Comprehensive Unit 2 Study Guide: Advanced Probability, Normal Distributions, and Statistical Inference

Course Curriculum & Instructional Objectives

• Unit Scope and Focus Areas:

  • Probability Foundations: Empirical versus theoretical probabilities, counting principles, permutations, combinations, and visual representation models (two-way tables, tree diagrams, and Venn diagrams).

  • Probability Laws: Addition rules, multiplication rules, conditional probability, complements, independent events, and mutually exclusive events.

  • Normal Distributions: Properties of bell curves, the Empirical Rule ($68-95-99.7\%$ Rule), standardization and $z$-score calculation, cumulative area determination using tables and technology, and graphical evaluation of normality (normal probability plots, dotplots, and boxplots).

  • Statistical Inference: The Central Limit Theorem ($\text{CLT}$), sampling distributions of the sample mean ($\bar{x}$), sources of sampling error, confidence interval construction, and six-step one-sample $z$-tests of significance.

• Instructional Schedule and Milestone Dates:

  • Tue/Wed, Sept 29–30: Practice Topics 3–5; Introduction to Probability Vocabulary and Diagrams.

  • Thu/Fri, Oct 1–2: Unit 1 Test administration.

  • Mon/Tue, Oct 5–6: Knowledge Check on Vocabulary and Diagrams; Instruction on Conditional Probability and Mathematical Rules.

  • Wed, Oct 7: PSAT Day (no regular class instruction).

  • Thu/Fri, Oct 8–9: Mutually Exclusive, Independent, and Dependent Events.

  • Mon, Oct 12: Student Holiday.

  • Tue/Wed, Oct 13–14: Knowledge Check on Conditional Probability, Mathematical Rules, Mutually Exclusive, Independent, and Dependent Events; Instruction on Counting Techniques.

  • Thu/Fri, Oct 15–16: Instruction on Theoretical Probability.

  • Mon, Oct 19: Instruction on Topic 9 (Subtopics 9-4 and 9-5: Spread, Empirical Rule, Standardization).

  • Tue/Wed, Oct 20–21: Knowledge Check on Counting Techniques and Theoretical Probability; Instruction on Topic 12 (Normal Distributions).

  • Thu/Fri, Oct 22–23: Instruction on Introduction to Inference.

  • Mon, Oct 26: Knowledge Check on Practice Topics 9, 12, and Intro to Inference; Comprehensive Review of Unit 2.

  • Tue/Wed, Oct 27–28: Unit 2 Test administration.

• Visible Learning Objectives (VLO Tracking Framework):

  • 2.PVDa: Understand the foundational distinction between empirical and theoretical probability.

  • 2.PVDb: Calculate empirical probabilities of various outcomes given experimental trial data.

  • 2.PVDc: Calculate theoretical probabilities given a defined sample space of equally likely outcomes.

  • 2.PVDd: State and interpret the Law of Large Numbers and explain its application to random processes.

  • 2.PVDe: Calculate the expected value ($E(X)$) of a discrete random variable from a probability distribution.

  • 2.PVDf: Construct tree diagrams and apply branch multiplication to compute sequential probabilities.

  • 2.PVDg: Construct multi-set Venn diagrams and use set regions to calculate event probabilities.

  • 2.PVDh: Correctly utilize set and probability notation: event complement ($A^c$ or $A'$), set union ($\cup$), and set intersection ($\cap$).

  • 2.CPMRa: Apply formal conditional probability notation: $P(B|A)$.

  • 2.CPMRb: Calculate conditional probabilities directly from two-way contingency tables by restricting the sample space.

  • 2.CPMRc: Apply the Complement Rule: $P(A^c) = 1 - P(A)$.

  • 2.CMPRd: Apply the General Addition Rule: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.

  • 2.CMPRe: Apply the General Multiplication Rule: $P(A \cap B) = P(A) \cdot P(B|A)$.

  • 2.CMPRf: Rearrange the General Multiplication Rule to solve algebraically for conditional probabilities: $P(B|A) = \frac{P(A \cap B)}{P(A)}$.

  • 2.MEIDa: Determine whether two events are mutually exclusive (disjoint) conceptually and algebraically.

  • 2.MEIDb: Apply the Addition Rule for Mutually Exclusive Events: $P(A \cup B) = P(A) + P(B)$.

  • 2.MEIDc: Use contextual reasoning to assess whether events are independent or dependent.

  • 2.MEIDd: Verify independence algebraically by demonstrating $P(A|B) = P(A)$ or $P(A \cap B) = P(A) \cdot P(B)$.

  • 2.MEIDe: Apply the Special Multiplication Rule for Independent Events.

  • 2.CTa: Apply the Fundamental Counting Principle to multi-stage decision processes.

  • 2.CTb: Evaluate factorial expressions ($n!$), including simplifying factorial fractions.

  • 2.CTc: Differentiate between scenarios requiring permutations (order matters) and combinations (order is irrelevant).

  • 2.CTd: Compute permutations ($_n P_r$) and combinations ($_n C_r$) using formal algebraic formulas.

  • 2.TPa: Use combinations and permutations to determine theoretical probabilities for complex configurations.

  • 2.TPb: Combine counting techniques with probability rules to solve multi-event compound probability problems.

  • 2.9a: Apply the Empirical Rule ($68-95-99.7\%$ Rule) to symmetric, mound-shaped distributions.

  • 2.9b: Compute standard scores ($z$-scores) and apply them to execute relative comparisons across disparate datasets.

  • 2.9c: Use inverse cumulative normal distribution functions (invNorm) to identify raw data cutoffs from percentiles.

  • 2.12d: Describe the geometric and structural properties of normal density curves.

  • 2.12e: Calculate probabilities and percentiles for arbitrary normal distributions using standard tables and technology.

  • 2.12f: Compute bounded normal interval probabilities using normalcdf(lower, upper, \mu, \sigma).

  • 2.12g: Assess sample normality using normal probability plots, dotplots, and boxplots.

Probability Foundations & Expected Value

• Definitions of Core Probability Concepts:

  • Chance Process (Random Process): Any activity or phenomenon with an uncertain outcome, but exhibiting a regular, predictable distribution of outcomes over a large number of repetitions.

  • Outcome: A single specific result that can occur from an experiment or chance process.

  • Sample Space ($S$): The comprehensive set of all possible outcomes of a random process. For a single fair six-sided die, $S = \{1, 2, 3, 4, 5, 6\}$.

  • Event ($E$): Any specific collection or subset of outcomes within a sample space ($E \subseteq S$). An example is rolling a prime number: $E = \{2, 3, 5\}$.

  • Equally Likely Outcomes: Outcomes that share an identical theoretical likelihood of occurring. When all outcomes in $S$ are equally likely, the probability of an event $E$ is given by:     $P(E) = \frac{n(E)}{n(S)} = \frac{\text{number of outcomes favorable to } E}{\text{total number of outcomes in } S}$

  • Fundamental Axioms of Probability:

  • For any event $E$, $0 \le P(E) \le 1$.

  • The probability of an impossible event is exactly $0$: $P(\emptyset) = 0$.

  • The probability of an absolute certainty is exactly $1$: $P(S) = 1$.

• Empirical versus Theoretical Probability:

  • Empirical (Experimental) Probability: The observed relative frequency with which an event occurs throughout actual physical trials or experiments. It is determined exclusively through direct observation:     $P_{\text{empirical}}(E) = \frac{\text{observed frequency of } E}{\text{total number of experimental trials}}$

  • Theoretical (Exact) Probability: The ratio of favorable outcomes to total possible outcomes derived through mathematical reasoning, combinatorial counting, or established physical properties of the system, without running physical trials.

  • Comparison of Probability and Statistics:

  • Probability: Operates from a known population to deduce the likelihood of sample outcomes. If a box contains $5$ blue, $5$ red, and $5$ white poker chips ($N = 15$), probability determines the exact likelihood of drawing a blue chip: $P(\text{Blue}) = \frac{5}{15} = \frac{1}{3}$.

  • Statistics: Operates in reverse, drawing an observed sample from an unknown population to infer the characteristics of that population. Data is summarized using descriptive statistics, and inferences regarding population parameters are made using inferential statistics.

• The Law of Large Numbers ($\text{LLN}$):

  • Formal Principle: While the outcome of any individual trial in a chance process is impossible to predict in the short term, as the number of repetitions increases indefinitely, the relative proportion of times that a specific outcome occurs approaches a single, constant numerical value (its theoretical probability).

  • Short-Term Misconception (Gambler's Fallacy): Probability is strictly a long-run property. In the short term, empirical proportions fluctuate significantly away from theoretical values. For a fair coin, $P(\text{Heads}) = 0.5$ indicates that heads will appear half the time over thousands of tosses; it does not guarantee $5$ heads and $5$ tails in $10$ tosses.

  • Coin Toss Demonstration:

* In the first $20$ tosses (graph a), the cumulative proportion of heads oscillates sharply between $0.2$ and $0.7$.
* Across $500$ tosses (graph b), the cumulative proportion flattens out, converging smoothly onto the theoretical value of $0.5$.

• Probability Distributions & Expected Value ($E(X)$):

  • Probability Distribution: A complete listing of all possible numerical values of a random process paired with their respective probabilities, satisfying $\sum P(x_i) = 1$.

  • Expected Value Definition: The long-run weighted average value achieved by a numerical random variable over an infinite series of repetitions:     $E(X) = \sum x_i \cdot P(x_i) = x_1 \cdot P(x_1) + x_2 \cdot P(x_2) + \dots + x_k \cdot P(x_k)$

  • Standard Die Calculation:

  • Outcomes: $x \in \{1, 2, 3, 4, 5, 6\}$, each with probability $P(x) = \frac{1}{6}$.

  • Expected Value Computation:       $E(X) = 1\left(\frac{1}{6}\right) + 2\left(\frac{1}{6}\right) + 3\left(\frac{1}{6}\right) + 4\left(\frac{1}{6}\right) + 5\left(\frac{1}{6}\right) + 6\left(\frac{1}{6}\right) = \frac{21}{6} = 3.5$

  • Interpretation: Over many rolls, the average outcome per roll approaches $3.5$, which lies at the center of the distribution despite never being obtainable on a single roll.

  • Insurance Underwriting Problem:

  • Policy Premium: An insurer charges an annual premium of $\text{\$}900$ per policy.

  • Payout Schedule: Minor accident payout = $\text{\$}1{,}500$ with $P = 0.15$; Major accident payout = $\text{\$}8{,}000$ with $P = 0.05$; No accident payout = $\text{\$}0$ with $P = 1 - (0.15 + 0.05) = 0.80$.

  • Expected Claim Payout per Policy:       $E(\text{Payout}) = 1500(0.15) + 8000(0.05) + 0(0.80) = 225 + 400 + 0 = 625$

  • Expected Net Profit per Policy:       $E(\text{Profit}) = \text{Premium} - E(\text{Payout}) = 900 - 625 = 275$

  • Expected Earnings for $10$ Policies:       $\text{Total Profit} = 10 \times 275 = 2750$

  • Automobile Dealership Hybrid Sales Problem:

  • Unit Economics: Purchase cost = $\text{\$}21{,}000$; Sale price = $\text{\$}24{,}000$; Profit per car = $24000 - 21000 = 3000$.

  • Probability Distribution for Weekly Cars Sold ($X$):

    • $P(0) = 0.32$

    • $P(1) = 0.28$

    • $P(2) = 0.15$

    • $P(3) = 0.11$

    • $P(4) = 0.08$

    • $P(5) = 0.06$

    • Check sum: $0.32 + 0.28 + 0.15 + 0.11 + 0.08 + 0.06 = 1.00$.

  • Expected Number of Hybrid Cars Sold per Week:       $E(X) = 0(0.32) + 1(0.28) + 2(0.15) + 3(0.11) + 4(0.08) + 5(0.06)$       $E(X) = 0 + 0.28 + 0.30 + 0.33 + 0.32 + 0.30 = 1.53$

  • Expected Weekly Profit:       $E(\text{Weekly Profit}) = 1.53 \times 3000 = 4590$

  • Restaurant Menu Profit Analysis:

  • Menu Items: Stew ($\text{\$}0.34$ profit, $12\%$ sold), Soup ($\text{\$}0.41$ profit, $7\%$ sold), Lasagna ($\text{\$}0.64$ profit, $45\%$ sold), Chili ($\text{\$}0.73$ profit, $36\%$ sold).

  • Expected Profit per Serving:       $E(\text{Profit}) = 0.34(0.12) + 0.41(0.07) + 0.64(0.45) + 0.73(0.36)$       $E(\text{Profit}) = 0.0408 + 0.0287 + 0.2880 + 0.2628 = 0.6203$

Visual & Organizational Probability Tools

• Tables and Two-Way Arrays:

  • Two-Dice Sum Sample Space ($6 \times 6$ Array):

  • Tossing two fair six-sided dice generates $6 \times 6 = 36$ equally likely ordered pairs $(d_1, d_2)$.

  • Complete sum matrix:

    • Row 1 ($d_1 = 1$): $2, 3, 4, 5, 6, 7$

    • Row 2 ($d_1 = 2$): $3, 4, 5, 6, 7, 8$

    • Row 3 ($d_1 = 3$): $4, 5, 6, 7, 8, 9$

    • Row 4 ($d_1 = 4$): $5, 6, 7, 8, 9, 10$

    • Row 5 ($d_1 = 5$): $6, 7, 8, 9, 10, 11$

    • Row 6 ($d_1 = 6$): $7, 8, 9, 10, 11, 12$

  • Total number of distinct possible sums: $11$ distinct values (integers from $2$ through $12$).

  • Equal Likelihood Assessment: The sums are not equally likely. Sums near the center have more permutations ($7$ has $6$ combinations: $1+6, 2+5, 3+4, 4+3, 5+2, 6+1$; probability $=\frac{6}{36} = \frac{1}{6}$), whereas extremes have only one ($2$ occurs via $1+1$, and $12$ via $6+6$; probability $=\frac{1}{36}$).

  • Calculations for Specific Events:

    • Event $E$ (Sum is a prime number: $2, 3, 5, 7, 11$):

    • Sum $= 2$: $(1,1)$ ($1$ outcome)

    • Sum $= 3$: $(1,2), (2,1)$ ($2$ outcomes)

    • Sum $= 5$: $(1,4), (2,3), (3,2), (4,1)$ ($4$ outcomes)

    • Sum $= 7$: $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$ ($6$ outcomes)

    • Sum $= 11$: $(5,6), (6,5)$ ($2$ outcomes)

    • Total favorable outcomes: $n(E) = 1 + 2 + 4 + 6 + 2 = 15$.

    • Probability: $P(E) = \frac{15}{36} = \frac{5}{12}$.

    • Event $F$ (Dice show the same number: doubles):

    • Pairs: $(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)$.

    • Probability: $P(F) = \frac{6}{36} = \frac{1}{6}$.

    • Event $G$ (Sum is odd: $3, 5, 7, 9, 11$):

    • Exactly half of the $36$ cells yield odd sums ($2 + 4 + 6 + 4 + 2 = 18$ outcomes).

    • Probability: $P(G) = \frac{18}{36} = \frac{1}{2}$.

    • Complement of Event $E$ ($E^c$: sum is not prime):

    • $P(E^c) = 1 - P(E) = 1 - \frac{15}{36} = \frac{21}{36} = \frac{7}{12}$.

  • Biological Punnett Squares:

  • Genetic systems model independent assortment and dominance relationships.

  • Monohybrid Heterozygous Cross ($Bb \times Bb$):

  * Alleles: Dominant $B$ (dimples), Recessive $b$ (no dimples).
  * Genotypes: $1\,BB$, $2\,Bb$, $1\,bb$.
  * Phenotypes: Dimples ($BB, Bb$) have probability $\frac{3}{4}$; no dimples ($bb$) has probability $\frac{1}{4}$.
* Heterozygous and Homozygous Dominant Cross ($Bb \times BB$):
  * Genotypes produced: $2\,BB$ and $2\,Bb$.
  * Probability of dimples: $P(\text{Dimples}) = \frac{4}{4} = 1$.
* Dihybrid Cross ($RrYy \times RrYy$):
  * Genes: Seed shape (Round $R$, Wrinkled $r$) and Seed color (Yellow $Y$, Green $y$).
  * Total combinations in $4 \times 4$ grid: $16$ equally likely cells.
  * $P(\text{Round and Yellow})$ (requires at least one $R$ and one $Y$): $\frac{9}{16}$.
  * $P(\text{Wrinkled and Yellow})$ (requires $rr$ and at least one $Y$): $\frac{3}{16}$.
  * $P(\text{Wrinkled and Green})$ (requires genotype $rryy$): $\frac{1}{16}$.

• Tree Diagrams:

  • Used when outcomes develop across sequential stages or conditional branches.

  • Three-Coin Toss Experiment:

* Sequential stages: 1st toss, 2nd toss, 3rd toss. Total outcomes: $2^3 = 8$ equally likely paths:

      $\{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$ * Exactly $2$ heads: $(HHT, HTH, THH) \implies P(\text{exactly } 2\text{ heads}) = \frac{3}{8}$. * At least $2$ heads: $(HHT, HTH, THH, HHH) \implies P(\ge 2\text{ heads}) = \frac{4}{8} = \frac{1}{2}$. * No heads: $(TTT) \implies P(\text{no heads}) = \frac{1}{8}$. * At most $2$ tails: Everything except $TTT \implies P(\le 2\text{ tails}) = 1 - \frac{1}{8} = \frac{7}{8}$.

• Three Bags Chip Selection:

  • Bag 1: Chips labeled $\{1, 2, 3, 4\}$ ($4$ outcomes).

  • Bag 2: Chips labeled $\{1, 2, 3\}$ ($3$ outcomes).

  • Bag 3: Chips labeled $\{1, 2\}$ ($2$ outcomes).

  • Total sample space size: $4 \times 3 \times 2 = 24$ equally likely triplets $(c_1, c_2, c_3)$.

  • Probability of just one $2$ in the outcome:

    • Case 1 ($c_1 = 2, c_2 \ne 2, c_3 \ne 2$): $1 \times 2 \times 1 = 2$

    • Case 2 ($c_1 \ne 2, c_2 = 2, c_3 \ne 2$): $3 \times 1 \times 1 = 3$

    • Case 3 ($c_1 \ne 2, c_2 \ne 2, c_3 = 2$): $3 \times 2 \times 1 = 6$

    • Total outcomes with exactly one $2$: $2 + 3 + 6 = 11$.

    • $P(\text{just one } 2) = \frac{11}{24}$.

  • Probability of at least one $3$ in the outcome:

    • Complement is no $3$s chosen: Bag 1 has $3$ non-$3$s ($1, 2, 4$); Bag 2 has $2$ non-$3$s ($1, 2$); Bag 3 has $2$ non-$3$s ($1, 2$).

    • Total outcomes without any $3$: $3 \times 2 \times 2 = 12$.

    • $P(\ge 1\text{ three}) = 1 - \frac{12}{24} = \frac{12}{24} = \frac{1}{2}$.

  • Probability that the sum of outcomes is greater than $7$:

    • Maximum possible sum: $4 + 3 + 2 = 9$.

    • Minimum possible sum: $1 + 1 + 1 = 3$.

    • Sum $= 8$: $(4, 3, 1), (4, 2, 2), (3, 3, 2)$ ($3$ outcomes).

    • Sum $= 9$: $(4, 3, 2)$ ($1$ outcome).

    • Total outcomes with sum $> 7$: $3 + 1 = 4$.

    • $P(\text{Sum} > 7) = \frac{4}{24} = \frac{1}{6}$.

  • Probability that the sum of outcomes is prime:

    • Possible prime sums: $3, 5, 7$.

    • Sum $= 3$: $(1, 1, 1)$ ($1$ outcome).

    • Sum $= 5$: $(1, 2, 2), (1, 3, 1), (2, 1, 2), (2, 2, 1), (3, 1, 1)$ ($5$ outcomes).

    • Sum $= 7$: $(4, 2, 1), (4, 1, 2), (3, 2, 2), (3, 3, 1), (2, 3, 2)$ ($5$ outcomes).

    • Total prime outcomes: $1 + 5 + 5 = 11$.

    • $P(\text{Sum is prime}) = \frac{11}{24}$.

• Currency Bills Selection Without Replacement:

  • Bills: \{\text{\}1, \text{\}2, \text{\}5, \text{\}10, \text{\$}20\}. Select $2$ bills without replacement.

  • Total ordered outcomes: $5 \times 4 = 20$ ordered pairs, or $\binom{5}{2} = 10$ unordered pairs.

  • Probability of choosing a bill worth at least $\text{\$}2$: Only the pair $(1, 1)$ contains no bill $\ge 2$, but since bills are unique, every single pair drawn contains at least one bill worth $\ge \text{\$}2$. Thus, $P = \frac{20}{20} = 1$.

  • Probability of choosing the $\text{\$}1$ and $\text{\$}2$ bill: Ordered outcomes $(1, 2)$ and $(2, 1)$ yield $2$ outcomes: $P = \frac{2}{20} = \frac{1}{10}$.

  • Probability of having a total of at least $\text{\$}10$:

    • Unordered pairs with sum $< 10$: $(1, 2)=3$, $(1, 5)=6$, $(2, 5)=7$. ($3$ unordered pairs, or $6$ ordered pairs).

    • Pairs with sum $\ge 10$: $10 - 3 = 7$ unordered pairs ($14$ ordered pairs).

    • $P(\text{Total} \ge 10) = \frac{14}{20} = \frac{7}{10}$.

  • Probability of having a total of at most $\text{\$}20$:

    • Pairs exceeding $\text{\$}20$: Pairs containing the $\text{\$}20$ bill: $(20, 1)=21$, $(20, 2)=22$, $(20, 5)=25$, $(20, 10)=30$. ($4$ unordered pairs, or $8$ ordered pairs).

    • Pairs with sum $\le 20$: $10 - 4 = 6$ unordered pairs ($12$ ordered pairs).

    • $P(\text{Total} \le 20) = \frac{12}{20} = \frac{3}{5}$.

• Sequential Flight Punctuality and Luggage Connection:

* Flight 1 leaves on time: $P(\text{On Time}) = 0.15$; Delayed: $P(\text{Not On Time}) = 0.85$.
* Conditional luggage probabilities:
  * $P(\text{Luggage Arrives} | \text{On Time}) = 0.95$
  * $P(\text{Luggage Lost} | \text{On Time}) = 0.05$
  * $P(\text{Luggage Arrives} | \text{Not On Time}) = 0.65$
  * $P(\text{Luggage Lost} | \text{Not On Time}) = 0.35$
* Joint path probabilities:
  * $P(\text{On Time} \cap \text{Luggage Arrives}) = 0.15 \times 0.95 = 0.1425$
  * $P(\text{On Time} \cap \text{Luggage Lost}) = 0.15 \times 0.05 = 0.0075$
  * $P(\text{Not On Time} \cap \text{Luggage Arrives}) = 0.85 \times 0.65 = 0.5525$
  * $P(\text{Not On Time} \cap \text{Luggage Lost}) = 0.85 \times 0.35 = 0.2975$
* Probability that flight is late and luggage does not arrive: $0.2975$.
* Total Probability of luggage arriving at final destination:

      $P(\text{Luggage Arrives}) = 0.1425 + 0.5525 = 0.6950$

• Venn Diagrams:

  • Used for set overlaps, unions, intersections, and complements.

  • Set Notation Conventions:

  • Universe ($S$): Represented by the surrounding rectangle.

  • Complement ($A^c$ or $A'$): "Not $A$", the area outside circle $A$.

  • Intersection ($A \cap B$): "$A$ and $B$", the central overlapping region.

  • Union ($A \cup B$): "$A$ or $B$", the combined area across both circles.

  • Prime and Even Numbers Venn Diagram:

* Set components: Prime only: $\{3, 5, 11\}$; Both Prime and Even: $\{2\}$; Even only: $\{4, 8, 10, 16\}$; Outside both: $\{9, 15\}$.
* Total elements in universe: $3 + 1 + 4 + 2 = 10$.
* $P(\text{Prime}) = \frac{4}{10} = \frac{2}{5}$.
* $P(\text{Even}) = \frac{5}{10} = \frac{1}{2}$.
* $P(\text{Prime and Even}) = P(\text{Prime} \cap \text{Even}) = \frac{1}{10}$.
* $P(\text{Neither Prime nor Even}) = \frac{2}{10} = \frac{1}{5}$.

• High School Sports Participation ($100$ Students):

* Regions:
  * Basketball only = $18$
  * Track only = $24$
  * Swimming only = $17$
  * Basketball and Track only = $11$
  * Basketball and Swimming only = $8$
  * Track and Swimming only = $1$
  * All three sports ($B \cap T \cap S$) = $7$
  * Non-athletes (outside all circles) = $14$
* Total students: $18 + 24 + 17 + 11 + 8 + 1 + 7 + 14 = 100$.
* $P(\text{Participates in sports}) = \frac{100 - 14}{100} = \frac{86}{100} = 0.86$.
* $P(B \cap T \cap S) = \frac{7}{100} = 0.07$.
* $n(B) = 18 + 11 + 8 + 7 = 44$; $n(T) = 24 + 11 + 1 + 7 = 43$.
* $P(B \cup T) = \frac{44 + 43 - (11 + 7)}{100} = \frac{87 - 18}{100} = \frac{69}{100} = 0.69$.
* $P((B \cup T)^c) = 1 - 0.69 = 0.31$ (Swimming only $17$ + None $14 = 31$).
* $P(T \cap S) = \frac{1 + 7}{100} = \frac{8}{100} = 0.08$.
* $P((T \cap S)^c) = 1 - 0.08 = 0.92$.

• Car Characteristics Probability Venn Diagram:

* Given: $64\%$ foreign-made ($P(F) = 0.64$), $12\%$ blue ($P(B) = 0.12$), $7.7\%$ blue and foreign ($P(F \cap B) = 0.077$).
* Disjoint regions:
  * Foreign only: $P(F \cap B^c) = 0.64 - 0.077 = 0.563$
  * Blue only: $P(F^c \cap B) = 0.12 - 0.077 = 0.043$
  * Neither foreign nor blue: $1 - (0.563 + 0.077 + 0.043) = 0.317$
* Calculations:
  * $P(F \cup B) = 0.563 + 0.077 + 0.043 = 0.683$
  * $P(F^c \cap B) = 0.043$
  * $P(F | B) = \frac{P(F \cap B)}{P(B)} = \frac{0.077}{0.077 + 0.043} = \frac{0.077}{0.12} \approx 0.6417$
  * $P(B^c | F^c) = \frac{P(B^c \cap F^c)}{P(F^c)} = \frac{0.317}{1 - 0.64} = \frac{0.317}{0.36} \approx 0.8806$

Conditional Probability & Two-Way Contingency Tables

• Mathematical Definition of Conditional Probability:

  • The probability that event $B$ will occur given that event $A$ has already occurred, denoted $P(B|A)$.

  • Restricts the effective sample space from the entire universe $S$ exclusively to the subset where event $A$ occurs:     $P(B|A) = \frac{P(A \cap B)}{P(A)}, \quad \text{provided } P(A) \ne 0$

  • Visualizing Sample Space Restriction:

  • Consider a collection of tiles containing circles and squares, shaded and unshaded.

  • If there are $5$ circles total and $2$ of them are shaded, then the probability of choosing a shaded tile given that it is a circle is:       $P(\text{Shaded} | \text{Circle}) = \frac{n(\text{Shaded} \cap \text{Circle})}{n(\text{Circle})} = \frac{2}{5}$

• Two-Way Contingency Table Calculations:

  • Utility Bill Payment Distribution ($N = 50$):

  • Cells: Male Online = $12$, Male Mail = $8$ (Total Male = $20$); Female Online = $24$, Female Mail = $6$ (Total Female = $30$).

  • $P(\text{Online} | \text{Male}) = \frac{12}{20} = 0.60$.

  • Higher Education Demographics (2005 Census Bureau Data in Thousands):

  • Data:

    • Two-year colleges: Males = $1866$, Females = $2462$

    • Four-year colleges: Males = $4324$, Females = $5517$

    • Graduate schools: Males = $1349$, Females = $1954$

  • Total Graduate School Students: $1349 + 1954 = 3303\text{ thousand}$.

  • $P(\text{Female} | \text{Graduate School}) = \frac{1954}{3303} \approx 0.5916 \approx 0.592$.

  • Camp Counselor Characteristics ($N = 80$):

  • Table:

    • Junior Males = $18$, Junior Females = $21$ (Total Juniors = $39$)

    • Senior Males = $25$, Senior Females = $16$ (Total Seniors = $41$)

    • Total Males = $43$, Total Females = $37$

  • Probabilities:

    • $P(\text{Junior}) = \frac{39}{80} = 0.4875$

    • $P(\text{Female}) = \frac{37}{80} = 0.4625$

    • $P(\text{Senior} \cap \text{Male}) = \frac{25}{80} = 0.3125$

    • $P(\text{Junior} | \text{Female}) = \frac{21}{37} \approx 0.5676$

    • $P(\text{Male} | \text{Senior}) = \frac{25}{41} \approx 0.6098$

  • Employee Education and Salary Distribution:

  • Categories: Under $\text{\$}20{,}000$, \text{\}20{,}000\text{ to }\text{\}30{,}000, Over $\text{\$}30{,}000$.

  • Less than High School: $69, 36, 2$ (Total = $107$)

  • High School: $112, 98, 14$ (Total = $224$)

  • Some College: $102, 193, 143$ (Total = $438$)

  • College Degree: $13, 173, 245$ (Total = $431$)

  • Overall Grand Total: $107 + 224 + 438 + 431 = 1200$

  • Column Totals: Under $\text{\$}20\text{k} = 296$; \text{\}20\text{k–}\text{\}30\text{k} = 500; Over $\text{\$}30\text{k} = 404$.

  • Probabilities:

    • $P(\text{Less than high school}) = \frac{107}{1200} \approx 0.0892$

    • $P(\text{Over }\text{\$}30\text{k} \cap \text{Less than HS}) = \frac{2}{1200} \approx 0.00167$

    • $P(\text{Over }\text{\$}30\text{k} | \text{High school only}) = \frac{14}{224} = 0.0625$

    • $P(\text{High school or less} | \text{Over }\text{\$}30\text{k}) = \frac{2 + 14}{404} = \frac{16}{404} \approx 0.0396$

  • Student Leisure Activities by Gender:

  • Female: Sports $39$, Hiking $48$, Reading $85$, Phoning $62$, Shopping $71$, Other $29$ (Total Females = $334$)

  • Male: Sports $67$, Hiking $58$, Reading $76$, Phoning $54$, Shopping $68$, Other $39$ (Total Males = $362$)

  • Column Totals: Sports = $106$, Hiking = $106$, Reading = $161$, Phoning = $116$, Shopping = $139$, Other = $68$.

  • Overall Total: $334 + 362 = 696$.

  • Calculations:

    • $P(\text{Sports} | \text{Female}) = \frac{39}{334} \approx 0.1168$

    • $P(\text{Female} | \text{Sports}) = \frac{39}{106} \approx 0.3679$

    • $P(\text{Reading} | \text{Male}) = \frac{76}{362} \approx 0.2099$

    • $P(\text{Male} | \text{Reading}) = \frac{76}{161} \approx 0.4720$

    • $P(\text{Hiking} | \text{Female}) = \frac{48}{334} \approx 0.1437$

    • $P(\text{Hiking} | \text{Male}) = \frac{58}{362} \approx 0.1602$

  • Red Dye #2 Laboratory Rat Carcinogenicity Study ($N = 88$):

  • Data breakdown across three variables (Tumor Status, Survival, Dosage):

    • Died Before End of Study ($131$ weeks): Low Dose: Tumor $= 4$, No Tumor $= 26$ (Total $= 30$); High Dose: Tumor $= 7$, No Tumor $= 16$ (Total $= 23$).

    • Survived to End of Study: Low Dose: Tumor $= 0$, No Tumor $= 14$ (Total $= 14$); High Dose: Tumor $= 7$, No Tumor $= 14$ (Total $= 21$).

    • Margin Totals: High Dose Total = $23 + 21 = 44$; Low Dose Total = $30 + 14 = 44$.

  • Conditional Analysis:

    • $P(\text{Tumor} | \text{High Dosage}) = \frac{7 + 7}{44} = \frac{14}{44} \approx 0.3182$

    • $P(\text{Tumor} | \text{Low Dosage}) = \frac{4 + 0}{44} = \frac{4}{44} \approx 0.0909$

    • Comparison: Rats receiving the high dosage develop cancerous tumors at more than triple the rate ($31.8\%$ vs. $9.1\%$) of those receiving the low dosage, suggesting that high dosages of Red Dye #2 correlate with tumor growth.

    • $P(\text{Tumor} | \text{Died Before Study Ended}) = \frac{4 + 7}{30 + 23} = \frac{11}{53} \approx 0.2075$

    • $P(\text{Tumor} | \text{Survived to End}) = \frac{0 + 7}{14 + 21} = \frac{7}{35} = 0.2000$

Mathematical Rules of Probability

• Summary of Probability Rules:

  • Complement Rule:     $P(A^c) = 1 - P(A)$

  • Application: Rolling two dice, $P(\text{Sum} \ge 3) = 1 - P(\text{Sum} = 2) = 1 - \frac{1}{36} = \frac{35}{36}$.

  • General Addition Rule (Any Two Events):     $P(A \cup B) = P(A) + P(B) - P(A \cap B)$

  • Dice Example: Rolling a single die, probability of rolling a prime ($2, 3, 5$) or even ($2, 4, 6$):       $P(\text{Prime} \cup \text{Even}) = P(\text{Prime}) + P(\text{Even}) - P(\text{Prime} \cap \text{Even}) = \frac{3}{6} + \frac{3}{6} - \frac{1}{6} = \frac{5}{6}$

  • Camp Counselor Example: Counselor is a junior or a female:       $P(\text{Junior} \cup \text{Female}) = \frac{39}{80} + \frac{37}{80} - \frac{21}{80} = \frac{55}{80} = \frac{11}{16} = 0.6875$

  • General Multiplication Rule:     $P(A \cap B) = P(A) \cdot P(B|A)$

  • Camp Counselor Example: Counselor is a junior and female:       $P(\text{Junior} \cap \text{Female}) = P(\text{Junior}) \cdot P(\text{Female}|\text{Junior}) = \left(\frac{39}{80}\right)\left(\frac{21}{39}\right) = \frac{21}{80} = 0.2625$

• Mutually Exclusive (Disjoint) Events:

  • Definition: Events that cannot occur simultaneously; they have no overlapping outcomes ($A \cap B = \emptyset$).

  • Intersection Probability:     $P(A \cap B) = 0$

  • Special Addition Rule for Mutually Exclusive Events:     $P(A \cup B) = P(A) + P(B)$

  • Evaluating Mutual Exclusivity:

  • Rolling a $2$ and rolling a $3$ on a single die: Mutually exclusive ($P = 0$).

  • Rolling an even number and a multiple of $3$: Not mutually exclusive (outcome $6$ belongs to both).

  • Choosing an integer from $1$ to $100$: Less than $40$ and greater than $40$ are mutually exclusive. Odd and multiple of $4$ are mutually exclusive (multiples of $4$ are even). Less than $50$ and greater than $40$ are not mutually exclusive (numbers $41$ through $49$ satisfy both).

  • Local High School Languages: $37\%$ take Spanish, $15\%$ take French, with each student allowed at most one foreign language per semester:       $P(\text{Spanish} \cup \text{French}) = 0.37 + 0.15 = 0.52$

  • Mutually Exclusive vs. Complementary Events:

  • Complementary events are mutually exclusive and their probabilities must sum to $1$ ($P(A) + P(A^c) = 1$).

  • All complementary events are mutually exclusive, but not all mutually exclusive events are complementary (e.g., Freshman and Sophomore are mutually exclusive, but not complementary if Juniors and Seniors also exist).

• Independent versus Dependent Events:

  • Conceptual Definition: Two events are independent if the occurrence of one event has no effect on the probability of the other event occurring. Otherwise, they are dependent.

  • Algebraic Definitions: Events $A$ and $B$ are independent if and only if:     $P(A|B) = P(A) \quad \text{or} \quad P(B|A) = P(B) \quad \text{or} \quad P(A \cap B) = P(A) \cdot P(B)$

  • Card Dealing Example (Independence Check):

  • Let $F$ be the event that the first card drawn from a $52$-card deck is an ace, and $S$ be the event that the second card is an ace (without replacement):       $P(S) = \frac{4}{52} = \frac{1}{13} \approx 0.0769$       $P(S|F) = \frac{3}{51} = \frac{1}{17} \approx 0.0588$       Since $P(S|F) \ne P(S)$, the events are dependent.

  • Special Multiplication Rule for Independent Events:     $P(A \cap B) = P(A) \cdot P(B)$

  • Picnic Snack Selection Example:

  • $10$ diet drinks and $5$ regular drinks (total $= 15$).

  • $8$ bags of fat-free chips and $12$ bags of regular chips (total $= 20$).

  • Because drink choice and chip choice do not influence each other, they are independent:       $P(\text{Diet} \cap \text{Fat-free}) = \left(\frac{10}{15}\right)\left(\frac{8}{20}\right) = \left(\frac{2}{3}\right)\left(\frac{2}{5}\right) = \frac{4}{15} \approx 0.267$       $P(\text{Regular Drink} \cap \text{Regular Chips}) = \left(\frac{5}{15}\right)\left(\frac{12}{20}\right) = \left(\frac{1}{3}\right)\left(\frac{3}{5}\right) = \frac{1}{5} = 0.200$

  • Dependent Banana Selection:

  • $6$ green and $8$ yellow bananas (total $= 14$). Two bananas selected without replacement:       $P(\text{Both Yellow}) = P(Y_1) \cdot P(Y_2|Y_1) = \left(\frac{8}{14}\right)\left(\frac{7}{13}\right) = \left(\frac{4}{7}\right)\left(\frac{7}{13}\right) = \frac{4}{13} \approx 0.3077$

  • Critical Mathematical Warnings:

  • Mutual Exclusivity versus Independence: Mutually exclusive events with non-zero probabilities can never be independent. If events $A$ and $B$ are mutually exclusive, knowing $B$ has occurred means $A$ cannot occur, setting $P(A|B) = 0$. Because $P(A) > 0$, $P(A|B) \ne P(A)$.

  • Multiplication Assumption Warning: Do not multiply unconditional probabilities unless independence has been formally verified.

• Worked Compound Probability Problems:

  • High School Calculus Class ($18$ Juniors, $10$ Seniors; Total $= 28$):

  • Subgroups: Seniors: $6$ females, $4$ males. Juniors: $12$ males, $6$ females.

  • Total Females = $6 + 6 = 12$; Total Males = $12 + 4 = 16$.

  • $P(\text{Junior} \cup \text{Female}) = \frac{18}{28} + \frac{12}{28} - \frac{6}{28} = \frac{24}{28} = \frac{6}{7} \approx 0.8571$

  • $P(\text{Senior} \cup \text{Female}) = \frac{10}{28} + \frac{12}{28} - \frac{6}{28} = \frac{16}{28} = \frac{4}{7} \approx 0.5714$

  • $P(\text{Junior} \cup \text{Senior}) = \frac{18 + 10}{28} = \frac{28}{28} = 1.0$

  • Standard Deck Card Draw (King or Heart):

  • Deck size $= 52$. Kings $= 4$, Hearts $= 13$, King of Hearts $= 1$.

  • $P(\text{King} \cup \text{Heart}) = \frac{4}{52} + \frac{13}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13} \approx 0.3077$

  • Abstract Probabilities ($P(A) = 0.6$, $P(B) = 0.3$, $P(B|A) = 0.2$):

  • $P(A \cap B) = P(A) \cdot P(B|A) = 0.6 \times 0.2 = 0.12$

  • $P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.6 + 0.3 - 0.12 = 0.78$

  • Animal Shelter Dogs:

  • Dogs: White = $24\%$, Brown = $56\%$, Black = $100\% - (24\% + 56\%) = 20\%$.

  • $P(\text{Black}) = 0.20$

  • $P(\text{Not White}) = 1 - 0.24 = 0.76$

  • $P(\text{Brown} \cup \text{White}) = 0.56 + 0.24 = 0.80$

  • Job Interview Problem (Google and DropBox):

  • $P(G) = 0.35$, $P(D) = 0.48$, $P(G \cap D) = 0.15$.

  • $P(\text{Either Google or DropBox, but not both}) = P(G \cup D) - P(G \cap D)$

  • Since $P(G \cup D) = 0.35 + 0.48 - 0.15 = 0.68$, the probability is:       $0.68 - 0.15 = 0.53$

  • Commuter Transit Punctuality Problem:

  • Transit method: Subway $80\%$ ($P(S) = 0.80$), Uber $20\%$ ($P(U) = 0.20$).

  • Punctuality: $P(\text{On Time}|S) = 0.70$, $P(\text{On Time}|U) = 0.90$.

  • $P(\text{Subway} \cap \text{On Time}) = 0.80 \times 0.70 = 0.56$

  • $P(\text{Uber} \cap \text{On Time}) = 0.20 \times 0.90 = 0.18$

  • 2011 U.S. Senate Independence Check:

  • Composition: $51$ Democrats, $2$ Independents, $47$ Republicans ($N = 100$). Women = $17$ ($12$ Democrat, $5$ Republican). Men = $83$.

  • Addition Rule check: $P(\text{Woman} \cup \text{Democrat}) = \frac{17}{100} + \frac{51}{100} - \frac{12}{100} = \frac{56}{100} = 0.56$.

  • Conditional probabilities:       $P(\text{Democrat} | \text{Woman}) = \frac{12}{17} \approx 0.7059$       $P(\text{Democrat}) = \frac{51}{100} = 0.5100$

  • Independence evaluation: Since $P(\text{Democrat} | \text{Woman}) \ne P(\text{Democrat})$ ($0.7059 \ne 0.5100$), political party affiliation and gender are dependent within this Senate.

  • Geranium Plant Sequential Flowering: