Comprehensive Study Notes on Basic Probability Concepts and Rules

Key Concepts and Fundamental Definitions in Probability

Probability: The quantitative measure representing the likelihood or chance that an uncertain event will occur, strictly bounded within the closed interval $[0, 1]$ .
Event: Any distinct possible outcome or type of occurrence resulting from an experiment or observational process.
Simple Event: An outcome in a sample space that is categorized or described by a single characteristic.
Sample Space: The complete collection or set of all possible outcomes or simple events resulting from an experiment.

Probability Scale and Outcome Continuum

Probability Scale

The probability continuum spans continuously from $0.00$ to $1.00$ :
- $0.00$ (Impossible / Cannot Happen): Represents absolute impossibility. Example: The probability that the sun will disappear this year.
- $0.20$ (Low Likelihood): Represents a small chance of occurrence. Example: The chance that the horse Slo Poke will win the Kentucky Derby.
- $0.50$ (Even Chance / Maybe): Represents an equal probability of occurring or not occurring. Example: The chance of landing heads on a single fair coin toss.
- $0.70$ (High Likelihood): Represents a substantial chance of occurrence. Example: The chance of an increase in federal taxes.
- $1.00$ (Certain / Sure to Happen): Represents absolute certainty. Example: The chance of rain in Florida during a calendar year.

Classifications and Types of Events

Simple Event: An outcome from a sample space defined by only one attribute or characteristic.
- Example: Drawing a red card from a standard deck of cards.
Complement of an Event $A$ : Denoted as $\bar{A}$ or $A'$ , this encompasses all elementary outcomes in the sample space that are not included in event $A$ .
- Example: If event $A$ is drawing a diamond, the complement $\bar{A}$ consists of all cards drawn that are not diamonds (i.e., spades, hearts, and clubs).
Joint Event: An event that comprises two or more distinct characteristics occurring simultaneously.
- Example: Drawing a card from a deck that is simultaneously an Ace and Red.

Marginal vs. Joint Probabilities

Marginal Probability (Simple Probability): Refers to the unconditional probability of a simple event occurring without regard to any other events.
- Notation: $P(A)$ .
- Example: The probability of drawing a King, denoted $P(\text{King})$ or $P(K)$ .
Joint Probability: Refers to the probability of the simultaneous occurrence of two or more events.
- Notation: $P(A \text{ and } B)$ or $P(A \bigcap B)$ .
- Example: The probability of drawing a card that is both a King and a Spade, denoted $P(\text{King and Spade})$ or $P(K \bigcap S)$ .

Tools for Visualizing Sample Spaces

Contingency Tables: Matrix-structured tables used to classify sample observations according to two or more qualitative, categorical criteria.

Contingency Table for Deck of Cards

Structure Example (Deck of 52 Playing Cards):
- Black Cards: Ace = $2$ , Not Ace = $24$ , Total = $26$
- Red Cards: Ace = $2$ , Not Ace = $24$ , Total = $26$
- Column Totals: Ace = $4$ , Not Ace = $48$ , Grand Total = $52$
- Tree Diagrams: Graphical representations displaying sequence and branching logic for combinations of events.

Tree Diagram for Deck of Cards

Structure Example (Deck of 52 Cards): First stage branches into Black Card ( $26$ ) and Red Card ( $26$ ); secondary stage branches each color into Ace ( $2$ ) and Not an Ace ( $24$ ).

Event Relationships: Mutually Exclusive vs. Collectively Exhaustive

Mutually Exclusive Events: Events that cannot occur at the same time in a single trial or experiment. The occurrence of one event precludes the occurrence of the other ( $P(A \bigcap B) = 0$ ).
- Example 1: Selecting a single card where Event $A = \text{queen of diamonds}$ and Event $B = \text{queen of clubs}$ . A single card selection cannot yield both.
- Example 2: A single live birth outcome where Event $B = \text{having a boy}$ and Event $G = \text{having a girl}$ .
Collectively Exhaustive Events: A set of events such that at least one of them must occur whenever an experiment is conducted. The union of collectively exhaustive events forms the entire sample space.
- Example 1 (Rolling a Die): The set of outcomes $\{1, 2, 3, 4, 5, 6\}$ is collectively exhaustive because every possible outcome of a standard die roll is included.
- Example 2 (Stock Market Movement): The outcome set $\{\text{Price goes up}, \text{Price goes down}, \text{Price stays the same}\}$ is collectively exhaustive because one of these outcomes must occur for tomorrow's trading session.

Classical Probability

Definition: Classical probability applies when an experiment has $n$ mutually exclusive and equally likely outcomes.

Classical Probability Formula

Formula: $P(A) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{\# \text{ of } A}{\text{Sample size}}$
Calculation Example (Statistics Enrollment):
- Data Matrix:
- Male: Taking Stats = $84$ , Not Taking Stats = $145$ , Total Male = $229$
- Female: Taking Stats = $76$ , Not Taking Stats = $134$ , Total Female = $210$
- Totals: Taking Stats = $160$ , Not Taking Stats = $279$ , Grand Total = $439$
- Problem: Calculate the probability of selecting a male taking statistics from this population.
- Calculation: $P(\text{Male Taking Stats}) = \frac{\text{number of males taking stats}}{\text{total number of people}} = \frac{84}{439} \approx 0.191$

Rules of Addition

General Rule of Addition: Applies when events are not mutually exclusive (i.e., they can occur simultaneously, meaning $P(A \bigcap B) \neq 0$ ).

Venn Diagram of Non-Mutually Exclusive Events

Formula: $P(A \cup B) = P(A) + P(B) - P(A \bigcap B)$
Worked Example (Florida Tourist Destinations):

Venn Diagram for Tourist Attractions Example

- Sample size =  $200$  tourists.
- Visitors to Disney =  $120 \implies P(\text{Disney}) = \frac{120}{200} = 0.60$ 
- Visitors to Busch Gardens =  $100 \implies P(\text{Busch}) = \frac{100}{200} = 0.50$ 
- Visitors to both =  $60 \implies P(\text{Disney and Busch}) = \frac{60}{200} = 0.30$ 
- Step 1: Compute marginal and joint probabilities.
  -  $P(\text{Disney}) = 0.60$ 
  -  $P(\text{Busch}) = 0.50$ 
  -  $P(\text{Disney and Busch}) = 0.30$ 
- Step 2: Calculate the probability of visiting Disney or Busch Gardens.

$P(\text{Disney or Busch}) = P(\text{Disney}) + P(\text{Busch}) - P(\text{Disney and Busch}) = 0.60 + 0.50 - 0.30 = 0.80$

Special Rule of Addition: Applies strictly when events are mutually exclusive ( $P(A \bigcap B) = 0$ ).

Venn Diagram of Mutually Exclusive Events

Formula: $P(A \cup B) = P(A) + P(B)$
Worked Example (Stock Price Movements):
- Event $A$ (Stock increases) $\implies P(A) = 0.6$
- Event $B$ (Stock decreases) $\implies P(B) = 0.4$
- Events $A$ and $B$ cannot occur simultaneously.
- $P(A \text{ or } B) = P(A) + P(B) = 0.6 + 0.4 = 1.0$

Complement Rule

Concept: Computes the probability of an event occurring by subtracting the probability of its complement (not occurring) from $1.0$ .
Formulas: $P(A') = 1 - P(A)$ $P(A) = 1 - P(A')$
Worked Example:
- Given probability that a stock price increases tomorrow: $P(\text{Increase}) = 0.7$
- Probability that the stock does not increase: $P(\text{Not Increase}) = 1 - P(\text{Increase}) = 1 - 0.7 = 0.3$

Statistical Independence and Multiplication Rules

Statistical Independence: Two events are independent if the occurrence of one event provides no information about the likelihood of the occurrence of the other event.
Special Rule of Multiplication (Independent Events):
- Used strictly when events are independent.
- Formula: $P(A \bigcap B) = P(A) \times P(B)$
- Worked Example:
- Event $A$ (Stock A increases): $P(A) = 0.6$
- Event $B$ (Stock B increases): $P(B) = 0.5$
- Assuming independence between Stock A and Stock B: $P(A \bigcap B) = 0.6 \times 0.5 = 0.30$
Conditional Probability: The probability of event $A$ occurring given that event $B$ has already occurred.
- Formulas: $P(A|B) = \frac{P(A \bigcap B)}{P(B)}$ $P(B|A) = \frac{P(A \bigcap B)}{P(A)}$
Formal Proof of Independence:
- Events $A$ and $B$ are independent if and only if: $P(A|B) = P(A) \quad \text{or} \quad P(B|A) = P(B)$
General Rule of Multiplication (Dependent or General Events):
- Refers to events that are not independent.
- Formulas: $P(A \bigcap B) = P(A) \times P(B|A)$ $P(A \bigcap B) = P(B) \times P(A|B)$

Contingency Table Analysis and Class Exercises

Contingency Table Structure:
- Rows and columns represent discrete qualitative categories.
- Interior cells present joint frequencies and joint probabilities: $P(A_i \bigcap B_j)$ .
- Row and column totals present marginal (simple) probabilities: $P(A_i)$ and $P(B_j)$ .
Example 1: Adults and Facebook Accounts ( $150$ Adults):
- Data Table:
- Men: $0$ Accounts = $20$ , $1$ Account = $40$ , $2+$ Accounts = $10$ , Total = $70$
- Women: $0$ Accounts = $40$ , $1$ Account = $30$ , $2+$ Accounts = $10$ , Total = $80$
- Totals: $0$ Accounts = $60$ , $1$ Account = $70$ , $2+$ Accounts = $20$ , Grand Total = $150$
Example 2: Student Enrollment Status ( $14,262$ Students):
- Data Table:
- Male ( $M$ ): Full-Time ( $FT$ ) = $3808$ , Part-Time ( $PT$ ) = $2535$ , Total = $6343$
- Female ( $F$ ): Full-Time ( $FT$ ) = $4321$ , Part-Time ( $PT$ ) = $3598$ , Total = $7919$
- Totals: Full-Time ( $FT$ ) = $8129$ , Part-Time ( $PT$ ) = $6133$ , Grand Total = $14262$
- Calculations:
- $P(FT) = \frac{8129}{14262} \approx 0.5700$
- $P(FT \cup F) = P(FT) + P(F) - P(FT \bigcap F) = \frac{8129}{14262} + \frac{7919}{14262} - \frac{4321}{14262} = \frac{11727}{14262} \approx 0.8223$
- $P(F \cup M) = P(F) + P(M) = \frac{7919}{14262} + \frac{6343}{14262} = \frac{14262}{14262} = 1.0$
- $P(PT \bigcap M) = \frac{2535}{14262} \approx 0.1777$
Example 3: Exercise Preferences Across Age Groups ( $150$ Individuals):
- Data Table:
- Under 40: Prefers Morning ( $Y$ ) = $38$ , Does Not Prefer Morning ( $N$ ) = $54$ , Total = $92$
- 40 and Over: Prefers Morning ( $Y$ ) = $25$ , Does Not Prefer Morning ( $N$ ) = $33$ , Total = $58$
- Totals: Prefers Morning ( $Y$ ) = $63$ , Does Not Prefer Morning ( $N$ ) = $87$ , Grand Total = $150$
- Calculations:
- Problem 1: Find the probability that a selected individual is Under 40 and prefers morning exercise. $P(\text{Under 40} \bigcap Y) = \frac{38}{150} \approx 0.2533$
- Problem 2: Find the probability that a selected individual is Under 40 or does not prefer morning exercise. $P(\text{Under 40} \cup N) = P(\text{Under 40}) + P(N) - P(\text{Under 40} \bigcap N) = \frac{92}{150} + \frac{87}{150} - \frac{54}{150} = \frac{125}{150} \approx 0.8333$
- Problem 3: Suppose the individual chosen is 40 and Over. What is the probability that he/she does not prefer morning exercise? $P(N | \text{40 and Over}) = \frac{P(N \bigcap \text{40 and Over})}{P(\text{40 and Over})} = \frac{33/150}{58/150} = \frac{33}{58} \approx 0.5690$

Decision Rules for Organizing Data: Tree Diagrams vs. Contingency Tables

Selection Criteria Rule of Thumb:
- Use a Contingency Table if the available information consists of two marginal probabilities and at least one joint probability.
- Use a Tree Diagram if the available information consists of two marginal probabilities and at least one conditional probability.

Comprehensive Case Studies: Smartphone Retailer Analysis

Case Study 1: Constructing a Contingency Table & Testing Independence:
- Given Data:
- $60\%$ of customers buy a phone case ( $C \implies P(C) = 0.60$
- $35\%$ of customers buy a screen protector ( $S \implies P(S) = 0.35$
- $25\%$ of customers buy both ( $C \bigcap S \implies P(C \bigcap S) = 0.25$
- Constructed Contingency Table:
- Case ( $C$ ): Protector ( $S$ ) = $0.25$ , No Protector ( $S'$ ) = $0.35$ , Total = $0.60$
- No Case ( $C'$ ): Protector ( $S$ ) = $0.10$ , No Protector ( $S'$ ) = $0.30$ , Total = $0.40$
- Totals: Protector ( $S$ ) = $0.35$ , No Protector ( $S'$ ) = $0.65$ , Grand Total = $1.00$
- Conditional Probability Calculation:
- Find $P(S|C)$ , the probability that a customer buys a screen protector given that they bought a phone case: $P(S|C) = \frac{P(S \bigcap C)}{P(C)} = \frac{0.25}{0.60} \approx 0.4167$
- Independence Proof:
- Compare $P(S|C)$ to $P(S)$ : $P(S|C) = 0.4167$ and $P(S) = 0.35$ .
- Since $P(S|C) \neq P(S)$ , the events of buying a phone case and buying a screen protector are not independent.
Case Study 2: Constructing a Tree Diagram & Total Probability:
- Given Data:
- $60\%$ of customers buy a phone case ( $P(C) = 0.60$
- Among case buyers, $40\%$ buy a screen protector ( $P(S|C) = 0.40$
- Among non-case buyers, $20\%$ buy a screen protector ( $P(S|C') = 0.20$
- Tree Diagram Branch Breakdown:
- Primary Branch 1: Customer buys phone case ( $C = 0.60$ )
  - Sub-branch 1a: Buys screen protector ( $P(S|C) = 0.40$
  - Outcome: $P(C \bigcap S) = P(C) \times P(S|C) = 0.60 \times 0.40 = 0.24$
  - Sub-branch 1b: Does not buy screen protector ( $P(S'|C) = 0.60$
  - Outcome: $P(C \bigcap S') = P(C) \times P(S'|C) = 0.60 \times 0.60 = 0.36$
- Primary Branch 2: Customer does NOT buy phone case ( $C' = 0.40$ )
  - Sub-branch 2a: Buys screen protector ( $P(S|C') = 0.20$
  - Outcome: $P(C' \bigcap S) = P(C') \times P(S|C') = 0.40 \times 0.20 = 0.08$
  - Sub-branch 2b: Does not buy screen protector ( $P(S'|C') = 0.80$
  - Outcome: $P(C' \bigcap S') = P(C') \times P(S'|C') = 0.40 \times 0.80 = 0.32$
- Calculating Total Probability $P(S)$ :
- Find the overall probability that a customer buys a screen protector $P(S)$ , using the Law of Total Probability: $P(S) = P(C \bigcap S) + P(C' \bigcap S)$ $P(S) = [P(C) \times P(S|C)] + [P(C') \times P(S|C')]$ $P(S) = (0.60 \times 0.40) + (0.40 \times 0.20)$ $P(S) = 0.24 + 0.08 = 0.32$