Difficulty: Introductory to Intermediate | Prerequisites: Chapters 1–3 study notes
Chapter 4 introduces the language and rules of probability: how to assign numbers to events, combine them with complements, unions, and intersections, and update them when you learn new information (conditional probability). The big payoff is the multiplication rule and the concept of independence, which underpin every probability distribution in the chapters that follow.
Sample space
The complete list of every possible outcome of an experiment or process. Think of it as the full menu of things that could happen.
Event
A subset of the sample space: one or more outcomes you are interested in.
P(A) – probability of event A
The number of times A occurs divided by the total number of outcomes (in equally likely setups). Always between 0 and 1, inclusive.
Complement – P(A’)
Everything in the sample space that is not A. Calculated as 1 – P(A). In simple terms, the probability that A does not happen.
Union – P(A ∪ B)
The probability that A or B (or both) occurs. Calculated using the addition rule.
Intersection – P(A ∩ B)
The probability that both A and B occur together.
Disjoint (mutually exclusive)
Two events that cannot happen at the same time. If A and B are disjoint, P(A ∩ B) = 0, and P(A or B) simplifies to P(A) + P(B).
Conditional probability – P(A|B)
The probability of A given that B has already occurred. Formally, P(A ∩ B) / P(B). Think of it as zooming in on only the outcomes where B is true and then asking how often A also shows up.
Independence
Two events are independent when knowing one tells you nothing about the other. Formally: P(A|B) = P(A), or equivalently P(A ∩ B) = P(A) × P(B).
0 ≤ P(A) ≤ 1 for any event A
The probabilities of all outcomes in the sample space sum to 1
P(empty set) = 0
P(A’) = 1 – P(A)
Useful shortcut: when it is easier to calculate the probability of something not happening, find that and subtract from 1
General form: P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
If A and B are disjoint (mutually exclusive): P(A or B) = P(A) + P(B)
The subtraction of P(A ∩ B) prevents double-counting outcomes that belong to both events
P(A|B) = P(A ∩ B) / P(B)
Read as "the probability of A, given B"
You are restricting the sample space to only those outcomes where B is true
General form: P(A ∩ B) = P(A) × P(B|A) = P(B) × P(A|B)
For three events: P(A ∩ B ∩ C) = P(A) × P(B|A) × P(C|A and B)
Chain the conditionals: each new event is conditioned on everything before it
A and B are independent if and only if any one of these holds (they are equivalent):
P(A|B) = P(A)
P(B|A) = P(B)
P(A ∩ B) = P(A) × P(B)
If events are independent, the multiplication rule simplifies: just multiply their individual probabilities
Disjoint events are almost never independent (unless one of them has probability 0)
Rule | Formula |
|---|---|
Complement | P(A’) = 1 – P(A) |
Addition (general) | P(A ∪ B) = P(A) + P(B) – P(A ∩ B) |
Addition (disjoint) | P(A or B) = P(A) + P(B) |
Conditional | P(A|B) = P(A ∩ B) / P(B) |
Multiplication (general) | P(A ∩ B) = P(A) × P(B|A) |
Multiplication (3 events) | P(A ∩ B ∩ C) = P(A) × P(B|A) × P(C|A and B) |
Independence test | P(A ∩ B) = P(A) × P(B) |
Conditional probability is the engine behind medical screening. When a test comes back positive, doctors use P(disease|positive test) to judge how likely the patient actually has the condition, which depends heavily on how common the disease is in the population (base rate).
Independence matters in quality control. If defects on a production line are independent, the probability of two defective items in a row is simply the product of the individual defect rates.
Students frequently confuse P(A ∩ B) with P(A|B). The intersection is a joint probability; the conditional is that joint probability divided by P(B).
Disjoint and independent are not the same thing. Disjoint events cannot occur together, so knowing one happened tells you the other did not. That is the opposite of independence.
The addition rule requires subtracting P(A ∩ B) to avoid double-counting. Forgetting that subtraction is one of the most common arithmetic errors on the exam.
P(A|B) is not the same as P(B|A). This confusion is sometimes called the "prosecutor's fallacy."
⚠️ Be able to set up and solve a conditional probability problem from a two-way table or a word problem.
⚠️ Expect a question asking you to determine whether two events are independent, using any of the three equivalent conditions.
⚠️ Know how to apply the multiplication rule to chains of three events.
⚠️ The complement rule often appears as the fastest path to an answer: "at least one" problems are almost always solved as 1 – P(none).
True or false: If P(A) = 0.3 and P(B) = 0.5, and A and B are disjoint, then P(A or B) = 0.8.
Fill in the blank: P(A|B) = P(A ∩ B) / ___.
True or false: If A and B are independent, then P(A ∩ B) = P(A) + P(B).
P(A) = 0.4 and P(A’) = ___.
True or false: Disjoint events are always independent.
Answers: 1. True. 2. P(B). 3. False (it equals P(A) × P(B)). 4. 0.6. 5. False.
Q: P(A) = 0.6, P(B) = 0.3, P(A ∩ B) = 0.18. Are A and B independent?
A: Check whether P(A ∩ B) = P(A) × P(B). Here 0.6 × 0.3 = 0.18, which equals P(A ∩ B). Yes, A and B are independent.
Q: In a class of 100 students, 40 study maths, 30 study physics, and 10 study both. What is the probability a randomly chosen student studies maths or physics?
A: P(M ∪ P) = P(M) + P(P) – P(M ∩ P) = 0.40 + 0.30 – 0.10 = 0.60.
Q: You draw two cards from a deck without replacement. What is P(both are aces)?
A: P(1st ace) = 4/52. P(2nd ace | 1st ace) = 3/51. By the multiplication rule: (4/52) × (3/51) = 12/2652 = 1/221 ≈ 0.0045.
Q: A disease affects 1% of a population. A test detects it 95% of the time (sensitivity) and gives a false positive 3% of the time. What is P(disease | positive test)?
A: P(positive) = P(pos|disease) × P(disease) + P(pos|no disease) × P(no disease) = 0.95 × 0.01 + 0.03 × 0.99 = 0.0095 + 0.0297 = 0.0392. Then P(disease|positive) = 0.0095 / 0.0392 ≈ 0.242. Roughly 24%, despite a 95%-accurate test, because the disease is rare.
Conditional probability and independence are the foundation for every distribution in Ch. 5 and Ch. 6. The multiplication rule for independent events leads directly to the Poisson distribution. Bayes-style reasoning (flipping P(A|B) to P(B|A)) returns in inferential statistics when you evaluate how strong evidence from a sample really is.
STA 101, Purdue, probability, sample space, event, complement, union, intersection, disjoint, mutually exclusive, addition rule, conditional probability, multiplication rule, independence, Bayes, joint probability, marginal probability, two-way table, tree diagram, at least one, prosecutor's fallacy, base rate