Exhaustive Study Notes: Discrete Probability Distributions

Chapter Learning Objectives

• LO1: Define and identify discrete and continuous random variables.

• LO2: Develop, graph, and interpret valid probability distributions for discrete random variables.

• LO3: Calculate and interpret the expected value, variance, and standard deviation for discrete random variables.

• LO4: Develop and interpret the bivariate probability distribution for two discrete random variables.

• LO5: Calculate and interpret the covariance and correlation coefficient for two discrete random variables.

• LO6: Calculate and interpret the expected value and variance of a linear combination of two random variables.

• LO7: Identify and use the binomial probability function to compute the probability of a specified number of successes out of a given number of trials.

• LO8: Calculate and interpret the expected value, variance, and standard deviation for binomial experiments.

• LO9: Identify and use the Poisson probability function to compute the probability of a specified number of occurrences over a given interval.

• LO10: Calculate and interpret the expected value, variance, and standard deviation for Poisson experiments.

• LO11: Identify and use the hypergeometric probability function to compute the probability of a specified number of successes out of a given number of trials.

• LO12: Calculate and interpret the expected value, variance, and standard deviation for hypergeometric experiments.

Introduction to Discrete Random Variables and Distributions

• Probability distributions serve as statistical models for populations of data.

• A random variable provides a numerical description of experimental outcomes by associating a specific numerical value with each possible outcome.

• Discrete probability distributions fall into two primary structural types:

  • Tabular Discrete Distributions: Presented as a two-column table where the first column lists possible values of the random variable and the second column lists their corresponding probabilities (assigned using classical, subjective, or relative frequency methods).

  • Mathematical Functions: Uses a specific mathematical formula to compute probabilities for each value of the random variable (e.g., binomial, Poisson, hypergeometric distributions).

Classification of Random Variables

• Random variables are classified based on the numerical values they assume.

• Discrete Random Variables:

  • May assume either a finite number of values or an infinite sequence of integer values ($0, 1, 2, \dots$).

  • Example 1 (Finite Values): An accountant taking the Certified Public Accountant (CPA) examination. The exam consists of 4 parts. Defining $X = \text{number of parts passed}$, $X$ assumes finite values $x \in \{0, 1, 2, 3, 4\}$.

  • Example 2 (Infinite Sequence): Patient arrivals at a hospital emergency room during a 1-day period. Defining $X = \text{number of patient arrivals}$, $X$ assumes values from the infinite sequence $x \in \{0, 1, 2, \dots\}$.

  • Example 3 (Qualitative Outcomes Numerically Coded): A commercial recall survey asking whether an individual remembers a TV commercial message. Outcomes are qualitative (Cannot recall / Can recall). Arbitrarily defining $X = 0$ for cannot recall and $X = 1$ for can recall converts qualitative outcomes into a discrete random variable.

• Continuous Random Variables:

  • May assume any numerical value in an interval or collection of intervals.

  • Typically associated with experimental outcomes based on measurement scales such as time, weight, distance, and temperature.

  • Example 1 (Time Scale): Monitoring incoming calls to an insurance company claims office. Defining $X = \text{time between consecutive incoming calls in minutes}$, $X$ can assume any value in the interval $X \ge 0$ (e.g., $1.26\,\text{min}$, $2.751\,\text{min}$, $4.3333\,\text{min}$).

  • Example 2 (Distance Scale): Emergency ambulance service monitoring a 90-mile section of Interstate Highway I-75 north of Atlanta, Georgia. Defining $X = \text{number of miles to the location of the next traffic accident}$, $X$ assumes any value in the interval $0 \le X \le 90$.

Developing Discrete Probability Distributions

• The probability distribution for a random variable describes how probabilities are distributed across its values.

• For a discrete random variable $X$, a probability function $f(x)$ provides the probability for each value $x$.

• Probability Assignment Methods:

  • Classical Method: Applicable when experimental outcomes generate equally likely values of the random variable. For rolling a single die, $X = \text{number on upward face}$, $X \in \{1, 2, 3, 4, 5, 6\}$, $f(x) = \frac{1}{6}$.

  • Subjective Method: Relies on individual judgment; different analysts develop different probability distributions.

  • Relative Frequency Method: Used when large historical datasets exist (e.g., scanner data, credit card records, social media data). Produces an empirical discrete distribution.

• Case Study: DeCarlo Motors (Saratoga, New York) Empirical Distribution:

  • Historical data over $300$ operational days for daily automobile sales ($X$):

  • $0$ cars sold: $54$ days $\rightarrow f(0) = \frac{54}{300} = 0.18$

  • $1$ car sold: $117$ days $\rightarrow f(1) = \frac{117}{300} = 0.39$

  • $2$ cars sold: $72$ days $\rightarrow f(2) = \frac{72}{300} = 0.24$

  • $3$ cars sold: $42$ days $\rightarrow f(3) = \frac{42}{300} = 0.14$

  • $4$ cars sold: $12$ days $\rightarrow f(4) = \frac{12}{300} = 0.04$

  • $5$ cars sold: $3$ days $\rightarrow f(5) = \frac{3}{300} = 0.01$

  • Decision-making applications:

  • Most probable daily sales level: $1$ car ($f(1) = 0.39$).

  • Probability of selling $3$ or more cars in a day:       $f(3) + f(4) + f(5) = 0.14 + 0.04 + 0.01 = 0.19$

• Required Conditions for any Valid Discrete Probability Function:

  1. Non-negativity condition:      $f(x) \ge 0 \quad \text{for all values of } x$

  2. Summation condition:      $\sum f(x) = 1$

• Discrete Uniform Probability Distribution Formula:   $f(x) = \frac{1}{n}$   Where $n$ is the total number of possible values for the random variable.

Expected Value, Variance, and Standard Deviation

• Expected Value ($E(X)$ or $\mu$):

  • A measure of central location or expected long-run average.

  • Formula:     $E(X) = \mu = \sum x f(x)$

  • DeCarlo Motors Calculation:     $\mu = (0)(0.18) + (1)(0.39) + (2)(0.24) + (3)(0.14) + (4)(0.04) + (5)(0.01) = 1.50\,\text{automobiles/day}$

  • Forecasting Monthly Sales ($30$ operating days):     $\text{Expected Monthly Sales} = 30 \times 1.50 = 45\,\text{automobiles}$

• Variance (\text{Var}(X) or \sigma^2):

  • A measure of variability or dispersion that weights squared deviations from the mean by their probability.

  • Formula:     $\text{Var}(X) = \sigma^2 = \sum (x - \mu)^2 f(x)$

  • DeCarlo Motors Calculation:     $(0 - 1.50)^2 (0.18) = 2.25 \times 0.18 = 0.4050$     $(1 - 1.50)^2 (0.39) = 0.25 \times 0.39 = 0.0975$     $(2 - 1.50)^2 (0.24) = 0.25 \times 0.24 = 0.0600$     $(3 - 1.50)^2 (0.14) = 2.25 \times 0.14 = 0.3150$     $(4 - 1.50)^2 (0.04) = 6.25 \times 0.04 = 0.2500$     $(5 - 1.50)^2 (0.01) = 12.25 \times 0.01 = 0.1225$     $\sigma^2 = 0.4050 + 0.0975 + 0.0600 + 0.3150 + 0.2500 + 0.1225 = 1.2500$

• Standard Deviation (\sigma):

  • Measured in the same original units as the random variable, making interpretation easier than variance.

  • Formula:     $\sigma = \sqrt{\sigma^2}$

  • DeCarlo Motors Calculation:     $\sigma = \sqrt{1.25} = 1.118\,\text{automobiles}$

• Excel Implementation using SUMPRODUCT:

  • Inputs: Random variable values in range A2:A7, probabilities in B2:B7.

  • Expected Value (cell B9): =SUMPRODUCT(A2:A7, B2:B7) $\rightarrow 1.500$

  • Squared Deviations (cells C2:C7): =(A2-$B$9)^2

  • Variance (cell B11): =SUMPRODUCT(C2:C7, B2:B7) $\rightarrow 1.250$

  • Standard Deviation (cell B13): =SQRT(B11) $\rightarrow 1.118$

Bivariate Discrete Distributions, Covariance, Correlation, and Financial Portfolios

• A bivariate experiment yields outcomes consisting of pairs of values for two random variables $X$ and $Y$.

• Bivariate Empirical Discrete Probability Distribution Case Study (DeCarlo Motors Dealerships):

  • Dealership locations: Geneva ($X$) and Saratoga ($Y$) over $300$ days.

  • Total experimental outcomes: $4 \text{ (Geneva values)} \times 6 \text{ (Saratoga values)} = 24$ outcomes.

  • Joint probabilities $f(x, y)$: Given by dividing cell frequencies by $300$. E.g., $33$ days with Geneva selling $1$ and Saratoga selling $2$ $\rightarrow f(1, 2) = \frac{33}{300} = 0.1100$; $f(0, 1) = 0.1000$; $f(1, 4) = 0.0067$.

  • Marginal Distributions: Right margin gives marginal distribution for Geneva ($X$); bottom margin gives marginal distribution for Saratoga ($Y$).

  • Total Sales $S = X + Y$:

  • $f(S = 0) = 0.0700$

  • $f(S = 1) = 0.0700 + 0.1000 = 0.1700$

  • $f(S = 2) = 0.0300 + 0.1200 + 0.0800 = 0.2300$

  • Expected total sales $E(S) = 2.6433$

  • Variance of total sales $\text{Var}(S) = 2.3895$

  • Geneva ($X$): $E(X) = 1.1433$, $\text{Var}(X) = 0.8696$

  • Saratoga ($Y$): $E(Y) = 1.5000$, $\text{Var}(Y) = 1.2500$

• Covariance (\sigma_{xy}):

  • Formula:     $\sigma_{xy} = \frac{\text{Var}(X + Y) - \text{Var}(X) - \text{Var}(Y)}{2}$

  • DeCarlo Motors Calculation:     $\sigma_{xy} = \frac{2.3895 - 0.8696 - 1.2500}{2} = 0.1350$

  • Positive covariance indicates daily sales at the two dealerships move in the same direction.

• Correlation Coefficient (\rho_{xy}):

  • Formula:     $\rho_{xy} = \frac{\sigma_{xy}}{\sigma_x \sigma_y}$

  • DeCarlo Motors Calculation:     $\sigma_x = \sqrt{0.8696} = 0.9325$     $\sigma_y = \sqrt{1.2500} = 1.1180$     $\rho_{xy} = \frac{0.1350}{(0.9325)(1.1180)} = 0.1295$

  • Value near $0$ indicates a weak positive linear association. Independent random variables have a correlation coefficient of $0$.

• Financial Portfolio Applications:

  • Probability distribution across 4 economic scenarios for Large Cap Stock Fund ($X$) and Long-Term Government Bond Fund ($Y$):

  • Scenario 1 ($p = 0.10$): Stock return = $-40\%$, Bond return = $30\%$

  • Scenario 2 ($p = 0.25$): Stock return = $5\%$, Bond return = $5\%$

  • Scenario 3 ($p = 0.50$): Stock return = $15\%$, Bond return = $4\%$

  • Scenario 4 ($p = 0.15$): Stock return = $30\%$, Bond return = $2\%$

  • Individual Fund Expected Values & Variances:

  • $E(X) = 0.10(-40) + 0.25(5) + 0.50(15) + 0.15(30) = 9.25\%$

  • $E(Y) = 0.10(30) + 0.25(5) + 0.50(4) + 0.15(2) = 6.55\%$

  • $\text{Var}(X) = 0.10(-40-9.25)^2 + 0.25(5-9.25)^2 + 0.50(15-9.25)^2 + 0.15(30-9.25)^2 = 328.1875$

  • $\sigma_x = \sqrt{328.1875} = 18.1159\%$

  • $\text{Var}(Y) = 0.10(30-6.55)^2 + 0.25(5-6.55)^2 + 0.50(4-6.55)^2 + 0.15(2-6.55)^2 = 61.9475$

  • $\sigma_y = \sqrt{61.9475} = 7.8707\%$

• Linear Combinations of Two Random Variables ($R = aX + bY$):

  • Expected Value Formula:     $E(aX + bY) = a E(X) + b E(Y)$

  • Variance Formula:     $\text{Var}(aX + bY) = a^2 \text{Var}(X) + b^2 \text{Var}(Y) + 2ab \sigma_{xy}$

• Evaluation of Portfolio Alternatives:

  • Given $\text{Var}(X + Y) = 119.46$:     $\sigma_{xy} = \frac{119.46 - 328.1875 - 61.9475}{2} = -135.3375$

  • Alternative 1 (100% Stock Fund): Expected Return = $9.25\%$, Standard Deviation = $18.1159\%$

  • Alternative 2 (100% Bond Fund): Expected Return = $6.55\%$, Standard Deviation = $7.8707\%$

  • Alternative 3 (50/50 Portfolio, $a = 0.5, b = 0.5$):

  • $E(R) = 0.5(9.25) + 0.5(6.55) = 7.90\%$

  • $\text{Var}(R) = (0.5)^2(328.1875) + (0.5)^2(61.9475) + 2(0.5)(0.5)(-135.3375) = 29.865$

  • $\sigma_R = \sqrt{29.865} = 5.4650\%$

  • Key Insight: The 50/50 portfolio dominates the 100% bond fund because it achieves both a higher return ($7.90\% > 6.55\%$) and lower risk ($5.4650\% < 7.8707\%$) due to negative covariance between stock and bond returns.

  • Alternative 4 (25/75 Portfolio, $a = 0.25, b = 0.75$):

  • $E(R) = 0.25(9.25) + 0.75(6.55) = 7.225\%$

  • $\text{Var}(R) = (0.25)^2(328.1875) + (0.75)^2(61.9475) + 2(0.25)(0.75)(-135.3375) = 4.6056$

  • $\sigma_R = \sqrt{4.6056} = 2.1461\%$

Binomial Probability Distribution

• Four Properties of a Binomial Experiment:

  1. The experiment consists of a sequence of $n$ identical trials.

  2. Two outcomes are possible on each trial: Success ($S$) or Failure ($F$).

  3. The probability of success, denoted by $p$, remains constant from trial to trial (Stationarity assumption). The probability of failure is $1 - p$.

  4. The trials are independent (the outcome of any trial does not affect others).

  • If properties 2, 3, and 4 hold, the trials are generated by a Bernoulli process. Adding property 1 forms a binomial experiment.

• Difference between Stationarity and Independence:

  • Stationarity requires $p$ to remain constant. If a salesperson becomes tired over 10 calls and $p$ drops from $0.10$ to $0.05$ on the 10th call, stationarity fails even if each family's decision is independent.

• Binomial Formula Derivation:

  • Number of experimental outcomes with $x$ successes in $n$ trials:     $\binom{n}{x} = \frac{n!}{x!(n - x)!}$

  • Probability of any single sequence yielding $x$ successes in $n$ trials:     $p^x (1 - p)^{n - x}$

  • Binomial Probability Function $f(x)$:     $f(x) = \binom{n}{x} p^x (1 - p)^{n - x} \quad \text{for } x \in \{0, 1, 2, \dots, n\}$

• Case Study: Martin Clothing Store ($n = 3$ customer purchase decisions, $p = 0.30$, $1 - p = 0.70$):

  • Probability of $x = 2$ purchases out of $3$ customers:     $\binom{3}{2} = \frac{3!}{2!1!} = 3 \text{ outcomes } (SSF, SFS, FSS)$     $\text{Probability of each outcome} = (0.30)^2(0.70)^1 = 0.063$     $f(2) = 3 \times 0.063 = 0.189$

  • Complete Probability Distribution ($n = 3$):

  • $f(0) = \binom{3}{0}(0.30)^0(0.70)^3 = 1 \times 1 \times 0.343 = 0.343$

  • $f(1) = \binom{3}{1}(0.30)^1(0.70)^2 = 3 \times 0.30 \times 0.49 = 0.441$

  • $f(2) = 0.189$

  • $f(3) = \binom{3}{3}(0.30)^3(0.70)^0 = 1 \times 0.027 \times 1 = 0.027$

  • Example with $n = 10$, $x = 4$, $p = 0.30$:     $f(4) = \frac{10!}{4!6!}(0.30)^4(0.70)^6 = 210 \times 0.0081 \times 0.117649 = 0.2001$

• Expected Value and Variance Formulas for Binomial Distribution:

  • Expected Value:     $E(X) = \mu = n p$

  • Variance:     $\text{Var}(X) = \sigma^2 = n p (1 - p)$

  • Standard Deviation:     $\sigma = \sqrt{n p (1 - p)}$

  • Calculations for Martin Clothing Store:

  • For $n = 3$:       $\mu = 3(0.30) = 0.9$       $\sigma^2 = 3(0.30)(0.70) = 0.63$       $\sigma = \sqrt{0.63} = 0.79$

  • For $n = 1000$ customers:       $\mu = 1000(0.30) = 300$       $\sigma^2 = 1000(0.30)(0.70) = 210$       $\sigma = \sqrt{210} = 14.49$

• Excel BINOM.DIST Function:

  • Syntax: =BINOM.DIST(x, n, p, cumulative)

  • Point probability ($x = 0, n = 3, p = 0.3$): =BINOM.DIST(B5, $D$1, $D$2, FALSE) $\rightarrow 0.343$

  • Cumulative probability ($x \le 4, n = 10, p = 0.3$): =BINOM.DIST(B9, $D$1, $D$2, TRUE) $\rightarrow 0.8497$ ($0.6496 + 0.2001$)

Poisson Probability Distribution

• Used to estimate the number of occurrences over a specified interval of time or space (e.g., arrivals to emergency room, highway repairs per 10 miles, pipeline leaks per 100 miles).

• Two Required Properties for a Poisson Experiment:

  1. The probability of an occurrence is the same for any two intervals of equal length.

  2. The occurrence or non-occurrence in any interval is independent of occurrence or non-occurrence in any other interval.

• Poisson Probability Function Formula:   $f(x) = \frac{\mu^x e^{-\mu}}{x!} \quad \text{for } x \in \{0, 1, 2, \dots\}$   Where $\mu$ is the mean or expected number of occurrences in the interval, and e \approx 2.71828$.\n\n- Poisson Distribution Properties:\n - The mean and variance are equal:\n    E(X) = \mu\n    \text{Var}(X) = \sigma^2 = \mu\n    \sigma = \sqrt{\mu}\n\n- Case Study 1: Hospital Emergency Room Arrivals (Time Interval):\n - \mu = 10 arrivals per 15-minute period.\n - Probability of exactly 5 arrivals in 15 minutes:\n    f(5) = \frac{10^5 e^{-10}}{5!} = 0.0378\n - Variance \sigma^2 = 10$, Standard deviation$\sigma = \sqrt{10} = 3.16$.

  • Converting to a 3-minute interval:     $\mu_{1\text{ min}} = \frac{10}{15} = \frac{2}{3}$     $\mu_{3\text{ min}} = \left(\frac{2}{3}\right) \times 3 = 2 \text{ arrivals}$     $f(1) = \frac{2^1 e^{-2}}{1!} = 0.2707$

  • Note: Probability of $1$ arrival in 3 minutes ($0.2707$) is NOT equal to probability of $5$ arrivals in 15 minutes ($0.0378$), despite equal average arrival rates per unit time.

• Case Study 2: Major Highway Defects (Length/Distance Interval):

  • Average defect rate = $2$ defects per mile.

  • For a 3-mile highway section: $\mu = 2 \times 3 = 6$ defects.

  • Probability of no major defects ($x = 0$):     $f(0) = \frac{6^0 e^{-6}}{0!} = e^{-6} = 0.0025$

  • Probability of at least one major defect:     $1 - f(0) = 1 - 0.0025 = 0.9975$

• Excel POISSON.DIST Function:

  • Syntax: =POISSON.DIST(x, mu, cumulative)

  • Point probability for $\mu = 10, x = 5$: =POISSON.DIST(A9, $D$1, FALSE) $\rightarrow 0.0378$

  • Cumulative probability for $\mu = 10, x \le 5$: =POISSON.DIST(A9, $D$1, TRUE) $\rightarrow 0.0671$

  • Difference: $f(5) = P(X \le 5) - P(X \le 4) = 0.0671 - 0.0293 = 0.0378$

  • Interval Probability ($6 \le X \le 15$):     $P(X \le 15) - P(X \le 5) = 0.9513 - 0.0671 = 0.8842$

  • Probability of more than 20 arrivals:     $1 - P(X \le 20) = 1 - 0.9984 = 0.0016$

Hypergeometric Probability Distribution

• Key Differences from Binomial Distribution:

  1. Trials are NOT independent.

  2. Probability of success changes from trial to trial because sampling occurs without replacement from a finite population.

• Mathematical Parameters:

  • $N$ = Total population size

  • $r$ = Total number of successes in population

  • $N - r$ = Total number of failures in population

  • $n$ = Sample size

  • $x$ = Number of successes in sample ($n - x$ failures in sample)

• Hypergeometric Probability Function Formula:   $f(x) = \frac{\binom{r}{x} \binom{N - r}{n - x}}{\binom{N}{n}}$   Valid for $x \le r$ and n - x \le N - r$.\n\n- Case Study: Ontario Electric Fuses Quality Control:\n - Box of N = 12$electric fuses contains$r = 5$defective fuses ($N - r = 7$non-defective). Inspector selects$n = 3 fuses.\n - Probability of finding exactly 1$defective fuse ($x = 1):\n    f(1) = \frac{\binom{5}{1} \binom{7}{2}}{\binom{12}{3}} = \frac{(5)(21)}{220} = \frac{105}{220} = 0.4773\n - Probability of finding zero defective fuses (x = 0):\n    f(0) = \frac{\binom{5}{0} \binom{7}{3}}{\binom{12}{3}} = \frac{(1)(35)}{220} = 0.1591\n - Probability of finding at least one defective fuse:\n    1 - f(0) = 1 - 0.1591 = 0.8409\n\n- Hypergeometric Mean and Variance Formulas:\n - Mean:\n    E(X) = \mu = n \left(\frac{r}{N}\right)\n - Variance:\n    \text{Var}(X) = \sigma^2 = n \left(\frac{r}{N}\right) \left(1 - \frac{r}{N}\right) \left(\frac{N - n}{N - 1}\right)\n - Calculation for Fuse Example (n = 3, r = 5, N = 12):\n    \mu = 3 \left(\frac{5}{12}\right) = 1.25\n    \sigma^2 = 3 \left(\frac{5}{12}\right) \left(1 - \frac{5}{12}\right) \left(\frac{12 - 3}{12 - 1}\right) = 0.77\n\n- Excel `HYPGEOM.DIST` Function:\n - Syntax: `=HYPGEOM.DIST(x, n, r, N, cumulative)`\n - Point probability for x = 0, n = 3, r = 5, N = 12: `=HYPGEOM.DIST(A6, $D$1, $D$2, $D$3, FALSE)` \rightarrow 0.1581\n\n# Case Studies and Managerial Applications\n\n- Case Study 1: Go Bananas Cereal (Great Grasslands Grains, Inc. - GGG):\n - Scenario: 16 oz boxes containing rice flakes and banana-flavored marshmallows. Standard weight range for marshmallows is 1.6\,\text{oz}$to$2.4\,\text{oz}.\n - Quality target: Process failure rate p = 0.08$($8\% non-compliant boxes) when process operates properly.\n - Sampling plan: Weekly random sample of n = 25 boxes.\n - Shutdown criterion: Process shut down if at least 5$sampled boxes fail standard ($X \ge 5).\n - Managerial Report Tasks:\n 1. Calculate shutdown probability under normal operating conditions (n = 25, p = 0.08, P(X \ge 5)) and evaluate policy.\n 2. Determine maximum allowable defect threshold in sample of 25$boxes to achieve a false shutdown rate of$\le 1\%.\n 3. Determine target process failure rate p$required to ensure$P(X \ge 5) \le 0.01$.

• Case Study 2: McNeil's Auto Mall:

  • Scenario: Dealership staffing on Saturday mornings (8:00 AM to 12:00 PM). Customer arrival rate averages $\mu = 6.8$ customers per hour (Poisson process).

  • Service duration: Salesperson spends average of 1 hour per customer.

  • Management goal: Maintain optimal balance of high perceived demand by having 2 more customers on lot than salespeople.

  • Risk threshold: Accept customer overload (more than 2 unserved customers above staffing level) no more than $10\%$ of the time ($P(X > \text{salespeople} + 2) \le 0.10$).

  • Managerial Report Tasks:

  1. Identify arrival probability distribution (Poisson distribution with $\mu = 6.8$/hour).

  2. Evaluate current staffing of $5$ salespeople ($P(X > 7) = 1 - P(X \le 7)$) against the $10\%$ risk limit.

  3. Calculate minimum number of salespeople required to satisfy target criteria.

• Case Study 3: Tuggler Corporation Grievance Committee:

  • Scenario: Grievance committee composed of $10$ annual volunteers randomly drawn from total respondents. Accounting department observed zero representation over past 5 years.

  • Managerial Report Tasks:

  1. Identify selection probability distribution (Hypergeometric distribution with population $N$, accounting volunteers $r$, sample $n = 10$).

  2. Compute probability of zero accounting members selected for each individual year ($f(0)$).

  3. Compute combined 5-year probability of zero accounting representation (product of individual annual probabilities).

  4. Analyze root causes and recommend structural changes to increase accounting representation.

• Case Study 4: Sagittarius Casino - POSO Game:

  • Game Mechanics: Player chooses 2 distinct integers from $\{1, 2, 3, 4, 5, 6\}$. Two dice rolled.

  • Win $\$5$ if 1st chosen number appears on at least one die.

  • Win $\$10$ if 2nd chosen number appears on both dice.

  • Win $\$0$ otherwise.

  • Managerial Report Tasks:

  1. Calculate single-game winning probability $P(\text{Win} > 0)$.

  2. Calculate expected payout per play $E(X)$.

  3. Compute casino expected net profit per play given a $\$2$ play fee, and state minimum play fee required to guarantee positive casino profit.

Appendix: Discrete Probability Distributions with R

• R syntax and functions for discrete probability distributions:

• Binomial Distribution in R:

  • Case Study: Martens specialty clothing targeted email campaign ($n = 10$ emails sent, $p = 0.30$ click probability).

  • Probability function $f(x)$: dbinom(x, size, prob)

  • Example command: dbinom(0:10, size = 10, prob = 0.3)

  • Cumulative probability $P(X \le x)$: pbinom(q, size, prob)

• Poisson Distribution in R:

  • Probability function $f(x)$: dpois(x, lambda)

  • Example command for $\mu = 10, x = 5$: dpois(5, lambda = 10) $\rightarrow 0.0378$

  • Cumulative probability $P(X \le x)$: ppois(q, lambda)

• Hypergeometric Distribution in R:

  • Case Study: Ontario Electric fuses ($N = 12$, defective $m = 5$, non-defective $n = 7$, sampled $k = 3$).

  • Function syntax: dhyper(x, m, n, k)

  • Where $x$ = sample successes, $m$ = population successes, $n$ = population failures, $k$ = sample size.

  • Example command for $x = 1$ defective fuse: dhyper(1, m = 5, n = 7, k = 3) $\rightarrow 0.4773$

  • Cumulative probability $P(X \le x)$: phyper(q, m, n, k)