Statistics 1 for Economics - Probability and Hypergeometric Distributions

Combinatorics and Hypergeometric Probability

• Combinatorics Fundamentals

  • Permutations: Used when the order of selection matters. The formula for permutations of $r$ out of $n$ elements is:     $nPr = \frac{n!}{(n-r)!}$
  • Example: In a sports race with 12 athletes, the three winners are rewarded with a gold, silver, or bronze medal. The number of ways the medals can be distributed is a permutation because the rank (gold vs. silver) matters.
  • Combinations: Used when the order of selection does not matter. The formula for combinations of $r$ out of $n$ elements is:     $nCr = \binom{n}{r} = \frac{n!}{r!\,(n-r)!}$
  • Example: From a group of top-five athletes, two are randomly selected for the national team. The number of ways to compose the team is a combination because the order of selection within the team is irrelevant.

• Hypergeometric Probability

  • This probability distribution is used for drawing without replacement from a finite population.
  • Variables:
  • $N$: Total population size.
  • $k$: Total number of successes in the population.
  • $N - k$: Total number of failures in the population.
  • $n$: Sample size drawn from the population.
  • $x$: Exact number of successes in the sample.
  • $n - x$: Exact number of failures in the sample.
  • The Hypergeometric Formula:     $P(A) = \frac{\binom{k}{x} \binom{N-k}{n-x}}{\binom{N}{n}}$

• Case Study: Smokers vs. Non-Smokers

  • Population: 9 people (4 smokers, 5 non-smokers).
  • Experiment: Randomly select a sample of 3 persons.
  • Question: What is the probability of choosing exactly 1 smoker and 2 non-smokers?
  • Step 1: Total possible samples:     $\binom{9}{3} = \frac{9!}{3!\,(9 - 3)!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$
  • Step 2: Equally likely outcomes: Since drawing is random, each unique team has a probability of $\frac{1}{84}$.
  • Step 3: Successful outcomes: Number of ways to choose 1 smoker from 4 AND 2 non-smokers from 5:     $\binom{4}{1} \times \binom{5}{2} = 4 \times 10 = 40$
  • Step 4: Final Probability:     $P(A) = \frac{40}{84} \approx 0.476 \text{ (or } 47.6\%\text{)}$

• Case Study: Company Employees

  • Data: 80 employees total; 50 Full-time (FT), 30 Part-time (PT). Also categorized as 36 Junior (18 FT, 18 PT) and 44 Senior (32 FT, 12 PT).
  • Scenario: A council of 5 members is chosen at random.
  • Probability of exactly 1 part-time employee:     $P(1\text{ part-time}) = \frac{\binom{30}{1} \binom{50}{4}}{\binom{80}{5}} = 0.287$
  • Probability of at least 1 part-time employee:
  • This uses the complement rule: $P(\ge 1) = 1 - P(0)$.     $P(\ge 1) = 1 - \frac{\binom{50}{5} \binom{30}{0}}{\binom{80}{5}} = 1 - 0.088 = 0.912$

Basic Concepts of Probability Theory

• Random Experiment: An action where outcomes cannot be predicted with certainty, but it is repeatable under identical circumstances.
• Sample Space ($S$): The set of all possible outcomes. It must be:
  • Exhaustive: Contains all possible outcomes.
  • Mutually Exclusive: Outcomes do not overlap; only one outcome occurs at a time.
• Simple Event ($O_i$): A single specific outcome or realization from the sample space.
• Event ($A$): A collection or subset of simple events (e.g., rolling an even number on a die).
• Probability Definitions:
  • Classical (Laplace): Assumes all outcomes are equally likely.     P(A) = \frac{\text{# elements in A}}{\text{# elements in S}}
  • Relative Frequency: Based on long-run observation of repeated experiments.     P(A) = \frac{\text{# occurrences of A}}{\text{total number of experiments (n)}}
  • Example: Flipping a coin 10,000 times and getting tails 5,067 times results in $P(T) = 0.5067$.
  • Subjective: Used for non-repeatable events (e.g., predicting 2021 presidential election results).

Rules of Probability Theory

• Fundamental Rules:

  1. Every probability must be between 0 and 1: $0 \le P(A) \le 1$.
  2. The sum of all probabilities in a sample space is 1: $P(S) = 1$.

• Complement Rule:

  • The probability of an event not occurring: $P(A^c) = 1 - P(A)$.
  • Example: Rolling two dice. Probability that the numbers are different is easier calculated as $1 - P(\text{same numbers})$. Since there are 6 "same" outcomes $(1,1), (2,2), etc.$ out of 36, $P(A) = 1 - \frac{6}{36} = \frac{30}{36} = \frac{5}{6}$.

• Intersection and Union:

  • Intersection (AND): Denoted as $A \cap B$. Represents outcomes in both A and B.
  • Union (OR): Denoted as $A \cup B$. Represents outcomes in either A, B, or both.
  • Sum Rule: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.

• Disjoint Events:

  • Events are disjoint if they have no common outcomes ($P(A \cap B) = 0$).
  • For disjoint events, the sum rule simplifies to: $P(A \cup B) = P(A) + P(B)$.

• Conditional Probability:

  • The probability of event B occurring given that event A has already occurred.
  • Formula: $P(B|A) = \frac{P(A \cap B)}{P(A)}$.

• Product Rule:

  • Used to find the joint probability of two events.
  • Formula: $P(A \cap B) = P(A) \times P(B|A)$.

Independence of Events

• Definition of Independence: Two events A and B are independent if the occurrence of A does not change the probability of B.
  • Mathematical tests for independence:
  • $P(A) = P(A|B)$
  • $P(B) = P(B|A)$
  • $P(A \cap B) = P(A) \times P(B)$
• Dependency examples:
  • Rain ($R$) and Cloudy ($C$) are dependent because $P(R) \ne P(R|C)$.
  • Rain ($R$) and Rolling a Six ($S$) are independent because $P(R) = P(R|S)$.
• The Disjointness Paradox: Disjoint events can never be independent (unless one has a probability of 0). If events are disjoint, knowing A occurred means B cannot have occurred ($P(B|A) = 0$). Since this information changed the probability of B, they are dependent.

Probability Tables and Trees

• Probability Table: Organizes joint probabilities in cells and marginal probabilities in the margins.
  • Joint probabilities ($P(A \cap B)$) are inside the grid.
  • Marginal probabilities ($P(A), P(B)$, etc.) are the row and column totals.
• Probability Tree: Visualizes sequences of events.
  • Branch probabilities are conditional probabilities.
  • Multiplying along a path gives the joint probability (the intersection).
  • Path endings (Intersections) at the far right must sum to 1.

Bayes' Law

• The Concept: Used to "reverse the order" of conditional probabilities. It finds $P(A|B)$ when you only know $P(B|A)$.
• Formula:   $P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{P(A) \times P(B|A)}{P(B)}$
• Example calculation (Health Test):
  • Prior probabilities: $P(\text{Healthy}) = 0.60$, $P(\text{Sick}) = 0.40$.
  • Test Accuracy: $P(\text{Positive}|\text{Healthy}) = 0.10$, $P(\text{Positive}|\text{Sick}) = 0.80$.
  • To find the probability a person is sick given a positive test ($P(S|P)$):
  1. Find Joint Probabilities:
     • $P(H \cap P) = 0.60 \times 0.10 = 0.06$
     • $P(S \cap P) = 0.40 \times 0.80 = 0.32$
  2. Find Marginal Probability of a Positive Test ($P(P)$):
     • $P(P) = P(H \cap P) + P(S \cap P) = 0.06 + 0.32 = 0.38$
  3. Apply Bayes' Law:
     • $P(S|P) = \frac{0.32}{0.38} \approx 0.842$