Chapter 1B: Naive Probability  
MATH/STAT 394: Probability I  
Arman Jahangiri  
University of Washington Department of Mathematics  
Summer 2026  
Arman Jahangiri  
Chapter 1B: Naive Probability  
Summer 2026  
1
 
Learning Goals  
By the end of this lecture, students should be able to:  
▶ state the naive definition of probability;  
▶ identify when the naive definition is appropriate;  
▶ compute probabilities by counting favorable outcomes;  
▶ use complements to simplify probability calculations;  
▶ recognize common incorrect “equally likely” arguments;  
▶ distinguish between a probability model and reality.  
Arman Jahangiri  
Chapter 1B: Naive Probability  
Summer 2026  
2
Naive Probability  
Arman Jahangiri  
Chapter 1B: Naive Probability  
Summer 2026  
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From Events to Probabilities  
In the previous lecture:  
S = {all possible outcomes},  
Now we want to assign a number to an event:  
A ⊆ S.  
P(A) = probability that A occurs.  
Question  
If every outcome in a finite sample space is equally likely, how should we define P(A)?  
Arman Jahangiri  
Chapter 1B: Naive Probability  
Summer 2026  
4
Definition: Naive Probability  
Naive Definition of Probability  
Let A be an event for an experiment with a finite sample space S.  
If all outcomes in S are equally likely, then  
|A|  
number of outcomes favorable to A  
Pnaive(A) =  
=
.
|S|  
total number of outcomes in S  
Here |A| denotes the number of elements in the set A.  
Question: Why does this definition leads to equally-likely outcomes?  
Arman Jahangiri  
Chapter 1B: Naive Probability  
Summer 2026  
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Naive Probability - Properties  
Let S be finite, all outcomes equally likely, and let  
|A|  
P(A) =  
.
|S|  
Then naive probability satisfies:  
0 ≤ P(A) ≤ 1  
for every A ⊆ S.  
Why? Since A ⊆ S,  
0 ≤ |A| ≤ |S|,  
so  
|A|  
0 ≤  
≤ 1.  
|S|  
Also,  
|S|  
|S|  
0
P(S) =  
= 1,  
P(∅) =  
= 0.  
|S|  
Arman Jahangiri  
Chapter 1B: Naive Probability  
Summer 2026  
6
Additivity for Disjoint Events  
Suppose A, B ⊆ S and A ∩ B = ∅.  
Then A and B do not overlap, so  
|A ∪ B| = |A| + |B|.  
Therefore,  
|A ∪ B|  
|A| + |B|  
|A|  
|B|  
P(A ∪ B) =  
=
=
+
.
|S|  
|S|  
|S|  
|S|  
Thus  
P(A ∪ B) = P(A) + P(B)  
whenever A ∩ B = ∅.  
Looking ahead  
Later, we will define probability more generally using properties like these as axioms. Naive probability will  
then become a special case of the general theory.  
Arman Jahangiri  
Chapter 1B: Naive Probability  
Summer 2026  
7
What the Definition Means  
The naive definition says:  
how many outcomes make A happen  
P(A) =  
.
how many outcomes are possible  
This requires two assumptions:  
1
The sample space S is finite.  
All outcomes in S are equally likely.  
2
Important  
The naive definition is not wrong. It is just limited. It should only be used when its assumptions are justified.  
Arman Jahangiri  
Chapter 1B: Naive Probability  
Summer 2026  
8
Pebble World Interpretation  
Suppose each pebble has the same mass.  
number of pebbles in A  
P(A) =  
.
number of pebbles in S  
S
A
|A|  
P(A) =  
.
|S|  
Arman Jahangiri  
Chapter 1B: Naive Probability  
Summer 2026  
9
Example: One Fair Die  
Roll one fair six-sided die.  
S = {1, 2, 3, 4, 5, 6}.  
Let  
A = {roll is even} = {2, 4, 6}.  
Then  
|S| = 6,  
|A| = 3.  
Therefore  
|A|  
3
6
1
2
P(A) =  
=
=
.
|S|  
Arman Jahangiri  
Chapter 1B: Naive Probability  
Summer 2026 10  
Example: One Card  
Draw one card from a standard well-shuffled 52-card deck.  
Let  
A = {card is an ace}.  
There are 4 aces and 52 cards total, so  
|A|  
4
1
P(A) =  
=
=
.
|S|  
52  
13  
Let  
H = {card is a heart}.  
Then  
13  
52  
1
4
P(H) =  
=
.
Arman Jahangiri  
Chapter 1B: Naive Probability  
Summer 2026 11  
Example: Red or Ace  
Draw one card from a standard 52-card deck.  
Let  
R = {card is red},  
A = {card is an ace}.  
We want  
P(R ∪ A).  
There are:  
So  
26 red cards,  
4 aces,  
2 red aces.  
|R ∪ A| = 26 + 4 − 2 = 28.  
Therefore  
28  
52  
7
P(R ∪ A) =  
=
.
13  
Arman Jahangiri  
Chapter 1B: Naive Probability  
Summer 2026 12  
Why Did We Subtract?  
In the previous example,  
|R ∪ A| = |R| + |A|.  
̸
Why? The red aces were counted twice:  
R ∩ A = {Ace of Hearts, Ace of Diamonds}.  
Thus  
|R ∪ A| = |R| + |A| − |R ∩ A|.  
Preview  
This is the counting version of the addition rule:  
P(A ∪ B) = P(A) + P(B) − P(A ∩ B).  
Arman Jahangiri  
Chapter 1B: Naive Probability  
Summer 2026 13  
Complements  
Arman Jahangiri  
Chapter 1B: Naive Probability  
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Complement Rule Under the Naive Definition  
Let A ⊆ S.  
The complement of A is  
Ac = {s ∈ S : s ∈/ A}.  
Since every outcome is either in A or not in A,  
|Ac | = |S| − |A|.  
Therefore,  
|Ac |  
|S| − |A|  
|A|  
P(Ac ) =  
=
= 1 −  
.
|S|  
|S|  
|S|  
So  
P(Ac ) = 1 − P(A).  
Arman Jahangiri  
Chapter 1B: Naive Probability  
Summer 2026 15  
Why Complements Are Useful  
Sometimes it is easier to count when an event does not happen.  
Instead of computing P(A) directly, compute:  
P(A) = 1 − P(Ac ).  
Strategy  
When a probability asks for phrases like  
“at least one”,  
“at least two”,  
“not all”,  
first ask whether the complement is easier.  
Arman Jahangiri  
Chapter 1B: Naive Probability  
Summer 2026 16  
Example: At Least One Head  
Flip a fair coin 5 times.  
Let  
A = {at least one head occurs}.  
The complement is  
Ac = {no heads occur}.  
There is only one outcome with no heads: TTTTT.  
The total number of outcomes is 25 = 32.  
Therefore  
1
1
31  
32  
P(Ac ) =  
,
P(A) = 1 −  
=
.
32  
32  
Arman Jahangiri  
Chapter 1B: Naive Probability  
Summer 2026 17  
Example: At Least One Six  
Roll a fair die 4 times.  
Let  
A = {at least one six occurs}.  
It is easier to count  
Ac = {no six occurs}.  
Total outcomes: 64. Outcomes with no six: 54.  
Therefore,  
54  
64  
P(A) = 1 − P(Ac ) = 1 −  
.
Arman Jahangiri  
Chapter 1B: Naive Probability  
Summer 2026 18  
Common Mistakes  
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Chapter 1B: Naive Probability  
Summer 2026 19  
Biohazard: “Either It Happens or It Does Not”  
A common mistake is to argue:  
1
2
Either A happens or A does not happen, so P(A) =  
.
This is usually wrong.  
Why?  
A
and Ac  
are two events, but they do not have to be equally likely.  
Key point  
Having two possibilities does not imply that the two possibilities have equal probability.  
Arman Jahangiri  
Chapter 1B: Naive Probability  
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Example: Two Fair Dice  
Roll two fair dice.  
A correct sample space is  
S = {(i, j) : i, j ∈ {1, 2, 3, 4, 5, 6}}.  
Here (i, j) means:  
i = result of die 1,  
j = result of die 2.  
Thus  
|S| = 36.  
Important  
The outcomes are ordered pairs. For example, (5, 6) and (6, 5) are different outcomes.  
Arman Jahangiri  
Chapter 1B: Naive Probability  
Summer 2026 21  
Example: Sum 11 versus Sum 12  
For two fair dice:  
S = {(i, j) : i, j ∈ {1, . . . , 6}}.  
Let  
A = {sum is 11},  
B = {sum is 12}.  
Then  
So  
A = {(5, 6), (6, 5)},  
B = {(6, 6)}.  
2
1
,
P(A) =  
=
36  
18  
1
36  
while P(B) =  
.
Thus a sum of 11 is twice as likely as a sum of 12.  
Arman Jahangiri  
Chapter 1B: Naive Probability  
Summer 2026 22  
Why Labeling Matters  
If the dice are not labeled, someone might say:  
11 = 5 + 6,  
12 = 6 + 6.  
Then it may look as if each sum occurs in one way.  
But the dice are physically distinguishable:  
(5, 6) ̸= (6, 5).  
Lesson  
When using the naive definition, choose a sample space whose outcomes are equally likely. Labeling objects  
often prevents counting mistakes.  
Arman Jahangiri  
Chapter 1B: Naive Probability  
Summer 2026 23  
Mini-Exercises  
1
A fair die is rolled. What is the probability of rolling a number at least 5?  
2
A fair coin is flipped 4 times. What is the probability of getting all tails?  
3
A card is drawn from a standard deck. What is the probability that it is not a heart?  
4
Two fair dice are rolled. What is the probability that the sum is 7?  
5
Why is the statement “the probability of rain tomorrow is 1/2 because either it rains or it does not” invalid?  
Optional answers  
2
6
1
3
1
13  
52  
3
4
6
1
6
=
,
,
1 −  
=
,
=
.
16  
36  
Arman Jahangiri  
Chapter 1B: Naive Probability  
Summer 2026 24  
Summary  
▶ The naive definition is  
|A|  
P(A) =  
.
|S|  
▶ It applies when S is finite and all outcomes are equally likely.  
▶ Naive probability satisfies:  
0 ≤ P(A) ≤ 1,  
P(S) = 1,  
P(∅) = 0,  
If A ∩ B = ∅, then  
P(A ∪ B) = P(A) + P(B).  
▶ Complements are often useful:  
P(Ac ) = 1 − P(A).  
Next lecture  
Counting methods: multiplication rule, sampling with replacement, and sampling without replacement.  
Arman Jahangiri  
Chapter 1B: Naive Probability  
Summer 2026 25