Conditional Probability

Mathematical Statistics

Samir Orujov, PhD

ADA University, School of Business

Information Communication Technologies Agency, Statistics Unit

Overview

Today’s Journey

Conditional Probability Definition
Real-world Examples
Mathematical Applications
Problem-solving Strategies

Learning Objectives

Understand conditional probability concept
Apply the conditional probability formula
Solve complex probability problems
Connect theory to practical scenarios

Example 1: The Dice Problem

Imagine we are throwing two dice, and each possible outcome has the same chance of happening. So, each outcome has a probability of \(\frac{1}{36}\).

Question: We roll the dice, and we see that the first die shows a 3. With this new information, what’s the likelihood that the total of the two dice is 8?

🤔 Quick Poll: What do you think? - A) 1/36
- B) 1/6
- C) 1/3
- D) 5/36

Example 1: Building Intuition

Given information: First die = 3

Possible outcomes: (3,1), (3,2), (3,3), (3,4), (3,5), (3,6)

For sum = 8: We need 3 + ? = 8, so second die must be 5

Favorable outcome: Only (3,5)

Solution: \(P(\text{sum}=8|\text{first die}=3) = \frac{1}{6}\)

Definition: Conditional Probability

Conditional probabilities describe the likelihood of an event occurring given that another event has already occurred.

Mathematical Definition:

The conditional probability of event A occurring given event B is:

\[P(A | B) = \frac{P(A \cap B)}{P(B)} \quad \text{if } P(B) > 0\]

Key Insight: We’re updating our probability based on new information!

Interactive Exercise: Missing Keys

Scenario: Joe’s missing key problem

80% certain key is in jacket pockets
40% certain it’s in left pocket
40% certain it’s in right pocket

🔍 Search Result: Left pocket searched - no key found!

Question: What’s the probability it’s in the right pocket now?

Think-Write-Pair: Take 2 minutes to work this out with a partner

Example 2: Solution Steps

Initial probabilities: - P(left pocket) = 0.40 - P(right pocket) = 0.40
- P(elsewhere) = 0.20

After unsuccessful left search: - P(not in left) = 0.60 - P(right | not in left) = ?

Solution: \[P(\text{right}|\text{not left}) = \frac{P(\text{right})}{P(\text{not left})} = \frac{0.40}{0.60} = \frac{2}{3}\]

Student Activity: Coin Flips

Setup: Flip a coin twice
Sample Space: S = {(H,H), (H,T), (T,H), (T,T)}

🎯 Your Challenge: Find these conditional probabilities:

P(both heads | first flip is heads)
P(both heads | at least one head)

⏱️ Time: 3 minutes in small groups 📝 Show: Your reasoning process

Example 3: Solutions Revealed

Part (a): P(both heads | first flip is heads)

Given first flip is H: Sample space reduces to {(H,H), (H,T)}
Only (H,H) satisfies “both heads”
Answer: \(\frac{1}{2}\)

Part (b): P(both heads | at least one head)

At least one H: {(H,H), (H,T), (T,H)} - 3 outcomes
Both heads: {(H,H)} - 1 outcome
Answer: \(\frac{1}{3}\)

Key Insight: Subset Principle

Remark 1:

If each outcome of a finite sample space S is equally likely, then, conditional on the event that the outcome lies in a subset F ⊆ S, all outcomes in F become equally likely.

🤝 Discussion Moment: How does this principle apply to our coin example?

Bridge Card Challenge

Context: Bridge game with 52 cards dealt to 4 players (13 each)

Given: North and South have 8 spades total

Question: What’s the probability East has exactly 3 of the remaining 5 spades?

🧠 Strategy Session: - What type of probability distribution is this? - What are the key parameters?

Example 4: Bridge Solution

Problem Analysis: - East and West share 26 cards total - 5 spades and 21 non-spades available - East gets exactly 13 cards

Distribution Type: Hypergeometric!

Formula: \[P(X = 3) = \frac{\binom{5}{3} \times \binom{21}{10}}{\binom{26}{13}}\]

Solution: \(P(\text{East has 3 spades}) = \frac{5}{13} \approx 0.385\)

Example 5: Course Decision Solution

Events: - C: Takes chemistry - A: Gets an A grade

Given: - P(C) = 1/2 (fair coin) - P(A|C) = 2/3

Solution: \[P(\text{A in chemistry}) = P(C) \times P(A|C) = \frac{1}{2} \times \frac{2}{3} = \frac{1}{3}\]

Connection: Multiplication Principle

Remark 2: By our definition, we have:

\[P(EF) = P(F)P(E|F)\]

This extends to multiple events!

The Multiplication Rule

Formula 1:

\[P(E_1E_2E_3 \cdots E_n) = P(E_1)P(E_2|E_1)P(E_3|E_1E_2) \cdots P(E_n|E_1E_2E_3 \cdots E_{n-1})\]

💡 Memory Aid: Chain of conditional probabilities!

Proof: Multiplication Rule

Proof Strategy: Telescoping fractions

\[\text{R.H.S.} = P(E_1) \cdot \frac{P(E_1E_2)}{P(E_1)} \cdot \frac{P(E_1E_2E_3)}{P(E_1E_2)} \cdots \frac{P(E_1E_2 \cdots E_n)}{P(E_1E_2 \cdots E_{n-1})}\]

Cancellation: \[= P(E_1E_2E_3 \cdots E_n) = \text{L.H.S.}\]

Application: Urn Problems

Setup: 8 red balls, 4 white balls
Process: Draw 2 balls without replacement

Scenario 1: Equal probability selection
Scenario 2: Weighted selection (red weight = r, white weight = w)

🎯 Challenge: Find P(both red) for each scenario

⏱️ Time: Work in groups of 3-4 students

Example 6: Urn Solutions

Scenario 1 (Equal probability): \[P(\text{both red}) = \frac{8}{12} \times \frac{7}{11} = \frac{14}{33}\]

Scenario 2 (Weighted): \[P(\text{both red}) = \frac{8r}{8r + 4w} \times \frac{7r}{7r + 4w}\]

Special case: When r = w, we get the same result as Scenario 1!

Advanced Problem: The Matching Problem

Historical Context: N people randomly select from their own N hats

Known Result: \[P_N = \sum_{i=0}^{N} \frac{(-1)^i}{i!}\]

New Question: What’s the probability that exactly k people have matches?

🧠 Extension Challenge: Can you relate this to conditional probability?

Card Distribution Challenge

Problem: 52 cards divided into 4 piles of 13 each

Question: Probability that each pile has exactly 1 ace?

🤝 Strategy Discussion: - How many ways to distribute the aces? - What’s the total number of ways to divide cards?

Solution Approach: Multinomial coefficients + conditional probability

Example 8: Card Solution

Step 1: Ways to put one ace in each pile \[\binom{13}{1}^4 = 13^4\]

Step 2: Ways to distribute remaining 48 cards (12 to each pile) \[\binom{48}{12,12,12,12} = \frac{48!}{(12!)^4}\]

Step 3: Total ways to divide 52 cards \[\binom{52}{13,13,13,13} = \frac{52!}{(13!)^4}\]

Group Project: Football Tournament

The 2016 Champions League Scenario: - 8 teams in quarterfinals - 4 strong teams: Barcelona, Bayern Munich, Real Madrid, PSG - 4 weaker teams - Question: P(no strong team plays another strong team)?

🏆 Team Challenge: - Form groups of 4 - Each person tackles a different approach - Compare solutions at the end

Example 9: Tournament Analysis

Event Definition: Let E_i = team i plays a weak team (i = B, M, R, P)

Target: P(E_B ∩ E_M ∩ E_R ∩ E_P)

Approach 1 - Sequential: \[P = \frac{4}{7} \times \frac{3}{5} \times \frac{2}{3} \times 1 = \frac{8}{35}\]

Approach 2 - Combinatorial: \[P = \frac{\text{Favorable pairings}}{\text{Total pairings}} = \frac{4!}{\binom{8}{2,2,2,2}/4!} = \frac{8}{35}\]

Synthesis and Reflection

🎯 Key Takeaways: - Conditional probability updates our beliefs - Multiple solution approaches often exist - Real-world applications are everywhere

🤔 Reflection Questions: 1. When is conditional probability most useful? 2. How do we choose between different solution methods? 3. What connections do you see to other probability concepts?

Next Steps and Practice

📚 Homework Problems: - Complete Examples 7-9 with full solutions - Find one real-world conditional probability problem - Create your own dice/card problem

🔄 Coming Up: - Independence and conditional probability
- Bayes’ theorem - Total probability law

✨ Remember: Probability is about updating our understanding with new information!