Session 07-04 - Conditional Probability

Section 07: Probability & Statistics

Dr. Nikolai Heinrichs & Dr. Tobias Vlćek

Entry Quiz - 10 Minutes

Quick Review from Session 07-03

Test your understanding of Combinatorics

Calculate \(\binom{8}{3}\)
How many ways can 5 people line up in a row?
A committee of 3 is chosen from 10 people. How many ways?
What’s the probability of getting exactly 3 heads in 5 coin flips? (Use combinations)

Learning Objectives

What You’ll Master Today

Define conditional probability: \(P(A|B) = \frac{P(A \cap B)}{P(B)}\)
Apply the multiplication rule: \(P(A \cap B) = P(A|B) \cdot P(B)\)
Construct and use tree diagrams for sequential events
Apply the law of total probability
Test for independence using conditional probability

Conditional probability is heavily tested on the Feststellungsprüfung!

Part A: Conditional Probability Concept

Motivation

Question: A card is drawn from a deck. What’s the probability it’s a king?

\[P(\text{King}) = \frac{4}{52} = \frac{1}{13}\]

Question: Given that the card is a face card, what’s the probability it’s a king?

The condition changes the sample space!

\[P(\text{King} | \text{Face card}) = \frac{4}{12} = \frac{1}{3}\]

Conditional Probability Definition

Definition: Conditional Probability

The probability of A given that B has occurred:

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

where \(P(B) > 0\).

Read “\(P(A|B)\)” as “probability of A given B”

Visual Interpretation

Example Calculation

In a company: 60% work full-time, 30% work full-time AND have a degree.

Question: What’s the probability an employee has a degree, given they work full-time?

Let D = has degree, F = full-time

\[P(D|F) = \frac{P(D \cap F)}{P(F)} = \frac{0.30}{0.60} = 0.50\]

Given that someone works full-time, there’s a 50% chance they have a degree.

Part B: Multiplication Rule

From Conditional to Joint Probability

Rearranging the definition:

Multiplication Rule

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

or equivalently:

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

Both formulas give the same result!

Example: Sequential Selection

A box contains 5 red and 3 blue balls. Two balls are drawn without replacement.

Question: What’s the probability both are red?

Let \(R_1\) = first ball red, \(R_2\) = second ball red

\[P(R_1 \cap R_2) = P(R_2|R_1) \cdot P(R_1)\]

\[= \frac{4}{7} \times \frac{5}{8} = \frac{20}{56} = \frac{5}{14}\]

Extended Multiplication Rule

For three or more events:

\[P(A \cap B \cap C) = P(A) \cdot P(B|A) \cdot P(C|A \cap B)\]

Example: Drawing 3 cards without replacement

\[P(\text{3 aces}) = \frac{4}{52} \times \frac{3}{51} \times \frac{2}{50} = \frac{24}{132,600} = \frac{1}{5,525}\]

Part C: Tree Diagrams

Visual Tool for Sequential Events

Tree diagrams show:

Branches = outcomes at each stage
Branch labels = probabilities (conditional if sequential)
Multiply along paths = joint probability
Add across paths = union probability

Example: Quality Control

A machine produces items that pass through 2 inspections.

90% pass inspection 1
Of those passing inspection 1: 95% pass inspection 2
Of those failing inspection 1: 60% pass inspection 2

Reading Tree Diagrams

From the previous example:

P(item passes both inspections): Follow Pass-Pass path \[0.90 \times 0.95 = 0.855\]

P(item passes inspection 2): Add all paths ending with Pass I2 \[0.855 + 0.060 = 0.915\]

P(passed I1 | passed I2): (We’ll use Bayes for this later!) \[P(I_1|I_2) = \frac{P(I_1 \cap I_2)}{P(I_2)} = \frac{0.855}{0.915} \approx 0.934\]

Break - 10 Minutes

Part D: Law of Total Probability

Partitioning the Sample Space

Law of Total Probability

If events \(B_1, B_2, ..., B_n\) partition the sample space (mutually exclusive and exhaustive):

\[P(A) = \sum_{i=1}^{n} P(A|B_i) \cdot P(B_i)\]

Common case with two partitions:

\[P(A) = P(A|B) \cdot P(B) + P(A|B') \cdot P(B')\]

Example: Market Segments

A company sells to three market segments:

Segment A: 50% of customers, 20% buy premium
Segment B: 30% of customers, 35% buy premium
Segment C: 20% of customers, 50% buy premium

Question: What percentage of all customers buy premium?

\[P(\text{Premium}) = 0.20(0.50) + 0.35(0.30) + 0.50(0.20)\] \[= 0.10 + 0.105 + 0.10 = 0.305\]

30.5% of customers buy premium products.

Part E: Independence Revisited

Testing Independence

Independence Test

Events A and B are independent if and only if:

\[P(A|B) = P(A)\]

(Learning B occurred doesn’t change the probability of A)

Equivalently: \(P(B|A) = P(B)\) or \(P(A \cap B) = P(A) \cdot P(B)\)

Example: Testing Independence

Survey data shows:

\(P(\text{Exercise regularly}) = 0.40\)
\(P(\text{Healthy weight}) = 0.55\)
\(P(\text{Healthy weight}|\text{Exercise}) = 0.70\)

Question: Are “exercise regularly” and “healthy weight” independent?

Test: Is \(P(\text{Healthy}|\text{Exercise}) = P(\text{Healthy})\)?

\(0.70 \neq 0.55\)

Conclusion: Events are NOT independent - exercise is associated with healthy weight.

Guided Practice - 20 Minutes

Practice Problems

Work in pairs

Problem 1: 70% of students study math, 60% study economics, and 45% study both. a) Find \(P(\text{Econ}|\text{Math})\) b) Find \(P(\text{Math}|\text{Econ})\) c) Are these events independent?

Problem 2: Draw a tree diagram for: A bag has 3 red and 2 blue marbles. Two marbles are drawn without replacement. Find \(P(\text{both same color})\).

Wrap-Up & Key Takeaways

Today’s Essential Concepts

Conditional probability: \(P(A|B) = \frac{P(A \cap B)}{P(B)}\)
Multiplication rule: \(P(A \cap B) = P(A|B) \cdot P(B)\)
Tree diagrams: Multiply along paths, add across paths
Total probability: Sum over all branches
Independence: \(P(A|B) = P(A)\)

Next Session Preview

Coming Up: Bayes’ Theorem

Reversing conditional probabilities
Prior and posterior probabilities
Medical testing applications (sensitivity, specificity)
Business decision making

Homework

Complete Tasks 07-04 - focus on tree diagrams and conditional probability!