Chapter 1F: Axiomatic Probability and the Addition Rule  
MATH/STAT 394: Probability I  
Arman Jahangiri  
University of Washington Department of Mathematics  
Summer 2026  
Arman Jahangiri  
Chapter 1F: Axiomatic Probability and the Addition Rule  
Summer 2026  
1
 
Learning Goals  
By the end of this lecture, students should be able to:  
▶ state the axioms of probability;  
▶ understand probability as a mathematical function on events;  
▶ derive basic consequences of the axioms;  
▶ apply the addition rule;  
▶ use complements and inclusion–exclusion;  
▶ identify mutually exclusive events;  
▶ understand why axiomatic probability generalizes naive probability.  
Arman Jahangiri  
Chapter 1F: Axiomatic Probability and the Addition Rule  
Summer 2026  
2
Why Axiomatize Probability (or anything)?  
Arman Jahangiri  
Chapter 1F: Axiomatic Probability and the Addition Rule  
Summer 2026  
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Limitations of Naive Probability  
Previously, we defined (naive) probability by  
|A|  
P(A) =  
.
|S|  
This works well when:  
▶ the sample space is finite;  
▶ all outcomes are equally likely.  
But many important situations do not satisfy these assumptions.  
Examples:  
▶ infinitely many outcomes;  
▶ continuous measurements;  
▶ unequal likelihoods;  
▶ etc.  
Arman Jahangiri  
Chapter 1F: Axiomatic Probability and the Addition Rule  
Summer 2026  
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The Axiomatic Point of View  
Instead of defining probability by counting, we define probability abstractly using properties that probabilities  
should satisfy.  
Idea  
A probability measure assigns a number to each event:  
A −→ P(A),  
subject to certain axioms (i.e., principles, rules) .  
The naive definition will become a special case of the general theory.  
Arman Jahangiri  
Chapter 1F: Axiomatic Probability and the Addition Rule  
Summer 2026  
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The Axioms of Probability  
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Chapter 1F: Axiomatic Probability and the Addition Rule  
Summer 2026  
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Probability Space  
A probability model consists of:  
1
S: sample space (set of all possible outcomes in an experiment);  
2
F: a collection of subsets of S (aka ”events”) ;  
3
P(·): a probability function P : F → [0, 1].  
So for any event A ⊆ S (or A ∈ F), P(A) is the probability that the function P(·) assigns to A.  
These 3 elements together form a Probability Model:  
(S, F, P)  
Note that the probability function assigns numbers to sets (events), not to numbers (this might be new for us). So  
we will see things like P([1, 2]), P((−∞, 4]), or P({2, 4, 6})  
Arman Jahangiri  
Chapter 1F: Axiomatic Probability and the Addition Rule  
Summer 2026  
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Axiom 1: Nonnegativity  
Axiom 1  
For every event A ⊆ S,  
P(A) ≥ 0.  
Interpretations:  
▶ probabilities cannot be negative;  
▶ impossible events have probability 0;  
▶ larger probabilities correspond to more likely events.  
Arman Jahangiri  
Chapter 1F: Axiomatic Probability and the Addition Rule  
Summer 2026  
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Axiom 2: Normalization  
Axiom 2  
The probability that ”something happens”, is 1:  
P(S) = 1.  
In other words, P(·) assigns the number 1 to the event S ∈ F.  
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Chapter 1F: Axiomatic Probability and the Addition Rule  
Summer 2026  
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Axiom 3: Additivity  
Axiom 3  
If  
A1, A2, A3, . . .  
are pairwise disjoint events, then  
 
!
∞
∞
[
X
P
Ai = P(Ai ).  
i=1  
i=1  
Pairwise disjoint means:  
Ai ∩ Aj = ∅  
whenever i ̸= j.  
Arman Jahangiri  
Chapter 1F: Axiomatic Probability and the Addition Rule  
Summer 2026 10  
Finite Additivity  
Axiom 3 immediately implies the finite version:  
If  
A1, . . . , An  
are disjoint, then  
P(A1 ∪ · · · ∪ An) = P(A1) + · · · + P(An).  
In particular, if  
then  
A ∩ B = ∅,  
P(A ∪ B) = P(A) + P(B).  
Arman Jahangiri  
Chapter 1F: Axiomatic Probability and the Addition Rule  
Summer 2026 11  
Basic Consequences of the Axioms  
Arman Jahangiri  
Chapter 1F: Axiomatic Probability and the Addition Rule  
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Probability of the Empty Set  
Proposition  
P(∅) = 0.  
Reason:  
S = S ∪ ∅,  
and  
S ∩ ∅ = ∅.  
Using additivity,  
P(S) = P(S) + P(∅).  
Since P(S) = 1, we obtain P(∅) = 0.  
Arman Jahangiri  
Chapter 1F: Axiomatic Probability and the Addition Rule  
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Complements  
Proposition (Complement Rule)  
For every event A,  
P(Ac ) = 1 − P(A).  
Reason:  
A ∪ Ac = S,  
and  
A ∩ Ac = ∅.  
Thus,  
1 = P(S) = P(A) + P(Ac ).  
Arman Jahangiri  
Chapter 1F: Axiomatic Probability and the Addition Rule  
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Probabilities Are Between 0 and 1  
Proposition  
For every event A,  
0 ≤ P(A) ≤ 1.  
Why?  
Axiom 1 gives:  
P(A) ≥ 0.  
P(Ac ) ≥ 0.  
Also,  
Using  
P(Ac ) = 1 − P(A),  
1 − P(A) ≥ 0  
we obtain  
Arman Jahangiri  
Chapter 1F: Axiomatic Probability and the Addition Rule  
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Monotonicity  
Proposition (Monotonicity)  
For any event A, B ⊆ S if A ⊆ B, then  
P(A) ≤ P(B).  
Proof.  
If A ⊆ B, then B = A ∪ (B \ A), where the two events (A and B − A) are disjoint (why?). Thus,  
P(B) = P(A) + P(B \ A) ≥ P(A),  
since P(B − A) ≥ 0 according to Axiom 1.  
Arman Jahangiri  
Chapter 1F: Axiomatic Probability and the Addition Rule  
Summer 2026 16  
Monotonicity: Picture  
S
B
A
B \ A  
B = A ∪ (B \ A)  
A ∩ (B \ A) = ∅  
Since B contains all of A plus the extra part B \ A,  
P(B) = P(A) + P(B \ A) ≥ P(A).  
Arman Jahangiri  
Chapter 1F: Axiomatic Probability and the Addition Rule  
Summer 2026 17  
The Addition Rule  
Arman Jahangiri  
Chapter 1F: Axiomatic Probability and the Addition Rule  
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Why Finite Additivity Is Not Enough  
Suppose A and B are not disjoint.  
Then:  
P(A ∪ B)  
̸= P(A) + P(B).  
Why?  
Because outcomes in  
A ∩ B  
are counted twice.  
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Chapter 1F: Axiomatic Probability and the Addition Rule  
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The Addition Rule  
Proposition: Addition Rule  
For any events A and B,  
P(A ∪ B) = P(A) + P(B) − P(A ∩ B).  
Proof.  
Partition A ∪ B into three disjoint subsets and use a similar trick that was appkied to prove Monotonicity in Slide  
106.  
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Chapter 1F: Axiomatic Probability and the Addition Rule  
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Why the Addition Rule Works  
When adding  
outcomes in  
P(A) + P(B),  
A ∩ B  
are counted twice.  
Therefore we subtract one copy:  
P(A ∪ B) = P(A) + P(B) − P(A ∩ B).  
Arman Jahangiri  
Chapter 1F: Axiomatic Probability and the Addition Rule  
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Example: One Card  
Draw one card from a standard 52-card deck.  
Let  
R = {card is red},  
A = {card is an ace}.  
Then:  
Also,  
26  
4
P(R) =  
, P(A) =  
.
52  
52  
R ∩ A = {Ace of Hearts, Ace of Diamonds},  
so  
2
P(R ∩ A) =  
.
52  
Therefore:  
26  
52  
4
2
28  
52  
7
P(R ∪ A) =  
+
−
=
=
.
52  
52  
13  
Arman Jahangiri  
Chapter 1F: Axiomatic Probability and the Addition Rule  
Summer 2026 22  
Mutually Exclusive Events  
Definition  
Events A and B are called mutually exclusive (or disjoint) if  
A ∩ B = ∅.  
In that case,  
P(A ∩ B) = 0,  
so the addition rule becomes:  
P(A ∪ B) = P(A) + P(B).  
Arman Jahangiri  
Chapter 1F: Axiomatic Probability and the Addition Rule  
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Example: Die Roll  
Roll one fair die.  
Let  
A = {roll is even} = {2, 4, 6},  
B = {roll is odd} = {1, 3, 5}.  
Then  
A ∩ B = ∅.  
Thus  
P(A ∪ B) = P(A) + P(B).  
Indeed,  
1
2
1
2
+
= 1.  
Arman Jahangiri  
Chapter 1F: Axiomatic Probability and the Addition Rule  
Summer 2026 24  
Inclusion–Exclusion  
Arman Jahangiri  
Chapter 1F: Axiomatic Probability and the Addition Rule  
Summer 2026 25  
Three-Set Inclusion–Exclusion  
For three events:  
we first add:  
P(A ∪ B ∪ C)  
P(A) + P(B) + P(C).  
Then subtract pairwise intersections:  
−P(A ∩ B) − P(A ∩ C) − P(B ∩ C).  
But now the triple intersection has been subtracted too many times, so we add it back.  
Arman Jahangiri  
Chapter 1F: Axiomatic Probability and the Addition Rule  
Summer 2026 26  
Inclusion–Exclusion Formula  
Three-Set Inclusion–Exclusion  
P(A ∪ B ∪ C) = P(A) + P(B) + P(C)  
− P(A ∩ B) − P(A ∩ C) − P(B ∩ C)  
+ P(A ∩ B ∩ C).  
Arman Jahangiri  
Chapter 1F: Axiomatic Probability and the Addition Rule  
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Example: Survey Data  
Suppose:  
P(A) = 0.6, P(B) = 0.5, P(A ∩ B) = 0.2.  
What is  
P(A ∪ B)?  
Using the addition rule:  
P(A ∪ B) = 0.6 + 0.5 − 0.2 = 0.9.  
Arman Jahangiri  
Chapter 1F: Axiomatic Probability and the Addition Rule  
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Partitions  
Arman Jahangiri  
Chapter 1F: Axiomatic Probability and the Addition Rule  
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Partitions of the Sample Space  
Definition  
Events  
A1, . . . , An  
form a partition of S if:  
1
Ai ∩ Aj = ∅  
A1 ∪ · · · ∪ An = S.  
A partition divides the sample space into non-overlapping pieces.  
(i ̸= j),  
2
Arman Jahangiri  
Chapter 1F: Axiomatic Probability and the Addition Rule  
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Probability of a Partition  
If  
A1, . . . , An  
form a partition of S, then  
P(A1) + · · · + P(An) = 1.  
Reason:  
The events are disjoint and their union is S.  
Arman Jahangiri  
Chapter 1F: Axiomatic Probability and the Addition Rule  
Summer 2026 31  
Naive Probability as a Special Case  
Arman Jahangiri  
Chapter 1F: Axiomatic Probability and the Addition Rule  
Summer 2026 32  
Recovering Naive Probability  
Suppose:  
▶ S is finite;  
▶ all outcomes are equally likely.  
Let S = {s1, s2, . . . , sn}  
Assign equal probability weights to each outcome:  
1
P(s1) = P(s2) = · · · = P(sn) =  
.
n
Then for an event A, we have  
X
1
1
|A|  
P(A) = P(∪s ∈Asj ) =  
= |A| ×  
=
.
j
n
n
|S|  
i∈A  
Thus the naive definition satisfies the probability axioms.  
Arman Jahangiri  
Chapter 1F: Axiomatic Probability and the Addition Rule  
Summer 2026 33  
Mini-Exercises  
1
If  
P(A) = 0.3,  
what is  
P(Ac )?  
P(A) = 0.6, P(B) = 0.5, P(A ∩ B) = 0.1,  
P(A ∪ B).  
2
If  
compute  
3
Can two disjoint events both have probability 1?  
Answers  
0.7,  
1.0.  
Arman Jahangiri  
Chapter 1F: Axiomatic Probability and the Addition Rule  
Summer 2026 34  
Summary  
▶ Probability is defined axiomatically.  
▶ The axioms are:  
P(A) ≥ 0, P(S) = 1,  
and countable additivity.  
▶ Important consequences include:  
P(∅) = 0, P(Ac ) = 1 − P(A),  
0 ≤ P(A) ≤ 1.  
▶ The addition rule is:  
P(A ∪ B) = P(A) + P(B) − P(A ∩ B).  
▶ Naive probability is a special case of axiomatic probability.  
Next lecture  
Conditional probability and independence.  
Arman Jahangiri  
Chapter 1F: Axiomatic Probability and the Addition Rule  
Summer 2026 35