\) is easier.
Use \(P(A)=1-P(A^c)\) or \(P(A^c)=1-P(A)\).
Check that the result lies in \([0,1]\).

A Reliable Procedure for “Or” Problems

Define events \(A\) and \(B\).
Determine whether they overlap.
If disjoint, add directly.
If overlapping, compute or identify \(P(A\cap B)\).
Use \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\).
Check that the result does not exceed \(1\).

A Reliable Procedure for “And” Problems

Identify the stages or simultaneous conditions.
Determine whether later probabilities depend on earlier outcomes.
Use conditional probabilities when dependence is present.
Multiply probabilities along the required path.
If several disjoint paths satisfy the event, add their path probabilities.
Check that each conditional probability matches the state of the experiment at that stage.

Example: Full Combined Workflow

A bag contains \(5\) red and \(3\) blue objects. Two objects are drawn without replacement.

Find the probability of drawing one red and one blue in either order.

The event can occur through two disjoint paths:

\[ R\text{ then }B \]

\[ B\text{ then }R. \]

Their probabilities are

\[ \frac58\cdot\frac37 \]

and

\[ \frac38\cdot\frac57. \]

Therefore,

\[ P(\text{one of each}) = \frac58\cdot\frac37 + \frac38\cdot\frac57 = \boxed{\frac{15}{28}}. \]

Practice 1: Complement

\[ P(A)=0.72, \]

find \(P(A^c)\).

Practice 2: Disjoint Addition

Suppose \(A\cap B=\varnothing\),

\[ P(A)=0.31, \qquad P(B)=0.28. \]

Find \(P(A\cup B)\).

Practice 3: General Addition

Suppose

\[ P(A)=0.55, \qquad P(B)=0.40, \qquad P(A\cap B)=0.18. \]

Find \(P(A\cup B)\).

Practice 4: Neither

Using the probabilities in Practice 3, find the probability that neither \(A\) nor \(B\) occurs.

Practice 5: Independent Multiplication

Suppose \(A\) and \(B\) are independent,

\[ P(A)=0.6, \qquad P(B)=0.35. \]

Find \(P(A\cap B)\).

Practice 6: Conditional Multiplication

Suppose

\[ P(A)=0.45 \]

and

\[ P(B\mid A)=0.30. \]

Find \(P(A\cap B)\).

Practice 7: Two Cards Without Replacement

Two cards are drawn from a standard deck without replacement.

Find the probability that both are kings.

Practice 8: At Least One Six

A fair die is rolled three times independently.

Find the probability of rolling at least one \(6\).

Practice 9: Exactly One Head

Two fair coins are tossed independently.

Find the probability of getting exactly one head.

Practice 10: Test Independence

Suppose

\[ P(A)=0.25, \qquad P(B)=0.40, \qquad P(A\cap B)=0.10. \]

Are \(A\) and \(B\) independent?

Practice 11: Exactly One Type

A bag contains \(4\) red and \(6\) blue objects. Two objects are drawn without replacement.

Find the probability of drawing one red and one blue in either order.

Practice 12: Identify the Error

A student draws two aces without replacement and writes

\[ P(\text{two aces}) = \frac4{52}\cdot\frac4{52}. \]

Explain the error and write the correct product.

Practice Answers

Practice 1

\[ P(A^c)=1-0.72=\boxed{0.28}. \]

Practice 2

Because the events are disjoint,

\[ P(A\cup B) = 0.31+0.28 = \boxed{0.59}. \]

Practice 3

\[ P(A\cup B) = 0.55+0.40-0.18 = \boxed{0.77}. \]

Practice 4

\[ P(\text{neither}) = 1-0.77 = \boxed{0.23}. \]

Practice 5

\[ P(A\cap B) = 0.6(0.35) = \boxed{0.21}. \]

Practice 6

\[ P(A\cap B) = 0.45(0.30) = \boxed{0.135}. \]

Practice 7

\[ P(\text{two kings}) = \frac4{52}\cdot\frac3{51} = \boxed{\frac1{221}}. \]

Practice 8

The probability of no sixes is

\[ \left(\frac56\right)^3. \]

Therefore,

\[ \boxed{ P(\text{at least one }6) = 1-\left(\frac56\right)^3 = \frac{91}{216} }. \]

Practice 9

The two disjoint paths are \(HT\) and \(TH\):

\[ \frac12\cdot\frac12 + \frac12\cdot\frac12 = \boxed{\frac12}. \]

Practice 10

Compute

\[ P(A)P(B)=0.25(0.40)=0.10. \]

Since

\[ P(A\cap B)=0.10, \]

the events are independent.

Practice 11

The two paths are red then blue and blue then red:

\[ \frac4{10}\cdot\frac69 + \frac6{10}\cdot\frac49. \]

Therefore,

\[ \boxed{ P(\text{one red and one blue}) = \frac8{15} }. \]

Practice 12

Without replacement, the first ace is removed. After drawing one ace, \(3\) aces remain among \(51\) cards.

The correct product is

\[ \boxed{ \frac4{52}\cdot\frac3{51} }. \]

What to Remember

The complement rule is

\[ \boxed{P(A^c)=1-P(A)}. \]

The general addition rule is

\[ \boxed{ P(A\cup B) = P(A)+P(B)-P(A\cap B) }. \]

The general multiplication rule is

\[ \boxed{ P(A\cap B) = P(A)P(B\mid A) }. \]

For independent events,

\[ \boxed{ P(A\cap B)=P(A)P(B) }. \]

Use complements for “not,” “none,” and often “at least one.”
Add probabilities across mutually exclusive alternatives.
Subtract overlap when using the general addition rule.
Multiply along sequential paths.
Use conditional probabilities when later stages depend on earlier ones.
Do not confuse independence with mutual exclusivity.
For several disjoint successful paths, multiply within each path and add across paths.

The most important habit is: translate the event structure first, then choose the probability rule.

Where We Go Next

Complement, addition, and multiplication rules provide the algebra for combining events. Many finite probability problems still require an efficient way to determine how many outcomes satisfy those events.

The next lesson, Counting-Based Probability, combines probability with permutations, combinations, and other counting methods to handle sample spaces that are too large to list directly.