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C-N MATH 201 Lecture Notes (Part 1) – Probability Foundations and Applications

Classical, Subjective, and Relative Frequency Forms of Probability

  • Classical (theoretical) probability is the form of probability used in this course and is especially suited for consistency and mathematical treatment.

  • Relative frequency probability arises after many observations of a variable and is used to infer future occurrence patterns.

  • Subjective probability does not necessarily involve calculation; it reflects belief based on information and judgment.

  • The course stresses the classical form for academic rigor, but the other forms are acknowledged as real-world approaches.

Sets, Sample Space, and Venn Diagrams

  • The universal set (population) is denoted by W and represents the set of all possible outcomes in a given experiment.

  • A subset of W is a possible event (e.g., A, B, etc.).

  • An event’s probability under a uniform model is the ratio of its size to the size of the sample space: P(A) = \frac{|A|}{|W|} or, in a geometric setting, P(A) = \frac{Area(A)}{Area(W)}. When outcomes are equally likely, this definition applies.

  • The probability of an event lies in the interval P(A) \in [0,1]. The more certain an event, the closer P(A) is to 1; the more uncertain, the closer P(A) is to 0.

  • In practice, people’s probabilities are often within the open interval (0,1); probabilities of 0 or 1 are generally reserved for theological or idealized considerations in this context.

  • An event’s complement: for an event A, its complement is A^c = W \setminus A (i.e., all outcomes in W not in A). The Complementation Rule: P(A) = 1 - P(A^c).

  • Example: If the probability of rain tomorrow in Jefferson City is P(R) = 0.7, then the probability that it does not rain is P(R^c) = 1 - P(R) = 0.3.

  • In the superhero example, with a fixed W (the set of all 15 superheroes) we have: P( ext{Sidekick}) = \frac{8}{15}, \quad P( ext{Sidekick}^c) = 1 - \frac{8}{15} = \frac{7}{15}.

  • The union and intersection of sets: A \cup B\, (A ext{ or } B) \text{ and } A \cap B\, (A ext{ and } B).

  • The remove operator: A \setminus B = A \setminus B = { x \in A: x \notin B }. The empty set is denoted by \emptyset (phi). Basic identities: A \cup A^c = W, \quad A \cap A^c = \emptyset.

Examples with Venn Diagrams and Sets

  • A simplified Venn diagram with sets Sidekick, Orphaned, Wealthy, and Masked is used to illustrate the single-universal-space concept. The universal set is the set of all superheroes; each superhero is equally likely to be chosen.

  • A single Venn diagram with 3 defined sets illustrates how to compute the probability of a defined subset from visual regions; e.g., the probability that a nerf dart lands in set A is the area of A divided by the area of W when every point in W is equally likely to be hit by the nerf dart:
    P(A) = \frac{Area(A)}{Area(W)}.n

  • Key reminders about probability in the context of Venn diagrams:

    • The probability of any set lies in [0,1].

    • For a disjoint union of sets, probabilities add without double counting the intersection. In general, you must account for overlap using the addition rule (GAR).

The Random Variable and Real-World Examples

  • Random variable X is a function: X: W \to A \subset \mathbb{R}^n. Think of X as a “big cat” pouncing on a random member of W and reporting a feature of that member (e.g., diastolic blood pressure, college class, hair color).

  • Example with 9 students in a DSC classroom: If the rv X reports student characteristics, we can compute probabilities such as:

    • P( ext{Student is Female}) = \frac{5}{9} \approx 0.56.

    • Because Female and Male are complementary, P( ext{Male}) = 1 - 0.55 = 0.44.

    • P( ext{Student has a laptop open}) = \frac{1}{9} \approx 0.11.

    • Joint events: P( ext{Female} \cap ext{Row 1}) = \frac{1}{9} \approx 0.11.

    • Union: P( ext{Female} \cup ext{Row 1}) = \frac{7}{9} \approx 0.77.

    • Complements: P( ext{Row 1}^c) = \frac{6}{9} \approx 0.67.

    • The two gender categories are disjoint: P( ext{Female} \cap \text{Male}) = 0.

  • A full introduction to the random variable framework then connects to the probability of events A or B (A ∪ B) and the need to avoid double-counting intersections.

  • The General Addition Rule for two events (often called the GAR or Pirate Rule): P(A \cup B) = P(A) + P(B) - P(A \cap B).

    • This extends to n events, though the notation becomes more involved; the principle is to add probabilities and subtract overlaps.

  • A short video reference (Shaka Khan Academy) is used to illustrate GAR in the context of a 52-card deck.

A Contingency-Table Example: The C-N Band Case

  • A band contingency table is used where X (the wildcat) lands on band members with probabilities derived from the table:

    • Example table (illustrative):

    • Female: Auxiliary 14, Brass 12, Wind 24, Percussion 10; Total 60

    • Male: Auxiliary 0, Brass 14, Wind 16, Percussion 10; Total 40

    • Total: Brass 26, Wind 40, Percussion 20, Auxiliary 14; Overall Total 100

  • Some derived probabilities:

    • P( ext{Male}) = \frac{40}{100} = 0.4, and P( ext{Female}) = 1 - 0.4 = 0.6.

    • P( ext{Brass}) = \frac{26}{100} = 0.26.

    • P( ext{Brass} \cap \text{Male}) = 0.14.

    • By GAR: P( ext{Brass} \cup \text{Male}) = 0.26 + 0.40 - 0.14 = 0.52.

    • A counting check shows that the only way to get 52/100 is to count the double-counted Brass-Male pair only once, so the result is consistent: P( ext{Brass} \cup \text{Male}) = 0.52.

  • Important points:

    • In a given contingency table, the events like Male and Female are treated as disjoint so that their intersection is zero: P( ext{Male} \cap \text{Female}) = 0.

    • Conditional probability provides another way to interpret intersections: P(A \cap B) = P(A)P(B|A) = P(B)P(A|B).

    • The equation can illustrate the sequential nature of information; information can be useful (dependent) or not (independent).

  • Exercise idea: identify independent pairs in W and verify independence with the product rule: if A and B are independent, then P(A \cap B) = P(A)P(B) and P(A|B) = P(A).

  • Key takeaway: independence is not the same as commutativity; in general, P(A|B) \neq P(B|A).

Conditional Probability, Dependence, and Independence

  • Conditional probability defined: P(A|B) = \frac{P(A \cap B)}{P(B)}.

  • If A and B are independent, then P(A|B) = P(A) and equivalently P(A \cap B) = P(A)P(B).

  • If conditional information changes the probability of A, then A and B are dependent; otherwise, they are independent.

  • Example: In the C-N Band scenario, compute: P(Male|Brass) = \frac{P(Male \cap Brass)}{P(Brass)} = \frac{0.14}{0.26} \approx 0.5385, while P(Male) = 0.40.

  • Also, P(Brass|Male) = \frac{P(Brass \cap Male)}{P(Male)} = \frac{0.14}{0.40} = 0.35. Since these are not equal, Brass and Male are not independent.

  • Quick prompts: can you find a pair of independent events in W? How would you show independence?

  • Reference: 1 Samuel 20 as an example of conditional reasoning in a real context.

Quick Reference: Key Formulas and Concepts (recap)

  • Classical probability under uniform sample space: P(A) = \frac{|A|}{|W|} = \frac{Area(A)}{Area(W)}.

  • Probability range: P(A) \in [0,1].

  • Complement rule: P(A) = 1 - P(A^c).

  • Union and intersection: P(A \cup B) = P(A) + P(B) - P(A \cap B).

  • Conditional probability: P(A|B) = \frac{P(A \cap B)}{P(B)}.

  • Independence: P(A|B) = P(A) \iff P(A \cap B) = P(A)P(B).

  • If A and B are independent, then P(A \cup B) = P(A) + P(B) - P(A)P(B).

  • Random variable overview: X: W \to A \subset \mathbb{R}^n, and X reports a feature of a randomly chosen element of W.

  • GAR (General Addition Rule) for two events: P(A \cup B) = P(A) + P(B) - P(A \cap B).

  • Common notations: A \setminus B = { x \in A : x \notin B }, \quad \emptyset = \varnothing, \quad W = \text{universal set}.