Unit 5: Probability Rules, Simulations, and Independence

Basic Probability Rules

• Probability Model:     * A probability model must show all possible outcomes in the sample space.     * A probability model must show all probabilities for those outcomes.     * Rule 1: The probability of each individual outcome must be a value between 0 and 1 (inclusive, representing $0\%$ to $100\%$).     * Rule 2: The sum of all probabilities for the entire sample space must equal exactly 1 (expressing $100\%$ certainty that one of the outcomes in the sample space will occur).

• Complement:     * Definition: If $A$ is designated as event $A$, then $A^C$ is the complement of event $A$, also known as "not $A$."     * Conceptual Meaning: The complement is the event that event $A$ did not happen.     * Complement Rule Formula: $P(A^C) = 1 - P(A)$.     * Plain Language Explanation: The probability of the complement of $A$ is equivalent to "everything" ($100\%$ or 1) minus the probability of event $A$ actually occurring.

• Mutually Exclusive Events:     * Definition: Events that have no outcomes in common.     * Addition Rule for Mutually Exclusive Events: When events $A$ and $B$ are mutually exclusive, the probability of either event occurring is the sum of their individual probabilities: $P(A \text{ or } B) = P(A) + P(B)$.

Randomness, Probability, and Simulation

• Probability:     * Definition: The likelihood that something happens in the "long run."     * Nature of Randomness: Random phenomena are characterized as being unpredictable in the short run but becoming predictable in the long run.

• Law of Large Numbers:     * Definition: If a chance process is repeated many, many times, the proportion of desired outcomes obtained will approach the actual probability of that outcome.

• Simulation:     * Definition: The act of imitating a chance process, often used by statisticians to estimate probabilities when direct calculation is difficult or to verify theoretical models.

Two-Way Tables and Venn Diagrams

• Two-Way Table:     * A grid format (often $2 \times 2$) used to organize data for two categorical variables, arranged in rows and columns.

• Venn Diagram Components:     * Circle: Represents a specific event.     * Rectangle: Represents the entire sample space.     * Intersection: The overlapping space in the Venn diagram representing the event where both designated outcomes occur simultaneously ($A \cap B$ or "Both").     * Union: The total space covered by the events ($A \cup B$), representing the occurrence of either event $A$, event $B$, or both.

• General Addition Rule:     * This rule is used when events are not necessarily mutually exclusive (there is an overlap/intersection).     * Formula: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.     * Symbolism: The symbol $\cup$ denotes the union ("or"), and the symbol $\cap$ denotes the intersection ("and").     * Logic: One must subtract the intersection ($A \cap B$) because that probability is counted twice if you simply add $P(A)$ and $P(B)$.

Conditional Probability and Independence

• Conditional Probability:     * Definition: The probability that event $A$ will occur given that event $B$ has already occurred.     * Notation: Notated as $P(A|B)$, read as "probability of $A$ given $B$."     * Formula: $P(A|B) = \frac{P(A \cap B)}{P(B)}$ (the probability of both events occurring divided by the probability of the given condition).

• Independence:     * Test for Independence: Events $A$ and $B$ are independent if and only if $P(A) = P(A|B) = P(A|B^C)$.     * Interpretation: Independence means that knowing whether or not event $B$ occurred does not change the probability/chance of event $A$ occurring.

Questions & Discussion

• Application 5.2: How Prevalent is High Cholesterol?     * Context: American adults chosen at random. Event $A$: high cholesterol ($\ge 240\,mg/dl$); Event $B$: borderline high cholesterol ($200$ to $< 240\,mg/dl$). Data: $P(A) = 0.16$ and $P(B) = 0.29$.     * Question 1: Explain why events $A$ and $B$ are mutually exclusive.     * Answer: Events $A$ and $B$ are mutually exclusive because a randomly chosen American adult cannot have high cholesterol ($240\,mg/dl$ or above) and borderline high cholesterol ($200$ to $< 240\,mg/dl$) at the same time.     * Question 2: Say in plain language what the event "$A \text{ or } B$ " is, then find $P(A \text{ or } B)$.     * Plain Language: A randomly chosen American adult has either high or borderline high cholesterol.     * Calculation: $P(A \text{ or } B) = P(A) + P(B) = 0.16 + 0.29 = 0.45$.     * Question 3: Let $C$ be the event that the person chosen has normal cholesterol (less than $200\,mg/dl$). Find $P(C)$.     * logic: Normal is the complement of "borderline or high."     * Calculation: $P(C) = 1 - P(A \text{ or } B) = 1 - 0.45 = 0.55$. A randomly chosen American adult has a $55\%$ chance of having normal cholesterol.

• Application 5.1: Will the Train Arrive on Time?     * Context: NJ Transit claims its 8:00 a.m. train has probability $0.9$ of arriving on time.     * Question 1: Explain what probability $0.9$ means in this setting.     * Answer: The train has a $90\%$ chance of arriving on time in a large sample of many trips.     * Question 2: The train arrived on time 5 days in a row. What is the probability it arrives on time tomorrow?     * Answer: The probability remains $0.9$. A short streak of on-time arrivals does not change the long-term probability.     * Question 3: Describe a simulation using a 10-sided die for late arrivals ($3$ of $20$ days late).     * Answer: Assign digits $1$ through $9$ as "on time" ($90\%$) and digit $10$ as "late" ($10\%$). Roll the die $20$ times and record whether the train is on time or late for each roll.     * Question 4: Explain what the dot at $7$ on the dotplot represents.     * Answer: It represents one repetition of the simulation (out of $100$) where the train arrived late exactly $7$ times out of $20$ rolls of the die.     * Question 5: Estimate the probability that the train will arrive late on $3$ or more of $20$ days based on simulation results.     * Answer: According to the dotplot, there are $30$ dots at $3$ or higher. $P(\text{Late } \ge 3 \text{ out of } 20) = \frac{30}{100} = 0.3$ or $30\%$.     * Question 6: Is there convincing evidence that New Jersey Transit's claim is false given $3$ late arrivals in $20$ days?     * Answer: No. Because it is fairly likely ($30\%$ chance) that the train will arrive late $3$ or more days out of $20$ just by chance, there is not convincing evidence the claim is false.

• Application 5.3: Who Owns a Home?     * Context: Random sample of $500$ U.S. adults. Event $G$: High school graduate; Event $H$: Homeowner.     * Table Data:         * HS Grad ($G$) and Homeowner ($H$): $221$         * HS Grad ($G$) and Not Homeowner ($H^C$): $89$         * Not HS Grad ($G^C$) and Homeowner ($H$): $119$         * Not HS Grad ($G^C$) and Not Homeowner ($H^C$): $71$         * Total HS Grads: $310$; Total Not HS Grads: $190$; Total Homeowners: $340$; Total Not Homeowners: $160$.     * Question 1: Find $P(G^C)$.     * Answer: $P(G^C) = \frac{190}{500} = 0.38$, which is a $38\%$ chance.     * Question 2: Explain why $P(G \text{ or } H) \neq P(G) + P(H)$, and find $P(G \text{ or } H)$.     * Answer: They are not equal because it is possible to be both a high school graduate and a homeowner (the events overlap). Simple addition would double-count the $221$ individuals who are both.     * Calculation: $P(G \cup H) = \frac{310 + 340 - 221}{500} = \frac{429}{500} = 0.858$.     * Question 3: Structure of the Venn Diagram for this data.     * G region only: $89$     * H region only: $119$     * Overlap (G and H): $221$     * Outside regions (Neither): $71$     * Question 4: Find $P(\text{is not a high school graduate and is a homeowner})$.     * Answer: $P(G^C \cap H) = \frac{119}{500} = 0.238$.

• Application 5.4: Who Earns A's in College?     * Context: 10,000 grades from UNH categorized by School (Liberal Arts, EPS, Health) and Grade (A, B, Lower than B). Event $E$: grade from EPS; Event $L$: grade lower than a B.     * Data Table:         * EPS: $368$ (A), $432$ (B), $800$ (Lower than B); Total EPS = $1600$.         * Total Lower than B ($L$): $3666$.     * Question 1: Find $P(L|E)$ and describe it in words.     * Answer: $P(L|E) = \frac{P(L \cap E)}{P(E)} = \frac{800}{1600} = 0.5$ or $50\%$. There is a $50\%$ chance that a random course grade is lower than a B, given that it is from an E.P.S. course.     * Question 2: Are events $L$ and $E$ independent? Justify.     * Answer: For independence, $P(L)$ must equal $P(L|E)$.     * Check: $P(L) = \frac{3666}{10000} = 0.3666$ ($36.66\%$). $P(L|E) = 0.5$ ($50\%$). Since $0.3666 \neq 0.5$, the events are not independent.     * Question 3: Given the grade is not lower than a B, find the probability it came from an EPS course.     * Answer: Find $P(E | L^C)$. $P(L^C) = 10000 - 3666 = 6334$. $P(E \cap L^C) = 368 + 432 = 800$.     * Calculation: $P(E | L^C) = \frac{800}{6334} \approx 0.126$ or $12.6\%$.