Chapter 2: Introduction to Probability

Meike Niederhausen and Nicky Wakim

2024-10-02

Learning Objectives

  1. Define basic axioms and propositions in probability

  2. Assign probabilities to events

  3. Perform manipulations on probabilities to make calculations easier

Where are we?

Probabilities of equally likely events

Probabilities of equally likely events

  • “Equally likely” means the probability of any possible outcome is the same
    • Think: each side of die is equally likely or picking a card in a deck is equally likely

Pick an equally likely card, any equally likely card

Example 1

Suppose you have a regular well-shuffled deck of cards. What’s the probability of drawing:

  1. any heart

  2. the queen of hearts

  3. any queen

Let’s break down this probability

If \(S\) is a finite sample space, with equally likely outcomes, then

\[\mathbb{P}(A) = \frac{|A|}{|S|}\]

In human speak:

  • For equally likely outcomes, the probability that a certain event occurs is: the number of outcomes within the event of interest (\(|A|\)) divided by the total number of possible outcomes (\(|S|\))

\[\mathbb{P}(A) = \frac{\text{total number of outcomes in event A}}{\text{total number of outcomes in sample space}}\]

  • Thus, it is important to be able to count the outcomes within an event

A probability is a function…

  • \(\mathbb{P}(A)\) is a function with

    • Input: event \(A\) from the sample space \(S\), (\(A \subseteq S\))

      • \(A \subseteq S\) means “A contained within S” or “A is a subset of S”

    • Output: a number between 0 and 1 (inclusive)

 

  • The probability function maps an event (input) to value between 0 and 1 (output)

    • When we speak of the probability function, we often call the values between 0 and 1 “probabilities”

      • Example: “The probability of drawing a heart is 0.25” for \(P(\text{heart}) = 0.25\)

 

  • The probability function needs to follow some specific rules!

 

See Probability Axioms on next slide.

Probability Axioms

Probability Axioms

Axiom 1

For every event \(A\), \(0\leq\mathbb{P}(A)\leq 1\). Probability is between 0 and 1.

Axiom 2

For the sample space \(S\), \(\mathbb{P}(S)=1\).

Axiom 3

If \(A_1, A_2, A_3, \ldots\), is a collection of disjoint events, then \[\mathbb{P}\Big( \bigcup \limits_{i=1}^{\infty}A_i\Big) = \sum_{i=1}^{\infty}\mathbb{P}(A_i).\] The probability of at least one \(A_i\) is the sum of the individual probabilities of each.

Some probability properties

Some probability properties

Using the Axioms, we can prove all other probability properties! Events A, B, and C are not necessarily disjoint!

Proposition 1

For any event \(A\), \(\mathbb{P}(A)= 1 - \mathbb{P}(A^C)\)

Proposition 2

\(\mathbb{P}(\emptyset)=0\)

Proposition 3

If \(A \subseteq B\), then \(\mathbb{P}(A) \leq \mathbb{P}(B)\)

Proposition 4

\[\mathbb{P}(A \cup B) = \mathbb{P}(A) + \mathbb{P}(B) - \mathbb{P}(A \cap B)\] where \(A\) and \(B\) are not necessarily disjoint

Proposition 5

\(\begin{aligned} \mathbb{P}(A \cup B & \cup C) = \mathbb{P}(A) + \mathbb{P}(B) + \\ & \mathbb{P}(C) - \mathbb{P}(A \cap B) - \mathbb{P}(A \cap C) - \\ & \mathbb{P}(B \cap C) + \mathbb{P}(A \cap B \cap C) \end{aligned}\)

Proposition 1 Proof

Proposition 1

For any event \(A\), \(\mathbb{P}(A)= 1 - \mathbb{P}(A^C)\)

Use Axioms!

A1: \(0\leq\mathbb{P}(A)\leq 1\)

A2: \(\mathbb{P}(S)=1\)

A3: For disjoint \(A_i\),

\(\mathbb{P}\Big( \bigcup \limits_{i=1}^{\infty}A_i\Big) = \sum_{i=1}^{\infty}\mathbb{P}(A_i)\)

Proposition 2 Proof

Proposition 2

\(\mathbb{P}(\emptyset)=0\)

Use Axioms!

A1: \(0\leq\mathbb{P}(A)\leq 1\)

A2: \(\mathbb{P}(S)=1\)

A3: For disjoint \(A_i\),

\(\mathbb{P}\Big( \bigcup \limits_{i=1}^{\infty}A_i\Big) = \sum_{i=1}^{\infty}\mathbb{P}(A_i)\)

Proposition 3 Proof

Proposition 3

If \(A \subseteq B\), then \(\mathbb{P}(A) \leq \mathbb{P}(B)\)

Use Axioms!

A1: \(0\leq\mathbb{P}(A)\leq 1\)

A2: \(\mathbb{P}(S)=1\)

A3: For disjoint \(A_i\),

\(\mathbb{P}\Big( \bigcup \limits_{i=1}^{\infty}A_i\Big) = \sum_{i=1}^{\infty}\mathbb{P}(A_i)\)

Proposition 4 Visual Proof

Proposition 4

\(\mathbb{P}(A \cup B) = \mathbb{P}(A) + \mathbb{P}(B) - \mathbb{P}(A \cap B)\)

Proposition 5 Visual Proof

Proposition 5

\(\mathbb{P}(A \cup B \cup C) = \mathbb{P}(A) + \mathbb{P}(B) + \mathbb{P}(C) - \mathbb{P}(A \cap B) - \mathbb{P}(A \cap C) - \mathbb{P}(B \cap C) + \mathbb{P}(A \cap B \cap C)\)

Some final remarks on these proposition

  • Notice how we spliced events into multiple disjoint events
    • It is often easier to work with disjoint events

 

  • If we want to calculate the probability for one event, we may need to get creative with how we manipulate other events and the sample space
    • Helps us use any incomplete information we have

Partitions

Partitions

Definition: Partition

A set of events \(\{A_i\}_{i=1}^{n}\) create a partition of \(A\), if

  • the \(A_i\)’s are disjoint (mutually exclusive) and

  • \(\bigcup \limits_{i=1}^n A_i = A\)

Example 2

  • If \(A \subset B\), then \(\{A, B \cap A^C\}\) is a partition of \(B\).

  • If \(S = \bigcup \limits_{i=1}^n A_i\), and the \(A_i\)’s are disjoint, then the \(A_i\)’s are a partition of the sample space.

Creating partitions is sometimes used to help calculate probabilities, since by Axiom 3 we can add the probabilities of disjoint events.

Venn Diagram Probabilities

Weekly medications

Example 3

If a subject has an

  • 80% chance of taking their medication this week,

  • 70% chance of taking their medication next week, and

  • 10% chance of not taking their medication either week,

then find the probability of them taking their medication exactly one of the two weeks.

Hint: Draw a Venn diagram labelling each of the parts to find the probability.

Chapter 2 Slides