Counting and Probability Fundamentals, STAT 3021 Ch. 1–2 – Study Notes
offline

Source: Walpole Ch. 1–2, Lecture Weeks 1–2

Tags: probability, counting, sample space, events, permutations, combinations, addition rule, multiplication rule, complementary events, STAT 3021, UMN

Difficulty: Introductory | Prerequisites: Basic algebra, set notation

Big Picture

This is the foundation for everything else in STAT 3021. Probability gives you the mathematical framework for quantifying uncertainty, and counting methods give you the tools to compute probabilities in finite settings. If you cannot set up a sample space or apply the multiplication rule confidently, every later topic (random variables, distributions, inference) will feel like guessing. This material maps to Chapters 1 and 2 of Walpole and covers roughly Weeks 1 and 2 of the course.

TL;DR

Probability is a number between 0 and 1 that measures how likely an event is. You calculate it by defining a sample space (the full set of possible outcomes), identifying the event of interest, and counting. Counting methods (multiplication rule, permutations, combinations) let you handle large sample spaces without listing every outcome by hand.


Key Terms

Experiment

Any process that generates well-defined outcomes. Tossing a coin, rolling a die, and measuring a component's lifespan are all experiments. Think of it as any action where the result is not predetermined.

Sample space (S)

The complete set of all possible outcomes of an experiment. For a coin toss, S = {H, T}. For two dice, S contains 36 ordered pairs. In simple terms, this is your universe of "what could happen."

Event

A subset of the sample space. An event is any collection of outcomes you care about. "Rolling an even number" on a die is the event {2, 4, 6}. Think of it as a specific question you are asking about the experiment.

Mutually exclusive events (disjoint events)

Two events that cannot both occur at the same time. If A happens, B cannot, and vice versa. Rolling a 2 and rolling a 5 on a single die are mutually exclusive. In simple terms, there is no overlap between them.

Exhaustive events

A collection of events that together cover the entire sample space. At least one of them must occur. Think of it as a set of events with no gaps.

Permutation

An arrangement of objects where order matters. Choosing a president, vice-president, and treasurer from 10 people is a permutation problem. In simple terms, "ABC" and "CBA" count as different arrangements.

Combination

A selection of objects where order does not matter. Choosing 3 committee members from 10 people is a combination problem. In simple terms, {A, B, C} and {C, B, A} count as the same group.

Probability of an event, P(A)

A number between 0 and 1 (inclusive) representing the likelihood that event A occurs. P(A) = 0 means impossible, P(A) = 1 means certain. Think of it as the long-run relative frequency of A if you repeated the experiment indefinitely.

Complement of an event (A')

Everything in the sample space that is not in A. P(A') = 1 - P(A). In simple terms, the complement is "A does not happen."


Core Content

Sample Spaces and Events

  • Every probability problem starts with defining the sample space S

  • S can be finite (coin toss), countably infinite (count until first success), or uncountable (measuring time)

  • Events are subsets of S; the empty set is the impossible event, S itself is the certain event

  • Set operations apply directly: union (A ∪ B = A or B), intersection (A ∩ B = A and B), complement (A' = not A)

  • Venn diagrams are the standard visual tool for checking overlap, complement, and union relationships

Counting Methods

  • Multiplication rule: if one operation can be done in n₁ ways and a second in n₂ ways, together they can be done in n₁ × n₂ ways

    • Extends to any number of sequential operations: n₁ × n₂ × ... × nₖ

    • Example: 3 shirts and 4 trousers give 3 × 4 = 12 outfits

  • Permutations (order matters):

    • n distinct objects taken r at a time: P(n, r) = n! / (n - r)!

    • All n objects: n!

    • With repeated elements: n! / (n₁! × n₂! × ... × nₖ!) where nᵢ are the counts of each repeated type

  • Combinations (order does not matter):

    • n objects taken r at a time: C(n, r) = n! / [r!(n - r)!]

    • C(n, r) = C(n, n - r), which is useful for simplifying calculations

    • Example: choosing 3 people from 10 gives C(10, 3) = 120

Axioms of Probability

  • Axiom 1: For any event A, P(A) ≥ 0

  • Axiom 2: P(S) = 1

  • Axiom 3: If A₁, A₂, ... are mutually exclusive events, then P(A₁ ∪ A₂ ∪ ...) = P(A₁) + P(A₂) + ...

  • Everything else in probability theory follows from these three axioms

Addition Rule

  • General form: P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

  • Mutually exclusive case: if A ∩ B = ∅, then P(A ∪ B) = P(A) + P(B)

  • The subtraction of P(A ∩ B) corrects for double-counting the overlap

  • For three events: P(A ∪ B ∪ C) = P(A) + P(B) + P(C) - P(A ∩ B) - P(A ∩ C) - P(B ∩ C) + P(A ∩ B ∩ C)

Complement Rule

  • P(A') = 1 - P(A)

  • Often the easiest route: when "at least one" is hard to count directly, compute the complement ("none") and subtract from 1

  • Example: probability of at least one head in 3 tosses = 1 - P(all tails) = 1 - (1/2)³ = 7/8


Formulas

P(n, r) = \frac{n!}{(n-r)!}
C(n, r) = \binom{n}{r} = \frac{n!}{r!(n-r)!}
P(A \cup B) = P(A) + P(B) - P(A \cap B)
P(A') = 1 - P(A)

Real-World Applications

Counting and probability underpin quality control in manufacturing: if a factory produces 10,000 chips per day and the defect rate is 0.1%, you use these tools to calculate the probability of finding defective chips in a random sample. Combinations specifically show up whenever you need to know how many ways a team, committee, or test group can be assembled from a larger pool.


Common Misconceptions

  • Students often confuse permutations and combinations. If the problem says "arrange," "order," or "rank," use permutations. If it says "choose," "select," or "committee," use combinations.

  • Students sometimes forget to subtract the intersection when applying the addition rule, which double-counts outcomes in the overlap.

  • "Mutually exclusive" and "independent" are not the same thing. Mutually exclusive events cannot happen together; independent events can, but knowing one occurred does not change the probability of the other. These two concepts are covered more in the conditional probability notes.

  • When using the complement rule, students occasionally compute 1 - P(A) when they mean 1 - P(A'), or vice versa. Always confirm which event you are complementing.


Why It Matters / Exam Flags

⚠️ Expect problems that require you to decide whether to use permutations or combinations. The exam will not tell you which one; you must read the context.

⚠️ The complement approach ("at least one" problems) appears frequently on exams. Practise recognising when it is easier to compute 1 - P(none).

⚠️ You are allowed one 8.5" × 11" sheet of notes. Write down the permutation and combination formulas, the addition rule, and the complement rule. These are high-value items for your cheat sheet.


Quick Self-Test

  1. True or False: P(A ∪ B) = P(A) + P(B) is always valid. (False, only when A and B are mutually exclusive.)

  1. Fill in the blank: The number of ways to arrange 5 books on a shelf is ___. (5! = 120)

  1. True or False: C(8, 3) = C(8, 5). (True, by the symmetry property.)

  1. Fill in the blank: If P(A) = 0.7, then P(A') = ___. (0.3)

  1. True or False: Choosing a password of 4 digits (0 to 9, repetition allowed) is a combination problem. (False, order matters, and it is a multiplication rule problem: 10⁴ = 10,000.)


Practice Q&A

Q: A committee of 4 is to be chosen from 6 men and 5 women. How many committees can be formed if the committee must contain at least 2 women?

A: Count the cases: exactly 2 women (C(5,2) × C(6,2) = 10 × 15 = 150), exactly 3 women (C(5,3) × C(6,1) = 10 × 6 = 60), exactly 4 women (C(5,4) × C(6,0) = 5 × 1 = 5). Total = 150 + 60 + 5 = 215.

Q: A bag contains 5 red and 3 blue balls. Two balls are drawn without replacement. What is the probability that both are red?

A: P(both red) = (5/8) × (4/7) = 20/56 = 5/14 ≈ 0.357.

Q: If P(A) = 0.4, P(B) = 0.5, and P(A ∩ B) = 0.2, find P(A ∪ B).

A: P(A ∪ B) = 0.4 + 0.5 - 0.2 = 0.7.

Q: In how many ways can the letters of the word MISSISSIPPI be arranged?

A: 11! / (4! × 4! × 2!) = 34,650. There are 11 letters total, with M appearing once, I four times, S four times, and P twice.


Connections to Other Topics

Counting methods feed directly into the next topic, conditional probability and Bayes' theorem, because those formulas require you to count or compute joint and marginal probabilities. The combination formula C(n, r) reappears when you study the binomial distribution in Chapter 5. Understanding sample spaces and events is also the basis for defining random variables in Chapter 3.

Related Terms / Search Tags

sample space, event, outcome, permutation, combination, factorial, multiplication principle, addition rule, complement rule, mutually exclusive, disjoint, exhaustive, Venn diagram, counting methods, nPr, nCr, n choose r, probability axioms, set theory in probability, STAT 3021, Walpole Chapter 2