Comprehensive Study Notes on Contingency Tables, Tree Diagrams, Venn Diagrams, and Probability Topics

Contingency Tables

A contingency table provides a structured method for displaying sample data involving two categorical variables that may be dependent or contingent upon one another. Contingency tables facilitate the calculation of simple, joint, and conditional probabilities by organizing raw counts into clear rows and columns with corresponding totals.

Structure and Basic Probability Calculations

• Table Layout: Rows represent the outcomes of one variable, while columns represent the outcomes of a second variable. Row totals, column totals, and the overall grand total ($N$) are calculated along the margins.

• Marginal Probability: The probability of a single event occurring, calculated using the row or column total divided by the total sample size $N$.

• Joint Probability ($\text{AND}$): The probability that two events occur simultaneously, found at the intersection of a specific row and column divided by $N$.

• Union Probability ($\text{OR}$): The probability that either event occurs, calculated using the addition rule:

$P(A \text{ OR } B) = P(A) + P(B) - P(A \text{ AND } B)$

• Conditional Probability ($\text{GIVEN}$ or $\mid$): The probability of an event occurring given that another event has already occurred. This restricts the sample space to the total of the given condition:

$P(A \mid B) = \frac{P(A \text{ AND } B)}{P(B)}$

Worked Examples and Exercises

Example 3.20: Speeding Violations and Cell Phone Use

A study examining speeding violations in the last year and cell phone usage while driving produced the following data:

• Sample Totals:

  • Total sample size: $N = 755$

  • Row totals: $305$ and $450$ ($305 + 450 = 755$)

  • Column totals: $70$ and $685$ ($70 + 685 = 755$)

• Calculations:

  • a. Find $P(\text{Driver is a cell phone user})$:

$P(\text{Cell Phone User}) = \frac{305}{755}$

*   **b. Find $P(\text{Driver had no violation in the last year})$**:

$P(\text{No Violation}) = \frac{685}{755}$

*   **c. Find $P(\text{Driver had no violation in the last year AND was a cell phone user})$**:

$P(\text{No Violation AND Cell Phone User}) = \frac{280}{755}$

*   **d. Find $P(\text{Driver is a cell phone user OR driver had no violation in the last year})$**:

$P(\text{Cell Phone User OR No Violation}) = \frac{305}{755} + \frac{685}{755} - \frac{280}{755} = \frac{710}{755}$

*   **e. Find $P(\text{Driver is a cell phone user GIVEN driver had a violation in the last year})$**:

$P(\text{Cell Phone User} \mid \text{Violation}) = \frac{25}{70}$

*   **f. Find $P(\text{Driver had no violation last year GIVEN driver was not a cell phone user})$**:

$P(\text{No Violation} \mid \text{Non-Cell Phone User}) = \frac{405}{450}$

Try It 3.20: Athletic Stretching and Injury Rates

A study tracked athletes who stretch before exercising and whether they sustained injuries within the past year:

• Calculations:

  • a. Find $P(\text{Athlete stretches before exercising})$:

$P(\text{Stretches}) = \frac{350}{800}$

*   **b. Find $P(\text{Athlete stretches before exercising} \mid \text{No injury in the last year})$**:

$P(\text{Stretches} \mid \text{No Injury}) = \frac{295}{514}$

Example 3.21: Hiking Area Preference by Sex

A sample of $100$ hikers produced preferences for hiking areas:

• Completing Table 3.5:

  • Women Mountain Peaks = $45 - 18 - 16 = 11$

  • Men Coastline = $41 - 18 = 23$

  • Men Total = $100 - 45 = 55$

  • Men Lakes and Streams = $55 - 23 - 14 = 18$

  • Lakes and Streams Total = $16 + 18 = 34$

  • Mountain Peaks Total = $11 + 14 = 25$

• Testing Independence:

  • Events: Let $F = \text{being a woman}$, $C = \text{preferring the coastline}$.

$P(F \text{ AND } C) = \frac{18}{100} = 0.18$

$P(F) = \frac{45}{100} = 0.45$

$P(C) = \frac{41}{100} = 0.41$

$P(F)P(C) = (0.45)(0.41) = 0.1845$

*   Since $P(F \text{ AND } C) \neq P(F)P(C)$ ($0.18 \neq 0.1845$), the events "being a woman" and "preferring the coastline" are **not independent**.

• Conditional Probability:

  • Find the probability that a person is a man given that the person prefers hiking near lakes and streams ($M \mid L$):

    1. The word "given" indicates a conditional probability.

    2. Calculation:

$P(M \mid L) = \frac{18}{34}$

    3. The sample space is restricted to the $34$ hikers who prefer lakes and streams, not all $100$ hikers.

• Union Probability:

  • Find the probability that a person is a woman or prefers hiking on mountain peaks ($F \text{ OR } P$):

    1. $P(F) = \frac{45}{100}$

    2. $P(P) = \frac{25}{100}$

    3. $P(F \text{ AND } P) = \frac{11}{100}$

    4. Calculation:

$P(F \text{ OR } P) = P(F) + P(P) - P(F \text{ AND } P) = \frac{45}{100} + \frac{25}{100} - \frac{11}{100} = \frac{59}{100} = 0.59$

Try It 3.21: Cyclist Route Preference by Age Group

A random sample of $200$ cyclists was surveyed regarding route preferences:

• a. Find $P(H \mid O)$:

$P(H \mid O) = \frac{52}{90}$

• b. Test Independence of "being older" ($O$) and "preferring the hilly path" ($H$):

$P(O \text{ AND } H) = \frac{52}{200} = 0.26$

$P(O)P(H) = \left(\frac{90}{200}\right)\left(\frac{90}{200}\right) = (0.45)(0.45) = 0.2025$

Since $0.26 \neq 0.2025$, the events are not independent.

Example 3.22: Muddy Mouse and Door Choice

Muddy Mouse lives in a cage with three doors. Door selection is equally likely ($P(\text{Door One}) = P(\text{Door Two}) = P(\text{Door Three}) = \frac{1}{3}$).

• Door One: $P(\text{Caught} \mid \text{Door One}) = \frac{1}{5}$, $P(\text{Not Caught} \mid \text{Door One}) = \frac{4}{5}$

• Door Two: $P(\text{Caught} \mid \text{Door Two}) = \frac{1}{4}$, $P(\text{Not Caught} \mid \text{Door Two}) = \frac{3}{4}$

• Door Three: $P(\text{Caught} \mid \text{Door Three}) = \frac{1}{2}$, $P(\text{Not Caught} \mid \text{Door Three}) = \frac{1}{2}$

Probability Contingency Table (Table 3.8):

• Calculations:

  • b. Find $P(\text{Alissa does not catch Muddy})$:

$P(\text{Not Caught}) = \frac{41}{60}$

*   **c. Find $P(\text{Door One OR Door Two} \mid \text{Caught})$**:

$P(\text{Door One OR Door Two} \mid \text{Caught}) = \frac{P((\text{Door One OR Door Two}) \text{ AND Caught})}{P(\text{Caught})} = \frac{\frac{1}{15} + \frac{1}{12}}{\frac{19}{60}} = \frac{\frac{9}{60}}{\frac{19}{60}} = \frac{9}{19}$

Try It 3.22: Car Selection (Mutually Exclusive Choice)

Anna must buy a new car and can afford only one. $P(A) = 0.25$ and $P(B) = 0.65$.

• Because Anna can afford only one car, buying car A and buying car B are mutually exclusive ($P(A \text{ AND } B) = 0$).

• $P(A \text{ AND } B) = 0$

• $P(A \text{ OR } B) = P(A) + P(B) - P(A \text{ AND } B) = 0.25 + 0.65 - 0 = 0.90$

Example 3.23: United States Crime Index Rates

Crimes per $100,000$ inhabitants over four years (Table 3.10):

• Calculations:

  • a. Find $P(\text{Year 2 AND Robbery})$:

$P(\text{Year 2 AND Robbery}) = \frac{133.1}{4520.7}$

*   **b. Find $P(\text{Year 3 AND Burglary})$**:

$P(\text{Year 3 AND Burglary}) = \frac{701}{4520.7}$

*   **c. Find $P(\text{Year 3 OR Burglary})$**:

$P(\text{Year 3 OR Burglary}) = \frac{1087.1}{4520.7} + \frac{2853.0}{4520.7} - \frac{701}{4520.7} = \frac{3239.1}{4520.7}$

*   **d. Find $P(\text{Year 4} \mid \text{Vandalism})$**:

$P(\text{Year 4} \mid \text{Vandalism}) = \frac{26.8}{113.3}$

*   **e. Find $P(\text{Vehicle} \mid \text{Year 1})$**:

$P(\text{Vehicle} \mid \text{Year 1}) = \frac{314.7}{1222.2}$

Try It 3.23: Weight and Height Observational Study

Relationships between weight and height categories (Table 3.11):

• Calculations:

  • a. Totals: Row totals are $60$, $99$, $46$. Column totals are $50$, $104$, $51$. Grand Total $= 205$.

  • b. Find $P(\text{Tall})$:

$P(\text{Tall}) = \frac{50}{205}$

*   **c. Find $P(\text{Overweight AND Tall})$**:

$P(\text{Overweight AND Tall}) = \frac{18}{205}$

*   **d. Find $P(\text{Tall} \mid \text{Overweight})$**:

$P(\text{Tall} \mid \text{Overweight}) = \frac{18}{60}$

*   **e. Find $P(\text{Overweight} \mid \text{Tall})$**:

$P(\text{Overweight} \mid \text{Tall}) = \frac{18}{50}$

*   **f. Find $P(\text{Tall AND Underweight})$**:

$P(\text{Tall AND Underweight}) = \frac{12}{205}$

*   **g. Are Overweight and Tall independent?**

$P(\text{Overweight AND Tall}) = \frac{18}{205} \approx 0.0878$

$P(\text{Overweight})P(\text{Tall}) = \left(\frac{60}{205}\right)\left(\frac{50}{205}\right) = \frac{3000}{42025} \approx 0.0714$

Since $0.0878 \neq 0.0714$, the events Overweight and Tall are not independent.

Tree Diagrams

A tree diagram is a structural graph used to model sequential outcomes of an experiment. It consists of branches labeled with frequencies or probabilities.

Sampling Methods

• Sampling With Replacement: The selected item is returned to the population before the subsequent draw. The sample size and probabilities remain constant for each draw.

• Sampling Without Replacement: The selected item is not returned. The sample size decreases and probabilities change for each subsequent draw.

Worked Examples and Exercises

Example 3.24: Sampling With Replacement (Urn with 11 Balls)

An urn contains $11$ balls: $3$ red ($R$) and $8$ blue ($B$). Draw two balls, one at a time, with replacement.

• Sample Space Size: $11 \times 11 = 121$

• Outcome Frequencies:

  • $BB = 8 \times 8 = 64$

  • $BR = 8 \times 3 = 24$

  • $RB = 3 \times 8 = 24$

  • $RR = 3 \times 3 = 9$

  • Total frequency $= 64 + 24 + 24 + 9 = 121$

• Listing Specific Outcomes:

  • The $9$ $RR$ outcomes are: $R1R1, R1R2, R1R3, R2R1, R2R2, R2R3, R3R1, R3R2, R3R3$.

  • The $24$ $BR$ outcomes are: $B1R1, B1R2, B1R3, B2R1, B2R2, B2R3, \dots, B8R3$.

• Calculations:

  • b. Find $P(RR)$:

$P(RR) = \frac{9}{121}$

*   **c. Find $P(RB \text{ OR } BR)$**:

$P(RB \text{ OR } BR) = \frac{24 + 24}{121} = \frac{48}{121}$

*   **d. Find $P(R \text{ on 1st AND } B \text{ on 2nd})$**:

$P(RB) = \frac{24}{121}$

*   **e. Find $P(R \text{ on 2nd} \mid B \text{ on 1st})$**:

$P(R \text{ on 2nd} \mid B \text{ on 1st}) = \frac{3}{11}$

*   **f. Find $P(BB)$**:

$P(BB) = \frac{64}{121}$

*   **g. Find $P(B \text{ on 2nd} \mid R \text{ on 1st})$**:

$P(B \text{ on 2nd} \mid R \text{ on 1st}) = \frac{8}{11}$

Try It 3.24: Card Drawing With Replacement

A standard deck of $52$ cards contains $12$ face cards ($F$) and $40$ non-face cards ($N$). Draw two cards with replacement.

$P(FF) = \left(\frac{12}{52}\right)\left(\frac{12}{52}\right) = \left(\frac{3}{13}\right)\left(\frac{3}{13}\right) = \frac{9}{169}$

Example 3.25: Sampling Without Replacement (Urn with Marbles)

An urn contains $3$ red marbles ($R$) and $8$ blue marbles ($B$). Draw two marbles, one at a time, without replacement.

• Branch Probabilities:

  • First Draw: $P(R) = \frac{3}{11}$, $P(B) = \frac{8}{11}$

  • Second Draw Given Red First: $P(R \mid R) = \frac{2}{10}$, $P(B \mid R) = \frac{8}{10}$

  • Second Draw Given Blue First: $P(R \mid B) = \frac{3}{10}$, $P(B \mid B) = \frac{7}{10}$

• End-of-Branch Probabilities:

  • $P(RR) = \left(\frac{3}{11}\right)\left(\frac{2}{10}\right) = \frac{6}{110}$

  • $P(RB) = \left(\frac{3}{11}\right)\left(\frac{8}{10}\right) = \frac{24}{110}$

  • $P(BR) = \left(\frac{8}{11}\right)\left(\frac{3}{10}\right) = \frac{24}{110}$

  • $P(BB) = \left(\frac{8}{11}\right)\left(\frac{7}{10}\right) = \frac{56}{110}$

  • Total probability sum $= \frac{6 + 24 + 24 + 56}{110} = \frac{110}{110} = 1$

• Calculations:

  • a. Find $P(RR)$: $\frac{6}{110}$

  • b. Find $P(RB \text{ OR } BR)$:

$P(RB \text{ OR } BR) = \left(\frac{3}{11}\right)\left(\frac{8}{10}\right) + \left(\frac{8}{11}\right)\left(\frac{3}{10}\right) = \frac{24}{110} + \frac{24}{110} = \frac{48}{110}$

*   **c. Find $P(R \text{ on 2nd} \mid B \text{ on 1st})$**: $\frac{3}{10}$
*   **d. Find $P(R \text{ on 1st AND } B \text{ on 2nd})$**:

$P(RB) = \left(\frac{3}{11}\right)\left(\frac{8}{10}\right) = \frac{24}{110}$

*   **e. Find $P(BB)$**: $\frac{56}{110}$
*   **f. Find $P(B \text{ on 2nd} \mid R \text{ on 1st})$**: $\frac{8}{10}$

• General Tree Diagram Notation Rules:

  • $P(R \mid R)$ represents $P(R \text{ on 2nd} \mid R \text{ on 1st})$

  • $P(B \mid R)$ represents $P(B \text{ on 2nd} \mid R \text{ on 1st})$

  • $P(R \mid B)$ represents $P(R \text{ on 2nd} \mid B \text{ on 1st})$

  • $P(B \mid B)$ represents $P(B \text{ on 2nd} \mid B \text{ on 1st})$

Try It 3.25: Card Drawing Without Replacement

In a standard deck of $52$ cards ($12$ face cards $F$, $40$ non-face cards $N$), draw two cards without replacement.

• a. Find $P(FN \text{ OR } NF)$:

$P(FN \text{ OR } NF) = \left(\frac{12}{52}\right)\left(\frac{40}{51}\right) + \left(\frac{40}{52}\right)\left(\frac{12}{51}\right) = \frac{480}{2652} + \frac{480}{2652} = \frac{960}{2652}$

• b. Find $P(N \mid F)$:

$P(N \mid F) = \frac{40}{51}$

• c. Find $P(\text{at most one face card})$ (means $0$ or $1$ face card):

$P(\text{at most 1 face card}) = 1 - P(FF) = 1 - \left(\frac{12}{52}\right)\left(\frac{11}{51}\right) = 1 - \frac{132}{2652} = \frac{2520}{2652}$

• d. Find $P(\text{at least one face card})$ (means $1$ or $2$ face cards):

$P(\text{at least 1 face card}) = 1 - P(NN) = 1 - \left(\frac{40}{52}\right)\left(\frac{39}{51}\right) = 1 - \frac{1560}{2652} = \frac{1092}{2652}$

Example 3.26: Kitten Adoption (Humane Society)

A litter of kittens has $4$ tabby kittens and $5$ black kittens ($9$ total). A family selects two kittens randomly without replacement.

• a. Probability both kittens are tabby:

$P(TT) = \left(\frac{4}{9}\right)\left(\frac{3}{8}\right) = \frac{12}{72}$

• b. Probability one kitten of each coloring is selected:

$P(TB \text{ OR } BT) = \left(\frac{4}{9}\right)\left(\frac{5}{8}\right) + \left(\frac{5}{9}\right)\left(\frac{4}{8}\right) = \frac{20}{72} + \frac{20}{72} = \frac{40}{72}$

• c. Probability tabby chosen second when black chosen first:

$P(T \text{ on 2nd} \mid B \text{ on 1st}) = \frac{4}{8}$

• d. Probability of choosing two kittens of the same color:

$P(TT \text{ OR } BB) = \left(\frac{4}{9}\right)\left(\frac{3}{8}\right) + \left(\frac{5}{9}\right)\left(\frac{4}{8}\right) = \frac{12}{72} + \frac{20}{72} = \frac{32}{72}$

Try It 3.26: Ball Selection Without Replacement

Four red balls and three yellow balls are in a box ($7$ total). Two balls are drawn without replacement.

$P(\text{one ball of each color}) = P(RY \text{ OR } YR) = \left(\frac{4}{7}\right)\left(\frac{3}{6}\right) + \left(\frac{3}{7}\right)\left(\frac{4}{6}\right) = \frac{12}{42} + \frac{12}{42} = \frac{24}{42} = \frac{4}{7}$

Venn Diagrams

A Venn diagram visually displays outcomes of an experiment using a rectangular box to represent the sample space $S$ and circles/ovals inside to represent individual events.

Structure and Basic Concepts

• Sample Space ($S$ or $X$): Represented by the outer bounding box.

• Events: Represented by circles or ovals inside the rectangle.

• Intersection ($A \text{ AND } B$): The overlapping region shared by circles $A$ and $B$.

• Union ($A \text{ OR } B$): The total area covered by circle $A$, circle $B$, or their overlap.

• Complement / Outside: The area within the rectangle but outside the circles.

Worked Examples and Exercises

Example 3.27: Integer Outcomes 1 through 12

An experiment has outcomes $1, 2, 3, \dots, 12$ with equal probability. Event $A = \{1, 2, 3, 4, 5, 6\}$ and event $B = \{6, 7, 8, 9\}$.

• $A \text{ AND } B = \{6\}$

• $A \text{ OR } B = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$

• Elements outside both circles: $\{10, 11, 12\}$

Try It 3.27: Color Outcomes

Sample space $S = \{\text{black}, \text{white}, \text{red}, \text{orange}, \text{yellow}, \text{green}, \text{blue}, \text{purple}\}$. Event $C = \{\text{green}, \text{blue}, \text{purple}\}$ and event $P = \{\text{red}, \text{yellow}, \text{blue}\}$.

• $C \text{ AND } P = \{\text{blue}\}$

• $C \text{ OR } P = \{\text{green}, \text{blue}, \text{purple}, \text{red}, \text{yellow}\}$

• Elements outside both circles: $\{\text{black}, \text{white}, \text{orange}\}$

Example 3.28: Coin Flipping Sample Space

Flip two fair coins. Let $A = \text{tails on the first coin}$ and $B = \text{tails on the second coin}$.

• Sample Space $X = \{HH, HT, TH, TT\}$

• $A = \{TT, TH\}$

• $B = \{TT, HT\}$

• $A \text{ AND } B = \{TT\}$

• $A \text{ OR } B = \{TH, TT, HT\}$

• The outcome $HH$ is in neither $A$ nor $B$ and sits in the region outside both circles.

Try It 3.28: Six-Sided Die Rolling

Roll a fair six-sided die. Sample space $S = \{1, 2, 3, 4, 5, 6\}$. Let $A = \text{prime number rolled} = \{2, 3, 5\}$ and $B = \text{odd number rolled} = \{1, 3, 5\}$.

• $A \text{ AND } B = \{3, 5\}$

• $A \text{ OR } B = \{1, 2, 3, 5\}$

• Elements outside both circles: $\{4, 6\}$

Example 3.29: College Student Clubs and Part-Time Work

At a local college, $40\%$ of students belong to a club ($C$), $50\%$ work part time ($PT$), and $5\%$ work part time and belong to a club ($C \text{ AND } PT$).

• $P(C) = 0.40$

• $P(PT) = 0.50$

• $P(C \text{ AND } PT) = 0.05$

• Conditional Probability:

$P(C \mid PT) = \frac{P(C \text{ AND } PT)}{P(PT)} = \frac{0.05}{0.50} = 0.1$

• Union Probability:

$P(C \text{ OR } PT) = P(C) + P(PT) - P(C \text{ AND } PT) = 0.40 + 0.50 - 0.05 = 0.85$

Try It 3.29: Factory Workers Second Job and Spouse Work

At a factory, $50\%$ of workers work a second job ($W$), $25\%$ have a spouse who works ($S$), and $5\%$ work a second job and have a spouse who works ($W \text{ AND } S$).

• Circle $W$ only: $50\% - 5\% = 45\%$

• Circle $S$ only: $25\% - 5\% = 20\%$

• Intersection $W \text{ AND } S$: $5\%$

• Outside both circles: $100\% - (45\% + 20\% + 5\%) = 30\%$

Example 3.30: Blood Types (Type O and Rh- Factor)

In African Americans, $4\%$ have type O blood and negative Rh factor ($Rh^{-}$), $5\text{--}10\%$ have the $Rh^{-}$ factor, and $51\%$ have type O blood.

• Taking the average of $5\%$ and $10\%$, use $7.5\%$ for $Rh^{-}$.

• Let $O = \text{Type O blood}$ and $R = Rh^{-} \text{ factor}$.

  • a. $P(O) = 0.51$

  • b. $P(R) = 0.075$

  • c. $P(O \text{ AND } R) = 0.04$

  • d. $P(O \text{ OR } R) = P(O) + P(R) - P(O \text{ AND } R) = 0.51 + 0.075 - 0.04 = 0.545$

  • e. Overlapping Area Description: The overlapping area represents the percentage of African Americans who have both type O blood and the $Rh^{-}$ factor ($4\%$

  • f. Outside Area Description: The area inside the rectangle but outside both shapes represents the percentage of African Americans who have neither type O blood nor the $Rh^{-}$ factor ($1 - 0.545 = 0.455$ or $45.5\%$

Try It 3.30: Bookstore Customer Purchases

In a bookstore, $P(\text{Novel}) = 0.6$, $P(\text{Non-fiction}) = 0.4$, and $P(\text{Both}) = 0.2$.

• b. Find $P(\text{Novel OR Non-fiction})$:

$P(\text{Novel OR Non-fiction}) = 0.6 + 0.4 - 0.2 = 0.8$

• c. Overlapping Area Description: The overlapping area represents customers who buy both a novel and a non-fiction book ($20\%$

• d. Disjoint Event: Customers buying only compact disks are represented by a separate oval that does not overlap with books.

Stats Lab: Probability Topics

Learning Outcomes

• Use theoretical and empirical methods to estimate probabilities.

• Appraise the differences between theoretical and empirical estimates.

• Demonstrate an understanding of long-term relative frequencies.

Experimental Procedure

1. Population Count: Count out $40$ mixed-color M&Ms (approximately one small bag) and record quantities by color in Table 3.12 (Yellow $Y$, Green $G$, Blue $BL$, Brown $B$, Orange $O$, Red $R$).

2. Trial 1 (With Replacement): Place M&Ms in a cup. Pick two M&Ms one at a time, replacing the first M&M before picking the second. Record results in Table 3.14. Perform this process $12$ times.

3. Trial 2 (Without Replacement): Pick two M&Ms one at a time without replacing the first M&M before picking the second. Record results in Table 3.14, then return both M&Ms to the cup. Perform this process $12$ times.

4. Data Processing: Calculate empirical probabilities in Table 3.15 in unreduced fractional form.

Event Notation and Formulas

• $R_1$ = Red on first pick

• $R_2$ = Red on second pick

• $B_1$ = Brown on first pick

• $B_2$ = Brown on second pick

• $G_2$ = Green on second pick

• $\text{doubles}$ = Both picks are the same color

Discussion and Conceptual Questions

1. Difference Between Replacement Conditions: "With Replacement" and "Without Replacement" probabilities differ because sampling without replacement changes both the total population size (from $40$ to $39$) and the specific color composition available for the second draw.

2. Comparison of Probabilities: Converting $P(\text{no yellows})$ to decimal format rounded to four decimal places allows direct comparison between theoretical predictions and empirical experiment results.

3. Effect of Sample Size: Increasing the number of repeated trials (e.g., from $12$ to $240$) causes empirical probabilities to fluctuate and refine based on cumulative observed outcomes.

4. Law of Large Numbers: Increasing trial counts causes empirical probabilities to move closer to theoretical probabilities, as long-term relative frequencies stabilize toward theoretical values as $n \to \infty$.

5. Joint vs. Conditional Meaning: $P(G_1 \text{ AND } R_2)$ represents the joint probability of selecting green first AND red second out of the entire original sample space, whereas $P(R_1 \mid G_2)$ represents the conditional probability that red was selected given that green occurred, evaluating outcomes over a restricted sample space.