Lesson 4: Rules of probability

Author

Meike Niederhausen and Nicky Wakim

Published

October 8, 2025

Learning objectives

  1. Define independence of 2-3 events given probability notation

  2. Calculate whether two or more events are independent

  3. Use set process to calculate probability of event of interest

  4. Calculate the probability of an event occurring, given that another event occurred.

  5. Define keys facts for conditional probabilities using notation.

  6. Calculate conditional probability of an event using Bayes’ Theorem

  7. Utilize additional probability rules in probability calculations, specifically the Higher Order Multiplication Rule and the Law of Total Probabilities

Where are we?

Independent Events

Definition: Independence

Events \(A\) and \(B\) are independent if \[\mathbb{P}(A \cap B) = \mathbb{P}(A) \cdot \mathbb{P}(B).\]

Notation: For shorthand, we sometimes write \(A \mathrel{\unicode{x2AEB}} B,\) to denote that \(A\) and \(B\) are independent events.

 

  • Also note: \[\begin{aligned} \mathbb{P}(A \cap B) = \mathbb{P}(A) \cdot \mathbb{P}(B) & \implies A \mathrel{\unicode{x2AEB}} B \\ A \mathrel{\unicode{x2AEB}} B & \implies \mathbb{P}(A \cap B) = \mathbb{P}(A) \cdot \mathbb{P}(B) \end{aligned}\]

Example of two dice

Example 1

Two dice (red and blue) are rolled. Let \(A =\) event a total of 7 appears, and \(B =\) event red die is a six. Are events \(A\) and \(B\) independent?

Independence of 3 Events

Definition: Independence of 3 Events

Events \(A\), \(B\), and \(C\) are mutually independent if

    • \(\mathbb{P}(A \cap B) = \mathbb{P}(A) \cdot \mathbb{P}(B)\)

    • \(\mathbb{P}(A \cap C) = \mathbb{P}(A) \cdot \mathbb{P}(C)\)

    • \(\mathbb{P}(B \cap C) = \mathbb{P}(B) \cdot \mathbb{P}(C)\)

  2. \(\mathbb{P}(A \cap B \cap C) = \mathbb{P}(A) \cdot \mathbb{P}(B) \cdot \mathbb{P}(C)\)

Remark:

On your homework you will show that \((1) \not \Rightarrow (2)\) and \((2) \not \Rightarrow (1)\).

Probability at least one smoker

Example 2

Suppose you take a random sample of \(n\) people, of which people are smokers and non-smokers independently of each other. Let

  • \(A_i =\) event person \(i\) is a smoker, for \(i=1, \ldots ,n\), and

  • \(p_i =\) probability person \(i\) is a smoker, for \(i=1, \ldots ,n\).

Find the probability that at least one person in the random sample is a smoker.

Building geometric series

Example 3

\(A, B,\) and \(C\) toss a fair coin in order. The first to throw heads wins. What are their respective chances of winning?

Let

  • \(A_H\) and \(A_T\) be the events player A tosses heads and tails, respectively.

  • Similarly define \(B_H\), \(B_T\), \(C_H\), and \(C_T\).

Learning Objectives

  1. Use set process to calculate probability of event of interest

  2. Calculate the probability of an event occurring, given that another event occurred.

  3. Define keys facts for conditional probabilities using notation.

General Process for Probability Word Problems

  1. Clearly define your events of interest

  2. Translate question to probability using defined events OR Venn Diagram

  3. Ask yourself:

    • Are we sampling with or without replacement?

    • Does order matter?

  4. Use axioms, properties, partitions, facts, etc. to define the end probability calculation into smaller parts

    • If probabilities are given to you, Venn Diagrams may help you parse out the events and probability calculations

    • If you need to find probabilities with counting, pictures or diagrams might help here

  5. Write out a concluding statement that gives the probability context

  6. (For own check) Make sure the calculated probability follows the axioms. Is is between 0 and 1?

Let’s revisit our deck of cards

Example 1

Suppose we randomly draw 2 cards from a standard deck of cards. What is the probability that we draw a spade then a heart?

Let

  • Let \(A =\) event \(1^{st}\) card is spades

  • Let \(B =\) event \(2^{nd}\) card is heart

Conditional Probability facts (1/2)

Fact 1: General Multiplication Rule

\[\mathbb{P}(A\cap B)=\mathbb{P}(A)\cdot\mathbb{P}(B|A)\]

Fact 2: Conditional Probability Definition

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

Conditional Probability facts (2/2)

Fact 3

If \(A\) and \(B\) are independent events (\(A \unicode{x2AEB}B\)), then \[\mathbb{P}(A|B) = \mathbb{P}(A)\]

Fact 4

\(\mathbb{P}(A|B)\) is a probability, meaning that it satisfies the probability axioms. In particular, \[\mathbb{P}(A|B) + \mathbb{P}(A^C|B) = 1\]

Monty Hall Problem

Survivor Season 42

With the Wiki page on it!

Conditional probability with two dice

Example 2

Two dice (red and blue) are rolled. If the dice do not show the same face, what is the probability that one of the dice is a 1?

Learning Objectives

  1. Calculate conditional probability of an event using Bayes’ Theorem

  2. Utilize additional probability rules in probability calculations, specifically the Higher Order Multiplication Rule and the Law of Total Probabilities

Introduction

  • So we learned about conditional probabilities

    • We learned how the occurrence of event A affects event B (B conditional on A)

  • Can we figure out information on how the occurrence of event B affects event A?

  • We can use the conditional probability (\(\mathbb{P}(A|B)\)) to get information on the flipped conditional probability (\(\mathbb{P}(B|A)\))

Bayes’ Rule for two events

Theorem: Bayes’ Rule (for two events)

For any two events \(A\) and \(B\) with nonzero probabilties,

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

Calculating probability with Higher Order Multiplication Rule

Example 1

Suppose we draw 5 cards from a standard shuffled deck of 52 cards. What is the probability of a flush, that is all the cards are of the same suit (including straight flushes)?

Higher Order Multiplication Rule

\[\mathbb{P}(A_1\cap A_2 \cap \ldots \cap A_n)=\mathbb{P}(A_1)\cdot\mathbb{P}(A_2|A_1) \cdot \\ \mathbb{P}(A_3|A_1A_2)\ldots \cdot\mathbb{P}(A_n|A_1A_2\ldots A_{n-1})\]

Calculating probability with Law of Total Probability

Example 2

Suppose 1% of people assigned female at birth (AFAB) and 5% of people assigned male at birth (AMAB) are color-blind. Assume person born is equally likely AFAB or AMAB (not including intersex). What is the probability that a person chosen at random is color-blind?

Law of Total Probability for 2 Events

For events \(A\) and \(B\),

\[%\left( \begin{array}{ccl} \mathbb{P}(B)&=&\mathbb{P}(B \cap A) + \mathbb{P}(B \cap A^C)\\ &=& \mathbb{P}(B|A) \cdot \mathbb{P}(A)+ \mathbb{P}(B | A^C)\cdot \mathbb{P}(A^C) \end{array} %\right)\]

General Law of Total Proability

Law of Total Probability (general)

If \(\{A_i\}_{i=1}^{n} = \{A_1, A_2, \ldots, A_n\}\) form a partition of the sample space, then for event \(B\),

\[%\left( \begin{array}{ccl} \mathbb{P}(B)&=& \sum_{i=1}^{n} \mathbb{P}(B \cap A_i)\\ &=& \sum_{i=1}^{n} \mathbb{P}(B|A_i) \cdot \mathbb{P}(A_i) \end{array} %\right)\]

Calculating probability with generalized Law of Total Probability

Calculating probability with generalized Law of Total Probability

Example 3

Individuals are diagnosed with a particular type of cancer that can take on three different disease forms,* \(D_1\), \(D_2\), and \(D_3\). It is known that amongst people diagnosed with this particular type of cancer,

  • 20% of people will eventually be diagnosed with form \(D_1\),

  • 30% with form \(D_2\), and

  • 50% with form \(D_3\).

The probability of requiring chemotherapy (\(C\)) differs among the three forms of disease:

  • 80% with \(D_1\),

  • 30% with \(D_2\), and

  • 10% with \(D_3\).

Based solely on the preliminary test of being diagnosed with the cancer, what is the probability of requiring chemotherapy (the event C)?

Skipping in class! Let me know if you would like me to post solutions to this if you work through it!

Let’s revisit the color-blind example

Example 4

Recall the color-blind example (Example 2), where

  • a person is AMAB with probability 0.5,

  • AMAB people are color-blind with probability 0.05, and

  • all people are color-blind with probability 0.03.

Assuming people are AMAB or AFAB, find the probability that a color-blind person is AMAB.

Calculate probability with both rules

Example 5

Suppose

  • 1% of people who are AFAB aged 40-50 years have breast cancer,

  • an AFAB person with breast cancer has a 90% chance of a positive test from a mammogram, and

  • an AFAB person has a 10% chance of a false-positive result from a mammogram.

What is the probability that an AFAB person has breast cancer given that they just had a positive test?

Bayes’ Rule

Theorem: Bayes’ Rule

If \(\{A_i\}_{i=1}^{n}\) form a partition of the sample space \(S\), with \(\mathbb{P}(A_i)>0\) for \(i=1\ldots n\) and \(\mathbb{P}(B)>0\), then

\[\mathbb{P}(A_j | B) = \frac{\mathbb{P}(B|A_j) \cdot \mathbb{P}(A_j)} {\sum_{i=1}^{n} \mathbb{P}(B|A_i) \cdot \mathbb{P}(A_i)}\]