SDS CH4 - Probability Theory and Contingency Tables Lecture Notes

Course Administrative Timeline and Preparation

• Current Progress: The class is starting Tutorial 4, which covers Chapter 4 material.
• Schedule:     - Term 2 will encompass six weeks of instruction following the upcoming recess.     - Assessment 2 (A2) follows Term 2.     - Tutorial 4 will be split over two sessions: approximately half today, and the remainder in two weeks (after recess).     - Tutorials 5 and 6 will follow the completion of Tutorial 4.     - A dedicated revision week is scheduled before the A1 exam.
• Exam Context: Tutorial 4 is the final practical component that will be included in the A1 exam.
• Methodology: Exercises are performed using pen and paper rather than Excel templates, as Chapter 4 focuses on probability calculations best suited for manual notation.

Question 1: Contingency Tables and Basic Probability Rules

• Contingency Table Construction: A contingency table displays frequencies for two categorical variables (e.g., Event A and Event B).
• Initial Data (Counts):     - Event $A$ and $B$: $10$     - Event $A$ and $B^c$: $20$     - Event $A^c$ and $B$: $20$     - Event $A^c$ and $B^c$: $40$
• Marginal Totals (Calculated by summing rows and columns):     - Total for Event $A$: $10 + 20 = 30$     - Total for Event $A^c$: $20 + 40 = 60$     - Total for Event $B$: $10 + 20 = 30$     - Total for Event $B^c$: $20 + 40 = 60$     - Grand Total (Sample Space $S$): $90$
• Calculations:     - Simple Probability of A: Represents the marginal probability.         - $P(A) = \frac{30}{90} = 0.3333$     - Probability of A Complement:         - Method 1 (Direct): $P(A^c) = \frac{60}{90} = 0.6667$         - Method 2 (Complement Rule): Since $A$ and $A^c$ are mutually exclusive and exhaustive, $P(A^c) = 1 - P(A) = 1 - 0.3333 = 0.6667$     - Joint Probability (Intersection): The probability that both events occur simultaneously.         - $P(A \text{ and } B) = P(A \text{ intersection } B) = \frac{10}{90} = 0.1111$     - Addition Rule (Union): The probability that event A or event B (or both) occurs.         - Formula: $P(A \text{ or } B) = P(A \text{ union } B) = P(A) + P(B) - P(A \text{ and } B)$         - Calculation: $\frac{30}{90} + \frac{30}{90} - \frac{10}{90} = \frac{50}{90} = 0.5556$
• Theoretical Justification for the Addition Rule:     - When calculating the area of $A$ and the area of $B$, the intersection (overlap) is counted twice (once in $A$ and once in $B$).     - Subtracting the joint probability $P(A \text{ and } B)$ corrects this double-counting to provide the pure total area occupied by both events.

Question 3: Three Approaches to Probability

• A Priori Approach:     - Probability is determined based on prior knowledge or theoretical properties rather than data collection.     - Example: Tossing a fair coin. We know theoretically that the probability of heads is $0.5$ without needing to perform the experiment. Large number simulations will eventually even out to this theoretical probability.     - Example: Rolling two dice to get a sum of seven. Knowing the dice are not "loaded" and have six sides allows for a theoretical calculation prior to the roll.
• Subjective Approach:     - Probability is based on an individual's personal judgment, opinion, or confidence level.     - Used when there are too many variables to quantify empirically or when historical data does not guarantee future outcomes.     - Example: Predicting that Italy will win the next soccer World Cup. This relies on subjective confidence rather than a fixed physical law or perfect historical replicate.
• Empirical Approach:     - Probability is determined by gathering and analyzing historical data or experimental results.     - Example: Calculating the chance that a commuter train will be more than 10 minutes late. This requires gathering specific data on arrival times to determine the frequency of delays.

Question 4 and 11: Application of Workforce Demographics

• Event Definitions:     - Simple Event: Involves only one characteristic (e.g., an employee being younger than 40).     - Joint Event: Involves two or more characteristics occurring simultaneously (e.g., an employee being younger than 40 AND working in production).     - Complement: Every outcome in the sample space not included in the primary event (e.g., the complement of "younger than 40" is "40 or older").
• Data from Question 11 Table ($n = 450$):     - Workers < 40: 334 total (Production: 320, Sales: 14)     - Workers $\text{>=}$ 40: 116 total (Production: 80, Sales: 36)     - Total Production: 400     - Total Sales: 50
• Probabilistic Calculations:     - P(Production): Marginal probability.         - $\frac{400}{450} = 0.8889$     - P(Older and Production): Joint probability.         - $\frac{80}{450} = 0.1778$     - P(Older or Production): Addition rule.         - $P(\text{Older}) + P(\text{Production}) - P(\text{Older and Production}) = \frac{116}{450} + \frac{400}{450} - \frac{80}{450} = 0.9689$
• Conditional Probabilities (Question 11):     - P(Sales | >= 40): Probability of working in sales GIVEN they are 40 or older.         - $\frac{P(\text{Sales and Older})}{P(\text{Older})} = \frac{36}{116} = 0.3103$     - P(Sales | < 40): Probability of working in sales GIVEN they are younger than 40.         - $\frac{P(\text{Sales and Younger})}{P(\text{Younger})} = \frac{14}{334} = 0.0419$     - Contextual Note: There is a higher probability of sales placement for those with more experience ($>= 40$).

Question 6: Constructs from Probability Values

• Scenario: Car manufacturing and warranty repairs (US vs. Non-US companies).
• Given Probabilities:     - $P(\text{Repair}) = 0.04$     - $P(\text{USA}) = 0.60$     - $P(\text{Repair and USA}) = 0.025$
• Inferred Probabilities (Filling the Table):     - $P(\text{No Repair}) = 1 - 0.04 = 0.96$     - $P(\text{Non-USA}) = 1 - 0.60 = 0.40$     - $P(\text{Repair and Non-USA}) = P(\text{Repair}) - P(\text{Repair and USA}) = 0.04 - 0.025 = 0.015$     - $P(\text{No Repair and USA}) = P(\text{USA}) - P(\text{Repair and USA}) = 0.60 - 0.025 = 0.575$     - $P(\text{No Repair and Non-USA}) = P(\text{Non-USA}) - P(\text{Repair and Non-USA}) = 0.40 - 0.015 = 0.385$
• Unions and Complements:     - $P(\text{Repair or USA}) = 0.04 + 0.60 - 0.025 = 0.615$     - $P(\text{Repair or Non-USA}) = P(\text{Repair}) + P(\text{Non-USA}) - P(\text{Repair and Non-USA})$     - $0.04 + 0.40 - 0.015 = 0.425$

Question 7 and 9: Conditional Probability and Independence

• Conditional Probability Formula: The probability of A occurring given that B has already occurred.     - $P(A|B) = \frac{P(A \text{ and } B)}{P(B)}$
• Statistical Independence:     - Events A and B are independent if the occurrence of one does not affect the probability of the other.     - Condition 1: $P(A|B) = P(A)$     - Condition 2: $P(A \text{ and } B) = P(A) \times P(B)$
• Verification Example (Q7):     - Given $P(A|B) = 0.2857$ and $P(A) = 0.4$.     - Since $0.2857 \neq 0.4$, the events are not independent.
• Multiplication Rule (Q9):     - If $P(A) = 0.7$ and $P(B) = 0.6$ and they are independent:     - $P(A \text{ and } B) = P(A) \times P(B) = 0.7 \times 0.6 = 0.42$

Question 14: Sampling and Card Game Applications

• Standard Deck Parameters: 52 total cards, 4 suits (Hearts, Diamonds, Clubs, Spades), 13 faces per suit.
• Sampling Without Replacement: The sample size decreases with each draw.     - Example: Drawing two queens.     - Draw 1: $4$ queens out of $52$ cards.     - Draw 2: $3$ queens left out of $51$ cards left.     - $P(\text{Two Queens}) = \frac{4}{52} \times \frac{3}{51} = 0.0045$
• Sampling With Replacement: The sample size remains constant.     - Draw 1: $4/52$     - Draw 2: $4/52$ (the card was put back).     - $P(\text{Two Queens}) = \frac{4}{52} \times \frac{4}{52} = 0.0059$     - Note: Probability increases when sampling with replacement as the favorable outcomes are not depleted.
• Application: Blackjack Probability:     - Rules:         - Ace = 11 points (usually 1 or 11, but 11 for this calculation).         - Jack, Queen, King (Face cards) = 10 points.         - Cards 2–10 = Face value.         - Blackjack = Getting 21 points in the first two cards.     - 10-Valued Cards: There are 4 types (10, J, Q, K) across 4 suits, totaling $16$ cards.     - Blackjack Paths:         - Path 1: Ace followed by a 10-valued card: $\frac{4}{52} \times \frac{16}{51}$         - Path 2: 10-valued card followed by an Ace: $\frac{16}{52} \times \frac{4}{51}$     - Total Probability:         - $(\frac{4}{52} \times \frac{16}{51}) + (\frac{16}{52} \times \frac{4}{51}) = 0.0483$

Questions & Discussion

• Question: Why did we use the addition rule formula specifically in Question 1?

• Answer: Because calculating $P(A \text{ or } B)$ by simply adding the separate marginal probabilities results in double-counting the intersection. The rule subtracts the intersection once to yield the correct total area for the union.

• Question: How do we define "independent" versus "mutually exclusive"?

• Answer: Mutually exclusive means two events cannot happen at the same time ($P(A \text{ and } B) = 0$). Independence means the occurrence of one does not change the probability of the other ($P(A|B) = P(A)$).

• Question: Where did the 16 come from in the Blackjack calculation?

• Answer: Blackjack rules count 10s, Jacks, Queens, and Kings as 10 points. Since there are 4 suits, we have 4 types of 10-point cards times 4 suits, equaling 16 cards in total worth 10 points.