Probability Concepts, Counting Rules, and Event Relationships

Learning Objectives

• Calculate the number of outcomes (sample points) for random experiments using tree diagrams, combinations, and permutations, and list these outcomes.
• Assign probabilities to outcomes and events for a random experiment using the classical method, the relative frequency method, and the subjective method.
• Calculate and interpret the probability of the complement, the union, and the intersection of events.
• Calculate and interpret the conditional probability associated with two events.
• Identify and interpret mutually exclusive events and independent events.
• Create and interpret joint probability tables.
• Identify and interpret prior probabilities and posterior probabilities, and apply Bayes' theorem to calculate posterior probabilities.

Role of Probability in Managerial Decision-Making

• Managers frequently base decisions on an analysis of uncertainties, including:
  • The chances that sales will decrease if prices are increased.
  • The likelihood that a new assembly method will increase productivity.
  • The duration required to complete a project.
  • The chance that a new investment will be profitable.
• Probability is defined as a numerical measure of the likelihood that an event will occur.
• Probability values are assigned on a scale from $0$ to $1$:
  • A probability near $0$ indicates an event is unlikely to occur.
  • A probability near $1$ indicates an event is almost certain to occur.
  • A probability of $0.50$ indicates that an event is just as likely to occur as not.
  • Intermediate values represent varying degrees of likelihood.
• Example (Weather Forecasting):
  • A report indicating a near-zero probability of rain signifies almost no chance of rain.
  • A $0.90$ probability indicates rain is very likely to occur.
  • A $0.50$ probability indicates rain is equally likely to occur or not occur.

Experiments, Sample Spaces, and Sample Points

• An experiment is defined as a process that generates well-defined outcomes.
• On any single repetition of an experiment, one and only one of the possible experimental outcomes will occur.
• A sample point is an individual experimental outcome and represents an element of the sample space.
• The sample space ($S$) for an experiment is the set of all possible experimental outcomes.

Examples of Experiments and Sample Spaces

• Tossing a Coin:
  • Experimental outcome is determined by the upward face of the coin.
  • Sample space:

$S = \{\text{Head}, \text{Tail}\}$

• Selecting a Part for Inspection:
  • Experimental outcome is determined by whether the part is acceptable or defective.
  • Sample space:

$S = \{\text{Defective}, \text{Nondefective}\}$

• Rolling a Die:
  • Experimental outcome is defined by the number of dots appearing on the upward face.
  • Sample space:

$S = \{1, 2, 3, 4, 5, 6\}$

Counting Rules for Multiple-Step Experiments, Combinations, and Permutations

• Identifying and counting experimental outcomes is a necessary prerequisite to assigning probabilities.

Multiple-Step Experiments

• If an experiment can be described as a sequence of $k$ steps in which there are $n_1$ possible outcomes on the first step, $n_2$ possible outcomes on the second step, and so on, then the total number of experimental outcomes is given by:

$N = n_1 \times n_2 \times \dots \times n_k$

• Tossing Two Coins:
  • Step 1 is tossing the first coin ($n_1 = 2$: Head or Tail).
  • Step 2 is tossing the second coin ($n_2 = 2$: Head or Tail).
  • Total experimental outcomes: $2 \times 2 = 4$.
  • Sample space:

$S = \{(\text{H}, \text{H}), (\text{H}, \text{T}), (\text{T}, \text{H}), (\text{T}, \text{T})\}$

• Tossing Six Coins:

  • Total experimental outcomes: $2 \times 2 \times 2 \times 2 \times 2 \times 2 = 64$.

• Tree Diagram:

  • A graphical representation used to visualize a multiple-step experiment.
  • The sequence of steps moves from left to right through branches.
  • Each path through the tree from left to right corresponds to a unique sequence of step outcomes leading to a specific experimental outcome.

Combinations

• The counting rule for combinations allows one to count the number of experimental outcomes when the experiment involves selecting $x$ objects from a set of $n$ objects without regard to order.
• Combination Formula:

$C_x^n = \binom{n}{x} = \frac{n!}{x! (n - x)!}$

• Example 1: Quality Control Part Selection:
  • Selecting $x = 2$ parts to test for defects from a group of $n = 5$ parts (labeled A, B, C, D, E).
  • Calculation:

$C_2^5 = \frac{5!}{2! (5 - 2)!} = \frac{5!}{2! \times 3!} = \frac{5 \times 4 \times 3 \times 2 \times 1}{(2 \times 1)(3 \times 2 \times 1)} = \frac{120}{12} = 10$

• The 10 combinations (experimental outcomes) are: AB, AC, AD, AE, BC, BD, BE, CD, CE, and DE.

  • Example 2: State Lottery System:

• Selecting $x = 6$ integers randomly from a group of $n = 53$ integers.

• Calculation:

$C_6^{53} = \frac{53!}{6! (53 - 6)!} = \frac{53!}{6! \times 47!} = \frac{53 \times 52 \times 51 \times 50 \times 49 \times 48}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 22,957,480$

• An individual purchasing one lottery ticket has a $1$ in $22,957,480$ chance of winning.

Permutations

• The counting rule for permutations computes the number of experimental outcomes when $x$ objects are selected from a set of $n$ objects where the order of selection is important.
• Permutation Formula:

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

• An experiment yields more permutations than combinations for the same values of $n$ and $x$ because every selection of $x$ objects can be ordered in $x!$ different ways.
• Example: Quality Control Inspection with Order:
  • Selecting $x = 2$ parts from $n = 5$ parts where selection order matters.
  • Calculation:

$P_2^5 = \frac{5!}{(5 - 2)!} = \frac{5!}{3!} = \frac{5 \times 4 \times 3 \times 2 \times 1}{3 \times 2 \times 1} = \frac{120}{6} = 20$

Methods for Assigning Probabilities

Basic Requirements for Probability Assignments

• For every experimental outcome $E_i$, its probability $P(E_i)$ must satisfy:

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

• The sum of the probabilities for all $k$ experimental outcomes in a sample space must equal $1.0$:

$\sum_{i=1}^k P(E_i) = 1.0$

Classical Method

• Appropriate when all experimental outcomes are equally likely.
• If an experiment has $n$ possible outcomes, the probability assigned to each experimental outcome is $\frac{1}{n}$.
• Example: Rolling a fair six-sided die yields $P(E_i) = \frac{1}{6}$ for each outcome i \in \{1, 2, 3, 4, 5, 6\}$.\n\n## Relative Frequency Method\n\n- Appropriate when historical data are available to estimate the proportion of time an experimental outcome will occur if the experiment is repeated a large number of times.\n- **Example: Hospital X-Ray Department Waiting Times**:\n - Data collected at 09:00 AM across 20 successive days:\n - 0$patients waiting:$2$days$\implies P(0) = \frac{2}{20} = 0.10\n - 1$patient waiting:$5$days$\implies P(1) = \frac{5}{20} = 0.25\n - 2$patients waiting:$6$days$\implies P(2) = \frac{6}{20} = 0.30\n - 3$patients waiting:$4$days$\implies P(3) = \frac{4}{20} = 0.20\n - 4$patients waiting:$3$days$\implies P(4) = \frac{3}{20} = 0.15\n - Sum of assigned probabilities: 0.10 + 0.25 + 0.30 + 0.20 + 0.15 = 1.00\n\n## Subjective Method\n\n- Appropriate when outcomes are not equally likely and little or no relevant historical data exist.\n- Probability expresses an individual's degree of belief (on a scale from 0$to$1) that a specific outcome will occur.\n- Subjective probabilities are personal and vary from person to person.\n- Must satisfy the two basic requirements (0 \le P(E_i) \le 1$and sum equals$1.0).\n- **Example: House Purchase Offer**:\n - Tom and Judy Elsberg submit an offer to buy a house.\n - Experimental outcomes: E_1$(offer accepted),$E_2 (offer rejected).\n - Judy's subjective probabilities: P(E_1) = 0.80$,$P(E_2) = 0.20$.
  • Tom's subjective probabilities: $P(E_1) = 0.60$, $P(E_2) = 0.40$ (reflecting greater pessimism).

Probability Analysis for the Kentucky Power and Light (KP&L) Company

Capacity Expansion Case Details

• KP&L is expanding generating capacity at a Northern Kentucky plant.
• Two sequential stages:
  • Stage 1: Design ($n_1 = 3$ possible durations: 2, 3, or 4 months).
  • Stage 2: Construction ($n_2 = 3$ possible durations: 6, 7, or 8 months).
• Total experimental outcomes: $3 \times 3 = 9$ outcomes.
• Management completion goal: $10$ months or less.

Outcome Space and Completion Times

• Outcomes are expressed as $(x, y)$, where $x$ is design months and $y$ is construction months:
  • $(2,6) \implies 2 + 6 = 8\,\text{months}$
  • $(2,7) \implies 2 + 7 = 9\,\text{months}$
  • $(2,8) \implies 2 + 8 = 10\,\text{months}$
  • $(3,6) \implies 3 + 6 = 9\,\text{months}$
  • $(3,7) \implies 3 + 7 = 10\,\text{months}$
  • $(3,8) \implies 3 + 8 = 11\,\text{months}$
  • $(4,6) \implies 4 + 6 = 10\,\text{months}$
  • $(4,7) \implies 4 + 7 = 11\,\text{months}$
  • $(4,8) \implies 4 + 8 = 12\,\text{months}$
• Overall project completion range: $8$ to $12$ months.
• $6$ of the $9$ experimental outcomes satisfy management's goal of $10$ months or less.

Relative Frequency Probability Assignment

• Based on a study of $40$ similar projects conducted over the past three years:
  • $P(2,6) = \frac{6}{40} = 0.15$
  • $P(2,7) = \frac{6}{40} = 0.15$
  • $P(2,8) = \frac{4}{40} = 0.10$
  • $P(3,6) = \frac{8}{40} = 0.20$
  • $P(3,7) = \frac{2}{40} = 0.05$
  • $P(3,8) = \frac{2}{40} = 0.05$
  • $P(4,6) = \frac{2}{40} = 0.05$
  • $P(4,7) = \frac{4}{40} = 0.10$
  • $P(4,8) = \frac{6}{40} = 0.15$
• Sum of all probabilities: $0.15 + 0.15 + 0.10 + 0.20 + 0.05 + 0.05 + 0.05 + 0.10 + 0.15 = 1.00$

Events and Event Probabilities

• An event is a collection of sample points.
• Probability of an Event: The sum of the probabilities of the sample points that make up the event.

KP&L Event Analysis

• Event $C$ (Project completed in 10 months or less):

$C = \{(2,6), (2,7), (2,8), (3,6), (3,7), (4,6)\}$

$P(C) = P(2,6) + P(2,7) + P(2,8) + P(3,6) + P(3,7) + P(4,6)$

$P(C) = 0.15 + 0.15 + 0.10 + 0.20 + 0.05 + 0.05 = 0.70$

• Event $L$ (Project completed in less than 10 months):

$L = \{(2,6), (2,7), (3,6)\}$

$P(L) = P(2,6) + P(2,7) + P(3,6) = 0.15 + 0.15 + 0.10 = 0.40$

• Event $M$ (Project completed in more than 10 months):

$M = \{(3,8), (4,7), (4,8)\}$

$P(M) = P(3,8) + P(4,7) + P(4,8) = 0.05 + 0.10 + 0.15 = 0.30$

• Managerial summary: There is a $0.70$ probability of completing the project in $10$ months or less, a $0.40$ probability of completion in under $10$ months, and a $0.30$ probability of exceeding $10$ months.

Fundamental Probability Relationships

Complement of an Event

• Given an event $A$, the complement of $A$ (denoted $A^c$) is defined as the event consisting of all sample points that are not in $A$.
• In a Venn diagram, the rectangular area represents the sample space $S$, the circle represents event $A$, and the area outside the circle represents $A^c$
• Fundamental Complement Formula:

$P(A) + P(A^c) = 1.0$

$P(A) = 1 - P(A^c)$

• Examples:
  • Sales Contact Application: If a sales manager states that $80\%$ of new contacts result in no sale ($P(A^c) = 0.80$), the probability of making a sale ($A$) is:

$P(A) = 1 - 0.80 = 0.20$

• Supplier Quality Application: If a purchasing agent states a $0.90$ probability that a shipment is free of defective parts ($P(A) = 0.90$), the probability that the shipment contains defective parts ($A^c$) is:

$P(A^c) = 1 - 0.90 = 0.10$

Union and Intersection of Events

• Union of Events ($A \cup B$): The event containing all sample points belonging to $A$, $B$, or both.
• Intersection of Events ($A \cap B$): The event containing all sample points belonging to both $A$ and $B$.

Addition Law

• Used to compute the probability that at least one of two events occurs (the probability of the union of two events):

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

• The subtraction of $P(A \cap B)$ prevents double counting the sample points that belong to both $A$ and B$.\n\n- **Example 1: Software Engineering Performance Evaluation**:\n - Company with 50 software engineers writing online banking code.\n - Over a evaluation period:\n - 5$workers completed work late ($L$)$\implies P(L) = \frac{5}{50} = 0.10\n - 6$workers produced code containing errors ($E$)$\implies P(E) = \frac{6}{50} = 0.12\n - 2$workers both completed work late and had errors ($L \cap E$)$\implies P(L \cap E) = \frac{2}{50} = 0.04\n - A poor performance rating is given to any employee whose work was late OR contained errors (L \cup E):\n\nP(L \cup E) = P(L) + P(E) - P(L \cap E) = 0.10 + 0.12 - 0.04 = 0.18\n\n- **Example 2: Medical Center Employee Turnover**:\n - Study of employees leaving within two years:\n - 30\%$left primarily due to salary dissatisfaction ($S$)$\implies P(S) = 0.30\n - 20\%$left due to work assignment dissatisfaction ($W$)$\implies P(W) = 0.20\n - 12\%$left due to dissatisfaction with both ($S \cap W$)$\implies P(S \cap W) = 0.12\n - Probability an employee leaves due to salary, work assignments, or both:\n\nP(S \cup W) = P(S) + P(W) - P(S \cap W) = 0.30 + 0.20 - 0.12 = 0.38\n\n## Mutually Exclusive Events\n\n- Two events A$and$B are **mutually exclusive** if, when one event occurs, the other cannot occur.\n- Mutually exclusive events have no sample points in common; their intersection is empty:\n\nP(A \cap B) = 0\n\n- **Special Addition Law for Mutually Exclusive Events**:\n\nP(A \cup B) = P(A) + P(B)\n\n# Conditional Probability and Joint Probability Tables\n\n- **Conditional Probability** is the probability of an event A$occurring given that a related event$B$has already occurred, denoted$P(A|B).\n- General Formula for Conditional Probability:\n\nP(A|B) = \frac{P(A \cap B)}{P(B)}\n\nP(B|A) = \frac{P(A \cap B)}{P(A)}\n\n## Case Study: Police Force Promotion Analysis\n\n- Metropolitan police force consisting of 1,200 officers.\n- Demographics: 960$male officers ($M$),$240$female officers ($F).\n- Promotions over past 2 years: 324$total officers promoted ($A).\n - 288 male officers promoted.\n - 36 female officers promoted.\n- Non-promoted officers (A^c$):$876 total officers.\n - 960 - 288 = 672 male officers not promoted.\n - 240 - 36 = 204 female officers not promoted.\n\n### Joint Probability Table Development\n\n- Dividing specific category counts by the total population of 1,200 officers gives joint probabilities:\n - P(M \cap A) = \frac{288}{1200} = 0.24\n - P(M \cap A^c) = \frac{672}{1200} = 0.56\n - P(F \cap A) = \frac{36}{1200} = 0.03\n - P(F \cap A^c) = \frac{204}{1200} = 0.17\n\n- **Marginal Probabilities** (located in the margins of the joint probability table, computed by summing row or column joint probabilities):\n - Marginal probability of being male: P(M) = 0.24 + 0.56 = 0.80\n - Marginal probability of being female: P(F) = 0.03 + 0.17 = 0.20\n - Marginal probability of promotion: P(A) = 0.24 + 0.03 = 0.27\n - Marginal probability of non-promotion: P(A^c) = 0.56 + 0.17 = 0.73\n\n### Discrimination Analysis Calculations\n\n- Conditional probability of promotion given officer is male:\n\nP(A|M) = \frac{P(A \cap M)}{P(M)} = \frac{0.24}{0.80} = 0.30\n\n- Conditional probability of promotion given officer is female:\n\nP(A|F) = \frac{P(A \cap F)}{P(F)} = \frac{0.03}{0.20} = 0.15\n\n- **Conclusion**: Male officers had a 30\%$chance of promotion versus a$15\%$chance for female officers. Because$P(A|M) eq P(A|F), the conditional probability values support the discrimination argument raised by female officers.\n\n# Independent Events and the Multiplication Law\n\n## Independent Events\n\n- Two events A$and$B$are **independent** if the probability of event$A$is not altered or affected by the occurrence of event$B:\n\nP(A|B) = P(A)\n\n\text{or } P(B|A) = P(B)\n\n- If P(A|B) eq P(A), the events are **dependent**.\n- In the police promotion example, P(A|M) = 0.30$while$P(A) = 0.27$; thus, promotion ($A$) and gender ($M$or$F) are dependent events.\n\n## Multiplication Law\n\n- Used to compute the probability of the intersection of two events (P(A \cap B)):\n\nP(A \cap B) = P(B) P(A|B)\n\n\text{or } P(A \cap B) = P(A) P(B|A)\n\n- **Example: Streaming Service Subscriptions**:\n - Household subscription to Netflix (N$):$P(N) = 0.84\n - Household subscribing to Disney+ given Netflix subscription (D|N$):$P(D|N) = 0.75\n - Probability a household subscribes to both Netflix and Disney+ (D \cap N):\n\nP(D \cap N) = P(N) P(D|N) = 0.84 \times 0.75 = 0.63\n\n## Special Case Multiplication Law for Independent Events\n\n- When events A$and$B are independent, the multiplication law reduces to:\n\nP(A \cap B) = P(A) P(B)\n\n- **Test for Independence**: If P(A \cap B) = P(A) P(B), the events are independent; otherwise, they are dependent.\n\n- **Example: Gasoline Service Station Credit Card Purchases**:\n - 80\%$of customers use a credit card for gasoline purchases ($P = 0.80).\n - Assuming choices of successive customers are independent:\n - Event A$= First customer uses credit card ($P(A) = 0.80).\n - Event B$= Second customer uses credit card ($P(B) = 0.80).\n - Probability both next two customers use a credit card:\n\nP(A \cap B) = P(A) P(B) = 0.80 \times 0.80 = 0.64$$