Statistical Methods in Business: Basic Probability and Discrete Probability Distributions

Foundations of Probability and Event Structures

Probability is defined as the numerical measure of the likelihood that a particular uncertain event will occur. The probability of any event $A$, denoted as $P(A)$, is strictly bounded between $0$ and $1$, inclusive:

$0 \le P(A) \le 1$

A sample space represents the collection of all possible outcomes of a variable.

Examples of sample spaces include:

• Rolling a single six-sided die: The sample space consists of all $6$ faces of the die: $\{1, 2, 3, 4, 5, 6\}$.   

  

• A standard bridge deck of playing cards: The sample space consists of all $52$ cards.

Events within a sample space are classified into distinct categories based on their operational definitions:

• Elementary Event: An event described by a single characteristic. Examples include drawing a red card from a deck, or a family spending $200 at a restaurant last week.

• Complementary Event: Comprises all outcomes that are not part of the specified initial event $A$, denoted as $A'$ or $A^c$. Examples include all cards in a deck that are not red, or all families who did not spend more than $200 at a restaurant last week.

• Joint Event: An event described by two or more characteristics occurring simultaneously. Examples include drawing a card that is both red and an Ace, or a family of three spending $200 at a restaurant.

Events are also categorized by their certainty of occurrence on the probability scale:

• Impossible Event (Null Event): An event that has zero chance of occurring ($P(\text{Event}) = 0$). Examples include rolling an $8$ on a standard six-sided die, or humans living $500$ years.

• Certain Event: An event that is guaranteed to occur ($P(\text{Event}) = 1$). An example is rolling a number between $1$ and $6$ inclusive on a standard six-sided die.

The qualitative spectrum of probability ranges as follows:

• $0$: Impossible

• Between $0$ and $0.5$: Unlikely

• $0.5$ ($\frac{1}{2}$): Even Chance

• Between $0.5$ and $1$: Likely

• $1$: Certain

Mutually Exclusive and Collectively Exhaustive Event Relationships

Relationships between multiple events dictate how their combined probabilities are evaluated:

• Mutually Exclusive Events: Two or more events are mutually exclusive if they cannot occur simultaneously. The occurrence of one event automatically precludes the occurrence of the other.

  • Example: Drawing a single card from a deck. Let Event $A$ be drawing an Ace and Event $B$ be drawing a King. Events $A$ and $B$ are mutually exclusive because a single card cannot be both an Ace and a King at the same time.   

    

• Collectively Exhaustive Events: A set of events is collectively exhaustive if it includes all possible outcomes in the sample space. At least one of the events must occur when an experiment is performed.

  • Example 1: Selecting a single card from a deck where suit events are Spades ($S$), Hearts ($H$), Clubs ($C$), and Diamonds ($D$). The set $\{S, H, C, D\}$ is collectively exhaustive (one suit must be drawn) and also mutually exclusive.

  • Example 2: Rolling a six-sided die where the outcomes $\{1, 2, 3, 4, 5, 6\}$ encompass the entire sample space, making them collectively exhaustive.

  • Example 3: Choosing a random day from the year 2022 where Event $A = \text{Weekday}$, Event $B = \text{Weekend}$, Event $C = \text{January}$, and Event $D = \text{Spring}$. Events $A, B, C, D$ are collectively exhaustive because any selected day falls into these categories; however, they are not mutually exclusive because a weekday can occur in January or Spring. In contrast, Events $A$ (Weekday) and $B$ (Weekend) are both mutually exclusive and collectively exhaustive.

Core Probability Axioms

1. For any event $A$:    $0 \le P(A) \le 1$

2. The sum of the probabilities of all mutually exclusive and collectively exhaustive events in a sample space is strictly equal to $1$:    $\sum_{i=1}^k P(E_i) = 1$

3. If events $A$, $B$, and $C$ are mutually exclusive and collectively exhaustive:    $P(A) + P(B) + P(C) = 1$

Approaches to Assessing Probability

Two distinct primary approaches exist for assigning probabilities to uncertain outcomes:

1. Classical Probability Approach

The classical approach determines probability prior to observing data, based on inherent knowledge of the underlying physical or conceptual process, assuming all outcomes are equally likely:

$P(A) = \frac{\text{Number of favorable outcomes to } A}{\text{Total number of possible outcomes}}$

• Card Example: Finding the probability of selecting a face card (Jack, Queen, or King) from a standard deck of 52 cards:   $P(\text{Jack, Queen, or King}) = \frac{12}{52} = \frac{3}{13} \approx 0.2308$

• Die Roll Example: Finding the probability of rolling a $2$ or a $4$ on a fair six-sided die:   $\text{Sample Space} = \{1, 2, 3, 4, 5, 6\}$   $\text{Favorable outcomes to } 2 = 1 \quad (\text{the face } 2)$   $\text{Favorable outcomes to } 4 = 1 \quad (\text{the face } 4)$   $P(2 \text{ or } 4) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3} \approx 0.3333$

2. Empirical Probability Approach

The empirical approach determines probability based on observed sample data or historical frequencies from an experiment or survey:

$P(A) = \frac{\text{Number of ways in which the event occurs}}{\text{Total number of possible outcomes}}$

• Voter Survey Example: A survey of $878$ registered voters reveals that $168$ support the governor's plan to raise taxes on gasoline. The empirical probability that a randomly chosen voter supports the plan is:   $P(\text{Support Plan}) = \frac{\text{Number supporting plan}}{\text{Total number in survey}} = \frac{168}{878} \approx 0.191$

• University Student Population Example: Evaluating a population of $439$ students categorized by gender and statistics course enrollment:

The probability of selecting a male student who is taking statistics is: $P(\text{Male Taking Stats}) = \frac{\text{Number of males taking stats}}{\text{Total number of people}} = \frac{84}{439} \approx 0.191$

Contingency Tables, Venn Diagrams, and Probability Calculations

Sample spaces with multi-dimensional categorical data are effectively summarized using contingency tables and Venn diagrams.

Consider a consumer market research survey ($N = 1000$) evaluating whether households planned to purchase a television and whether they actually purchased a television:

Contingency Table: M&R Survey Results

Venn Diagram Representation

Let:

• $A = \text{Planned to Purchase}$

• $A' = \text{Did Not Plan to Purchase}$

• $B = \text{Actually Purchased}$

• $B' = \text{Did Not Actually Purchase}$

From the Venn Diagram:

• $A \cap B = 200$ (Planned and Actually Purchased)

• $A \cap B' = 50$ (Planned but Did Not Purchase)

• $A' \cap B = 100$ (Did Not Plan but Actually Purchased)

• $A' \cap B' = 650$ (Did Not Plan and Did Not Purchase)

• Total union $A \cup B = 50 + 200 + 100 = 350$

Simple (Marginal) Probability

Simple probability refers to the probability of a single simple event occurring, found in the margins of a contingency table:

$P(\text{Actually Purchased}) = \frac{300}{1000} = 0.300$

$P(\text{Planned to Purchase}) = \frac{250}{1000} = 0.250$

Joint Probability

Joint probability refers to the probability of the simultaneous occurrence of two or more events (a joint event):

$P(\text{Planned to Purchase and Actually Purchased}) = \frac{200}{1000} = 0.200$

$P(\text{No Plan to Purchase and Actually Purchased}) = \frac{100}{1000} = 0.100$

General Contingency Table Structure for Joint vs. Marginal Probabilities

In a general $2 \times 2$ table with row events $A_1, A_2$ and column events $B_1, B_2$:

• The interior cells contain the Joint Probabilities: $P(A_1 \text{ and } B_1)$, $P(A_1 \text{ and } B_2)$, $P(A_2 \text{ and } B_1)$, $P(A_2 \text{ and } B_2)$.

• The margin totals contain the Marginal (Simple) Probabilities: $P(A_1)$, $P(A_2)$, $P(B_1)$, $P(B_2)$.

• The sum of all cell probabilities equals $1$.

Rules of Probability and Independence

General Addition Rule

The General Addition Rule computes the probability that Event $A$ or Event $B$ (or both) will occur:

$P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$

Applying this rule to the Television Purchase Survey data to find the probability that a household planned to purchase or actually purchased a TV:

$P(\text{Planned or Purchased}) = P(\text{Planned}) + P(\text{Purchased}) - P(\text{Planned and Purchased})$

$P(\text{Planned or Purchased}) = \frac{250}{1000} + \frac{300}{1000} - \frac{200}{1000} = \frac{350}{1000} = 0.350$

Simplification for Mutually Exclusive Events

If events $A$ and $B$ are mutually exclusive, then $P(A \text{ and } B) = 0$. The addition rule simplifies to:

$P(A \text{ or } B) = P(A) + P(B)$

Conditional Probability

A conditional probability is the probability of an event occurring given that another event has already occurred.

The conditional probability of Event $B$ given that Event $A$ has occurred is defined as:

$P(B | A) = \frac{P(A \text{ and } B)}{P(A)}$

Similarly, the conditional probability of Event $A$ given that Event $B$ has occurred is:

$P(A | B) = \frac{P(A \text{ and } B)}{P(B)}$

Applying this to the TV Purchase Survey data to find the probability that a household actually purchased a TV given that they planned to purchase one:

$P(\text{Purchased} | \text{Planned}) = \frac{P(\text{Purchased and Planned})}{P(\text{Planned})} = \frac{\frac{200}{1000}}{\frac{250}{1000}} = \frac{200}{250} = 0.800$

When a conditional probability $P(B | A)$ is calculated from a contingency table, the sample space shrinks to include only those observations satisfying condition $A$ (i.e., the row or column corresponding to $A$).

Statistical Independence

Two events $A$ and $B$ are independent if and only if the probability of occurrence of one event is completely unaffected by whether or not the other event has occurred.

Formal definition of independence:

$P(A | B) = P(A) \quad \text{or} \quad P(B | A) = P(B)$

Testing Independence in TV Purchase Example

To check whether "Planned to Purchase" and "Actually Purchased" are independent events, verify if $P(\text{Purchased} | \text{Planned}) = P(\text{Purchased})$:

$P(\text{Purchased} | \text{Planned}) = \frac{200}{250} = 0.800$

$P(\text{Purchased}) = \frac{300}{1000} = 0.300$

Since $0.800 \neq 0.300$, the two events are dependent.

Comprehensive Probability Problem Applications

Application 1: Used Car Lot Inventory

In a used car lot, $70\%$ of cars have Air Conditioning ($AC$) and $40\%$ have a CD player ($CD$). Additionally, $20\%$ of the cars have both features.

• Given probabilities: $P(AC) = 0.70$, $P(CD) = 0.40$, $P(AC \text{ and } CD) = 0.20$

• Probability Table:

To find the probability that a car has a CD player given that it has AC ($P(CD | AC)$):

$P(CD | AC) = \frac{P(AC \text{ and } CD)}{P(AC)} = \frac{0.20}{0.70} = \frac{2}{7} \approx 0.2857$

Application 2: Courier Package Delivery

A comparative study evaluates the delivery performance of two courier services across $439$ total deliveries:

1. Probability a package is delivered by Courier 1:

$P(\text{Courier 1}) = \frac{229}{439} \approx 0.522 = 52.2\%$

2. Probability a package is delivered late:

$P(\text{Late Delivery}) = \frac{160}{439} \approx 0.364 = 36.4\%$

3. Probability a package is delivered late by Courier 1 (Joint Probability):

$P(\text{Courier 1 and Late Delivery}) = \frac{84}{439} \approx 0.191 = 19.1\%$

4. Probability a package is delivered late given it was assigned to Courier 1:

$P(\text{Late Delivery} | \text{Courier 1}) = \frac{84}{229} \approx 0.367 = 36.7\%$

Using the formula: $P(\text{Late Delivery} | \text{Courier 1}) = \frac{P(\text{Courier 1 and Late Delivery})}{P(\text{Courier 1})} = \frac{\frac{84}{439}}{\frac{229}{439}} = \frac{0.191}{0.521} \approx 0.367$

5. Probability a package was handled by Courier 1 given that it was delivered late:

$P(\text{Courier 1} | \text{Late Delivery}) = \frac{84}{160} = 0.525 = 52.5\%$

Using the formula: $P(\text{Courier 1} | \text{Late Delivery}) = \frac{P(\text{Courier 1 and Late Delivery})}{P(\text{Late Delivery})} = \frac{\frac{84}{439}}{\frac{160}{439}} = \frac{0.191}{0.364} \approx 0.525$

Introduction to Discrete Random Variables and Probability Distributions

Variables are classified based on the nature of their data collection processes:

• Discrete Variables: Produce outcomes that arise from a counting process. They can only take on a countable number of distinct values. Examples include the number of university classes a student takes, the number of heads in $5$ coin tosses, or daily network interruptions.

• Continuous Variables: Produce outcomes that result from a measurement process over a continuous numerical scale. Examples include annual salary, height, or weight.

A probability distribution for a discrete variable is a mutually exclusive listing of all possible numerical outcomes for that variable, together with the probability of occurrence associated with each outcome.

Example: Computer Network Interruptions Per Day

A firm monitors network disruptions over a period of $200$ days:

Central Tendency and Dispersion of Discrete Distributions

Expected Value (Mean)

The Expected Value $\mu$ or $E(X)$ of a discrete random variable is the weighted average of its possible outcomes, weighted by the probability of each outcome:

$\mu = E(X) = \sum_{i=1}^N x_i P(X = x_i)$

Calculation for Network Interruptions Data:

The expected mean number of disruptions per day is $\mu = 1.40$

Variance and Standard Deviation

The variance $\sigma^2$ measures the dispersion of a discrete random variable around its mean:

$\sigma^2 = \sum_{i=1}^N [x_i - E(X)]^2 P(X = x_i)$

The standard deviation $\sigma$ is the square root of variance:

$\sigma = \sqrt{\sum_{i=1}^N [x_i - E(X)]^2 P(X = x_i)}$

Full Step-by-Step Dispersion Calculation ($\mu = 1.40$):

Taking the square root gives standard deviation: $\sigma = \sqrt{2.04} \approx 1.4283$

Categorization of Probability Distributions

Probability distributions are grouped based on whether the underlying random variable is discrete or continuous:

• Discrete Probability Distributions: Models discrete count data (e.g., Binomial Distribution, Poisson Distribution).

• Continuous Probability Distributions: Models continuous measurement data (e.g., Normal Distribution).

Binomial Probability Distribution

Assumptions and Properties

A process follows a Binomial Probability Distribution if it satisfies five strict requirements:

1. The experiment consists of a fixed number of observations or trials, denoted by $n$.

2. Each trial outcome is classified into one of two mutually exclusive and collectively exhaustive categories: the event of interest ("Success") or the non-event of interest ("Failure").

3. The probability of the event of interest, denoted by $\pi$, remains constant from observation to observation.

4. The probability of the non-event occurring is $1 - \pi$.

5. The observations or trials are strictly independent; the outcome of any trial is unaffected by the outcome of any other trial.

Business Applications

• Manufacturing Quality Control: Items inspected are classified as defective or non-defective.

• Contract Bidding: A construction firm bidding for jobs either wins a contract or does not.

• Market Research: Survey respondents answer either "Yes, I will buy" or "No, I will not buy".

• Human Resources: Job applicants either accept or reject employment offers.

Counting Rules: Rule of Combinations

To determine the number of ways $x$ success outcomes can be selected from $n$ total observations without regard to order, use the combination formula:

$_nC_x = \frac{n!}{x!(n - x)!}$

where:

• $n! = (n)(n - 1)(n - 2) \cdots (2)(1)$

• $x! = (x)(x - 1)(x - 2) \cdots (2)(1)$

• $0! = 1$ (by definition)

Ice Cream Combinations Example

Calculating the number of possible $3$--scoop combinations from $31$ available flavors, assuming no flavor is repeated ($n = 31, x = 3$):

$_{31}C_3 = \frac{31!}{3!(31 - 3)!} = \frac{31!}{3! 28!} = \frac{31 \times 30 \times 29 \times 28!}{(3 \times 2 \times 1) 28!} = \frac{31 \times 30 \times 29}{6} = 31 \times 5 \times 29 = 4,495$

Binomial Distribution Mathematical Formula

The probability of obtaining exactly $x$ success events in $n$ independent trials given constant success probability $\pi$ is:

$P(X = x | n, \pi) = \frac{n!}{x!(n - x)!} \pi^x (1 - \pi)^{n - x}$

where:

• $x = \text{number of events of interest in sample } (x = 0, 1, 2, \ldots, n)$

• $n = \text{sample size (number of trials)}$

• $\pi = \text{probability of event of interest}$

• $1 - \pi = \text{probability of non-event of interest}$

Worked Calculation Examples

Example 1

Find the probability of obtaining exactly $x = 1$ success in $n = 5$ observations when $\pi = 0.1$:

$P(X = 1 | 5, 0.1) = \frac{5!}{1!(5 - 1)!} (0.1)^1 (1 - 0.1)^{5 - 1} = \frac{5!}{1! 4!} (0.1)^1 (0.9)^4 = (5)(0.1)(0.6561) = 0.32805$

Example 2: Late Invoices

Suppose the probability of an invoice payment being late is $\pi = 0.10$. What is the probability of finding exactly $x = 1$ late invoice payment in a group of $n = 4$ invoices?

$P(X = 1 | 4, 0.10) = \frac{4!}{1!(4 - 1)!} (0.10)^1 (1 - 0.10)^{4 - 1} = \frac{4!}{1! 3!} (0.10)^1 (0.90)^3 = (4)(0.10)(0.729) = 0.2916$

Binomial Distribution Shapes

The shape of a binomial distribution depends on $\pi$ and $n$:

• When $\pi = 0.1$ ($n = 5$), the distribution is skewed to the right (positively skewed).

• When $\pi = 0.5$ ($n = 5$), the distribution is perfectly symmetrical.

Summary Parameters of Binomial Distribution

• Mean:   $\mu = E(X) = n\pi$

• Variance:   $\sigma^2 = n\pi(1 - \pi)$

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

Comparative Parameter Calculation:

1. For $n = 5, \pi = 0.1$:    $\mu = (5)(0.1) = 0.5$    $\sigma = \sqrt{(5)(0.1)(1 - 0.1)} = \sqrt{(5)(0.1)(0.9)} = \sqrt{0.45} \approx 0.6708$

2. For $n = 5, \pi = 0.5$:    $\mu = (5)(0.5) = 2.5$    $\sigma = \sqrt{(5)(0.5)(1 - 0.5)} = \sqrt{(5)(0.5)(0.5)} = \sqrt{1.25} \approx 1.118$

Cumulative Binomial Probabilities

For $n = 4, \pi = 0.10$:

• $P(X = 0) = 0.6561$

• $P(X = 1) = 0.2916$

• $P(X = 2) = 0.0486$

• $P(X = 3) = 0.0036$

• $P(X = 4) = 0.0001$

To compute cumulative conditions:

• Probability of at most 1 late invoice ($P(X \le 1)$):   $P(X \le 1) = P(X = 0) + P(X = 1) = 0.6561 + 0.2916 = 0.9477$

• Probability of more than 0 late invoices ($P(X > 0)$):   $P(X > 0) = 1 - P(X = 0) = 1 - 0.6561 = 0.3439$

Poisson Probability Distribution

Definitions and Assumptions

The Poisson distribution models the number of times an event occurs in a continuous area of opportunity (a specified interval of time, length, surface area, or volume in which multiple occurrences can happen).

Examples of areas of opportunity:

• Number of scratches on a car's paint

• Number of mosquito bites on a person

• Number of computer network crashes per day

• Number of bank customer arrivals per minute

Required Conditions for Poisson Application:

1. Interested in counting occurrences of an event within a defined area of opportunity.

2. The probability of an event occurring per unit area of opportunity is constant across all areas.

3. Occurrences in non-overlapping areas of opportunity are independent.

4. The probability of two or more occurrences in an area of opportunity approaches zero as the area becomes infinitely small.

Poisson Distribution Mathematical Formula

$P(X = x | \lambda) = \frac{e^{-\lambda} \lambda^x}{x!}$

where:

• $x = \text{number of events occurring in an area of opportunity } (x = 0, 1, 2, \ldots)$

• $\lambda = \text{expected (mean) number of events per area of opportunity}$

• $e = \text{base of the natural logarithm system } (\approx 2.71828)$

Poisson Distribution Summary Parameters

• Mean:   $\mu = \lambda$

• Variance:   $\sigma^2 = \lambda$

• Standard Deviation:   $\sigma = \sqrt{\lambda}$

Worked Bank Customer Arrival Example

Suppose the mean number of customers arriving per minute at a bank during the noon-to-1 PM hour is $\lambda = 3.0$. What is the probability that exactly $x = 2$ customers arrive in a given minute?

$P(X = 2 | 3.0) = \frac{e^{-3.0} (3.0)^2}{2!} = \frac{(2.71828)^{-3.0} (9)}{2} = \frac{9}{2.71828^3 \times 2} \approx \frac{9}{20.085537 \times 2} = \frac{9}{40.17107} \approx 0.2240$

Cumulative Poisson Probabilities

Given $\lambda = 3.0$ customer arrivals per minute:

• $P(X = 0) = 0.0498$

• $P(X = 1) = 0.1494$

• $P(X = 2) = 0.2240$

• $P(X = 3) = 0.2240$

• $P(X = 4) = 0.1680$

• $P(X = 5) = 0.1008$

• $P(X = 6) = 0.0504$

1. Probability of at most 3 arrivals ($P(X \le 3)$):

$P(X \le 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$ $P(X \le 3) = 0.0498 + 0.1494 + 0.2240 + 0.2240 = 0.6472$

2. Probability of more than 1 arrival ($P(X > 1)$):

$P(X > 1) = 1 - [P(X = 0) + P(X = 1)]$ $P(X > 1) = 1 - (0.0498 + 0.1494) = 1 - 0.1992 = 0.8008 \quad (\text{or } 0.8009 \text{ with exact template rounding})$

Poisson Distribution Shape Characteristics

The shape of a Poisson distribution is governed entirely by $\lambda$:

• For small values of $\lambda$ (e.g., $\lambda = 0.50$), the distribution is right-skewed.

• As $\lambda$ increases (e.g., $\lambda = 3.00$), the distribution becomes increasingly symmetrical and bell-shaped.