Basic Probability
Elementary Statistics
MTH-161D | Spring 2025 | University of Portland
February 19, 2025
Objectives
- Introduce basic probability rules
- Know how to compute probabilities using the basic
rules
- Develop an understanding of interpreting
probability
- Activity: Taking Samples from Dice Rolls
These slides are derived from Diez et al.
(2012).
Previously… (1/2)
The guiding principle of statistics is statistical thinking.
Previously… (2/2)
Probability is the Basis for Inference
- Probability provides a framework for drawing
conclusions about a population from a sample.
- It helps quantify uncertainty in estimates and
decisions.
The P-Value is a Probability
- The p-value measures the strength of
evidence against the null hypothesis.
Probability and Statistics
Probability
- A measure on how likely an event occurs
- Computing probabilities have specific rules
- Logical reasoning
- One answer
Statistics
- It’s an art and science
- Collecting, analyzing, interpreting, and presenting data
- Data-driven approach to make conclusions and prediction
- Multiple ways to solve problems
Basic Probability Definition
Probability is the branch of mathematics that deals
with randomness. The likelihood of an outcome happening.
An extent to which an outcome is likely to occur is \[\text{probability} = \frac{\text{number of
favorable outcomes}}{\text{total number of outcomes}}.\]
Coin
Fair Coin
- Possible outcomes: \(H\)
or \(T\) (\(H\) for heads, \(T\) for tails)
- Total number of possible outcomes: \(2\)
- Probabilities: \[
\begin{aligned}
\text{probability of } H & = \frac{1}{2} \\
\text{probability of } T & = \frac{1}{2}
\end{aligned}
\]
Dice
Fair Dice
- Possible outcomes: 1 (⚀), 2 (⚁), 3 (⚂), 4 (⚃), 5 (⚄), or 6
(⚅)
- Total number of possible outcomes: \(6\)
- Probabilities: \[
\begin{aligned}
\text{probability of } 1 & = \frac{1}{6} & \text{probability of
} 4 & = \frac{1}{6} \\
\text{probability of } 2 & = \frac{1}{6} & \text{probability of
} 5 & = \frac{1}{6} \\
\text{probability of } 3 & = \frac{1}{6} & \text{probability of
} 6 & = \frac{1}{6}
\end{aligned}
\]
Standard Deck of Cards
52-Card Deck
- Possible outcomes: The four suits are Hearts (♥), Diamonds
(♦), Clubs (♣), Spades (♠). Each suit has 13 ranks: Ace (A), 2, 3, 4, 5,
6, 7, 8, 9, 10, Jack (J), Queen (Q), and King (K).
- Total number of possible outcomes: \(52\)
- Probabilities: \[
\text{probability of Q of Hearts } = \frac{1}{52}
\] Actually, the probability of any card drawn once is \(\frac{1}{52}\).
Probability Notations (1/2)
We will use specific words for outcomes.
- A set of possible outcomes is called the sample
space.
- Any subset of \(S\) are called
events.
- An event space is a set all subsets of outcomes of
the sample space.
Fair Coin Example:
- Sample space: \(S = \{H,T\}\)
- Events space: \(\{H\}\), \(\{T\}\), \(\{H,T\}\), \(\emptyset\)
- Two events from \(S\): \(\{H\}\) and \(\{T\}\)
Probability Notations (2/2)
We will use specific notations for
probabilities.
Let \(A\) be an event with a finite
sample space \(S\). The probability of
\(A\) is \[P(A) = \frac{|A|}{|S|} \longrightarrow P(A) =
\frac{\text{number of outcome favorable to } A}{\text{total number of
outcomes in } S}.\]
Fair Coin Example:
\[
\begin{aligned}
\text{probability of } H & = \frac{1}{2} \longrightarrow P(H) =
\frac{1}{2} \\
\text{probability of } T & = \frac{1}{2} \longrightarrow P(T) =
\frac{1}{2}
\end{aligned}
\]
Independence
Two events, \(A\) and \(B\), are independent if
the occurrence of one does not affect the probability of the
other: \[P(A \text{ and } B) =
P(A)P(B)\]
If the event \(B\) is
dependent on \(A\),
then \[P(A \text{ and } B) \ne
P(A)P(B)\]
\(\star\) Key Idea:
Independent events is when one event happening does not affect the
other. Disjoint events is when one event happening prevents the
other.
Coin Flips
Suppose we conduct an experiment of flipping fair coins in sequence
and record the outcomes.
- One Coin: \(H\) or \(T\) (two possible outcomes)
- \(P(H) = \frac{1}{2}\) and \(P(T) = \frac{1}{2}\) but \(P(H \text{ and } T) = 0\) because they
can’t occur simultaneously
- Two Coins: \(HH\), \(HT\), \(TH\), or \(TT\) (four possible outcomes)
- \(P(H \text{ and } H) = P(H)P(H) = \left(
\frac{1}{2} \right) \left( \frac{1}{2} \right) = \left( \frac{1}{4}
\right)\) because each flip is independent
- \(P(H \text{ and } T) = P(H)P(T) = \left(
\frac{1}{2} \right) \left( \frac{1}{2} \right) = \left( \frac{1}{4}
\right)\)
- \(P(T \text{ and } H) = P(T)P(H) = \left(
\frac{1}{2} \right) \left( \frac{1}{2} \right) = \left( \frac{1}{4}
\right)\)
- \(P(T \text{ and } T) = P(T)P(T) = \left(
\frac{1}{2} \right) \left( \frac{1}{2} \right) = \left( \frac{1}{4}
\right)\)
\(\dagger\) How many possible
outcomes are there for three coins and what are the probabilities?
Disjoint and Joint Events
Two events, \(A\) and \(B\), are disjoint (or
mutually exclusive) if they cannot occur at the
same time: \[P(A \text{ and } B) =
0.\]
Two event, \(A\) and \(B\) are joint if they can
happen together: \[P(A \text{ and } B) \ne
0\]
Fair Coin Example:
- \(S = \{H,T\}\)
- \(P(H \text{ and } T) = 0\) since
\(H\) and \(T\) outcomes cannot occur simultaneously in
one flip
Union of Events
The union of two events, \(A\) and \(B\), is the event that at least one
of them occurs: \[P(A \text{ or } B)
= P(A) + P(B) - P(A \text{ and } B)\]
If \(A\) and \(B\) are disjoint, then \[P(A \text{ or } B) = P(A) + P(B)\]
\(\star\) Key Idea:
The probability of the union is the sum of individual probabilities
minus their intersection (to avoid double-counting).
Joint vs Disjoint Venn Diagram
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Drawing Cards
Suppose we conduct an experiment of drawing specific characteristics
of a card from a 52-card deck.
- Let \(A\) be the event that we draw
a Queen.
- Let \(B\) be the event that we draw
a Heart.
- Events \(A\) and \(B\) are joint.
- \(P(A \text{ and } B) =
\frac{1}{52}\) (Queen of Hearts).
- We know that there are 4 Queens and 13 Hearts.
- \(P(A) = \frac{4}{52}\), \(P(B) = \frac{13}{52}\), and
- \(P(A \text{ or } B) = P(A) + P(B) - P(A
\text{ and } B) = \frac{4}{52} + \frac{13}{52} - \frac{1}{52} =
\frac{16}{52}\)
\(\dagger\) Can you compute the
probability of drawing a face card (Ace, Jack, Queen, King) or a
Diamond?
Dice Rolls
Suppose we conduct an experiment of rolling two six-sided dice and
sum the outcomes.
- Each dice has six outcomes: 1 (⚀), 2 (⚁), 3 (⚂), 4 (⚃), 5 (⚄), or 6
(⚅)
- Let \(A\) be the event of the 1st
dice.
- Let \(B\) be the event of the 2nd
dice.
- The outcome we are interested in is the sum of \(A\) and \(B\). So, the sample space has 36 total
possible outcomes.
- The probability of getting a sum of 3 is \[\begin{aligned} P(3) & = P(1 \text{ and } 2)
+ P(2 \text{ and } 1) \\ & = P(1)P(2) + P(2)P(1) \\ & = \left(
\frac{1}{6} \right) \left( \frac{1}{6} \right) + \left( \frac{1}{6}
\right) \left( \frac{1}{6} \right) \\ & = 2 \times \left(
\frac{1}{6} \right) \left( \frac{1}{6} \right) \\ & = \frac{2}{36}
\end{aligned}\] because of independence and joint events.
\(\dagger\) Can you compute the
probability of rolling a sum of 4?
Summary of Basic Probability Rules
Basic Rules
Independence |
\(P(A \text{ and } B) =
P(A)P(B)\) |
Joint (Union) |
\(P(A \text{ or } B) = P(A) +
P(B) - P(A \text{ and } B)\) |
Disjoint |
\(P(A \text{ and } B) =
0\) |
Complement |
If \(P(A) + P(B) =
1\), then \(1-P(A)=P(B)\). |
Probability Axioms
\(P(S) = 1\) |
The sum of the probabilities for all outcomes in the
sample space is equal to 1. |
\(P \in [0,1]\) |
Probabilities are always positive and always between
\(0\) and \(1\). |
\(P(A \text{ or } B) = P(A) +
P(B)\) |
If events A and B are disjoint (mutually exclusive),
then their probabilities can be added. |
Interpreting Probability
Frequentist probability refers to the interpretation
of probability based on the long-run frequency of an event occurring in
repeated trials or experiments.
Coin Flipping Example
Suppose we conduct an experiment where we repeatedly flip a fair coin
(\(P(H) = 0.50\)), tracking the
cumulative count of \(H\) and its
proportion after each flip.

\(\star\) Key Idea:
As the number of flips (samples) increases the proportion of H gets
closer and closer to the true proportion of H, which is \(P(H)=0.50\).
Activity: Taking Samples from Dice Rolls
- Make sure you have a copy of the W 2/19 Worksheet. This
will be handed out physically and it is also digitally available on
Moodle.
- Work on your worksheet by yourself for 10 minutes. Please read the
instructions carefully. Ask questions if anything need
clarifications.
- Get together with another student.
- Discuss your results.
- Submit your worksheet on Moodle as a
.pdf
file.