
Probability I
Megan Ayers
Math 141 | Spring 2026
Friday, Week 2
Probability Theory: the study and quantification of uncertainty/randomness in outcomes of repeated experiments.
A random process is one which we know what results could happen, but don’t know which particular result will happen.
An event is a potential result of some particular random process.
We will soon define the word “probability.” Before that, consider the following questions in small groups:
Neither definition is “correct”. Statisticians are divided on how to interpret probability.
The “frequentist” definition of probability is…
Definition: Probability
The probability of a particular event is the proportion of times the event would occur, if we observed the random process an arbitrarily large number of times.
To say that a coin has 50% probability of landing heads, means that…
Since probabilities are defined as a proportion, they will always be values between 0 and 1.
For brevity, we’ll represent statements like the probability of the event “the coin lands heads” is 50% using the notation: \[ P(\textrm{Heads})=0.5 \quad \textrm{or} \quad P(H)=0.5 \]
The probability of an event refers to the long-run tendency of the proportion.

The proportion of heads deviates wildly from 0.5 during the first 200 flips
The probability of an event refers to the long-run tendency of the proportion.

The proportion of heads gets closer to 0.5 after 1000 flips
The probability of an event refers to the long-run tendency of the proportion.

The proportion of heads is very close to 0.5 after 2000 flips!
The probability of an event refers to the long-run tendency of the proportion.

This is the Law of Large Numbers in effect!
Suppose you’re interested in winning the Powerball Jackpot, where the chance of winning is 1 in 292 million. You buy a single lottery ticket and wait to see if you won.
We can formalize the connection between random processes/events with probabilities using a Probability Model
A probability model has two components:
Example: Probability Model for a Coin Toss:
When discussing probability, we always (explicitly or implicitly) define a probability model.
Two events \(A\) and \(B\) are said to be mutually exclusive (or disjoint) if it is not possible for both to occur at the same time.
Example: Single Roll of a Die
Theorem: Addition Rule
The probability that at least one event occurs in a pair of disjoint events is the sum of their individual probabilities: \[ P(A \textrm{ or } B) = P(A) + P(B) \]
Q: When a die is rolled, what is the probability that an odd number is rolled?
\[ \begin{aligned} P( \textrm{Odd} ) &= P(\textrm{roll a 1, 3, or 5}) \end{aligned} \]
Theorem: Addition Rule
The probability that at least one event occurs in a pair of disjoint events is the sum of their individual probabilities: \[ P(A \textrm{ or } B) = P(A) + P(B) \]
Q: When a die is rolled, what is the probability that an odd number is rolled?
\[ \begin{aligned} P( \textrm{Odd} ) &= P(\textrm{roll a 1, 3, or 5})\\ &= P(\textrm{roll a 1}) + P(\textrm{roll a 3}) + P(\textrm{roll a 5}) && \text{(by the Addition Rule)} \end{aligned} \]
Theorem: Addition Rule
The probability that at least one event occurs in a pair of disjoint events is the sum of their individual probabilities: \[ P(A \textrm{ or } B) = P(A) + P(B) \]
Q: When a die is rolled, what is the probability that an odd number is rolled?
\[ \begin{aligned} P( \textrm{Odd} ) &= P(\textrm{roll a 1, 3, or 5})\\ &= P(\textrm{roll a 1}) + P(\textrm{roll a 3}) + P(\textrm{roll a 5}) && \text{(by the Addition Rule)}\\ &= \frac{1}{6} + \frac{1}{6} + \frac{1}{6} \end{aligned} \]
Theorem: Addition Rule
The probability that at least one event occurs in a pair of disjoint events is the sum of their individual probabilities: \[ P(A \textrm{ or } B) = P(A) + P(B) \]
Q: When a die is rolled, what is the probability that an odd number is rolled?
\[ \begin{aligned} P( \textrm{Odd} ) &= P(\textrm{roll a 1, 3, or 5})\\ &= P(\textrm{roll a 1}) + P(\textrm{roll a 3}) + P(\textrm{roll a 5}) && \text{(by the Addition Rule)}\\ &= \frac{1}{6} + \frac{1}{6} + \frac{1}{6} \\ &= \frac{3}{6} \end{aligned} \]
Theorem: Addition Rule
The probability that at least one event occurs in a pair of disjoint events is the sum of their individual probabilities: \[ P(A \textrm{ or } B) = P(A) + P(B) \]
Q: In a die-roll, what’s the probability we get an even number?
Q: In a die-roll, what’s the probability we get at least a 3?
Q: What’s the probability you get an even number OR at least a 3?
The complement to an event \(A\) (denoted \(A^c\)) is the event that occurs exactly when the original does not.
Theorem: Complement Rule
The probability that the complement of an event occurs is 1 minus the probability of the event: \[ P(A^c) = 1 - P(A) \]
What is the probability that any number other than a 1 is rolled on a fair 6-sided die?
\[ P(\textrm{roll something other than a 1}) = 1 - P( \textrm{roll a 1}) = 1 - \frac{1}{6} = \frac{5}{6} \]
Theorem: Addition Rule
The probability that at least one event occurs in a pair of disjoint events is the sum of their individual probabilities: \[ P(A \textrm{ or } B) = P(A) + P(B) \]
Theorem: Complement Rule
The probability that the complement of an event occurs is 1 minus the probability of the event: \[ P(A^c) = 1 - P(A) \]
In the city of Portland, there’s rain on 60% of days and snow on 1% of days.
On a given day, what’s the probability that it doesn’t rain?
On a given day, what’s the probability that it rains OR doesn’t rain?
What if I told you it rains OR snows 61% of days in Portland. What’s the flaw in my reasoning?
In the city of Portland, there’s rain on 60% of days and snow on 1% of days.
\[P(\text{Doesn't Rain}) = P(\text{Rain}^c) = 1-P(\text{Rain}) = 1-0.6=0.4\]
\[P(\text{Rain or Doesn't Rain}) = P(\text{Rain}) + P(\text{Doesn't Rain}) = 0.6+0.4=1\]
It can rain and snow in one day! Thus, we can’t use the addition rule!
Today we defined: