Chapter 2 — Front Matter
Purpose and Overview
In Chapter 1 you encountered probability as the mathematics of uncertainty. You learned that a probability model assigns numbers to events, describes sample spaces, and provides a framework for reasoning about outcomes that are not predetermined. Now the work begins in earnest.
This chapter develops the rules of probability — the logical structure that governs how probabilities combine, relate, and transform. These rules are not tricks or formulas to memorize. They are consequences of a small number of foundational properties, derived through careful reasoning. Once you understand them as a coherent system, you will find that most probability calculations are simply disciplined applications of a few core ideas.
The most important idea in this chapter is conditional probability. Conditioning means restricting attention to a smaller part of the sample space — the part where some event has occurred. When you learn that something happened, the landscape of possibilities changes. Conditional probability captures that change precisely.
A close companion to conditional probability is independence. Two events are independent when knowing one occurred tells you nothing about whether the other occurred. Independence is a mathematical statement, not a casual one. Events do not become independent just because they feel unrelated. You will learn to verify independence rigorously.
From conditional probability grows Bayes' theorem — one of the most powerful reasoning tools in science, medicine, engineering, law, machine learning, and decision-making. Bayes' theorem answers the question: given the evidence I observe, how should I update what I believe?
This chapter concludes by preparing you for Chapter 3, where probability is extended to random variables — functions that assign numerical values to outcomes. The rules you develop here become the foundation for everything that follows in the course.
Opening Question
A diagnostic test for a disease is 95% accurate when the disease is present and 90% accurate when the disease is absent. The disease affects 1% of the population. If you test positive, what is the probability you actually have the disease? Most people guess the answer is close to 95%. The true answer, computed at the end of Section 2.11, will surprise you — and the reasoning behind it is one of the most important ideas in probability.
14Sections
12Objectives
28Key Terms
15+Worked Examples
Learning Objectives
- 1Use set notation to describe events, complements, unions, intersections, and apply De Morgan's laws.
- 2State the Kolmogorov axioms and derive basic probability rules from them.
- 3Apply the complement rule, including the "at least one" strategy.
- 4Apply the general addition rule and the inclusion-exclusion principle.
- 5Distinguish mutually exclusive events from independent events.
- 6Compute conditional probabilities from definitions, tables, and tree diagrams.
- 7Verify independence using the definition P(A ∩ B) = P(A)·P(B).
- 8Apply the general multiplication rule and the chain rule.
- 9Apply the law of total probability to compute overall probabilities by partitioning the sample space.
- 10Apply Bayes' theorem to compute posterior probabilities from priors and likelihoods.
- 11Construct and interpret tree diagrams for multi-stage probability problems.
- 12Identify and correct common errors in probability reasoning.
Sections 2.1 – 2.3
Events, Axioms, and the Complement Rule
Event Notation · Kolmogorov Axioms · De Morgan's Laws · Complement Strategy
Section 2.1
Events and Event Notation
In Chapter 1, you studied sample spaces — the set of all possible outcomes of a probability experiment. An event is a subset of the sample space. It is a collection of outcomes that share some property you care about.
For example, roll a standard six-sided die. The sample space is S = {1, 2, 3, 4, 5, 6}. The event "the result is even" is the subset A = {2, 4, 6}. Events can be simple — containing a single outcome — or compound, containing many outcomes. The sample space S itself is an event (something always happens), and the empty set ∅ is also an event (nothing happens).
Set Notation for Events
Because events are sets, all of set theory's notation applies to probability. The notation you will use throughout this course:
| Symbol | Name | Meaning |
|---|
Ac | Complement | All outcomes in S not in A |
A ∪ B | Union | Outcomes in A or B or both — "at least one" |
A ∩ B | Intersection | Outcomes in both A and B — "both occur" |
A \ B | Set difference | Outcomes in A but not B |
∅ | Empty set | No outcomes; the impossible event |
S | Sample space | All outcomes; the certain event |
Notation Note
Some textbooks write the complement of A as Ā or A′. This textbook uses Ac. All three notations mean the same thing: all outcomes in S that are not in A.
Example 2.1.1
Events from a card deck
- SetupA card is drawn at random from a standard 52-card deck. Let A = "the card is a heart" and B = "the card is a face card" (jack, queen, or king).
- A ∪ BThe card is a heart or a face card (or both). Includes all 13 hearts plus the jack, queen, and king of spades, clubs, and diamonds.
- A ∩ BThe card is both a heart and a face card. This is exactly {J♥, Q♥, K♥} — three cards.
- AcThe card is not a heart — it is a spade, club, or diamond.
- BcThe card is not a face card — it is an ace or a number card (2 through 10).
De Morgan's Laws
De Morgan's Laws
(A ∪ B)c = Ac ∩ Bc The complement of "at least one" is "neither."
(A ∩ B)c = Ac ∪ Bc The complement of "both" is "at least one is not."
Tip — Using De Morgan's Laws
When you need the complement of a union or intersection, De Morgan's laws give you the equivalent expression. This is useful when one form is easier to compute than the other. Sketch a Venn diagram whenever the structure of a problem is unclear.
- A fair die is rolled. Let A = "result is even" and B = "result is a multiple of 3." List the outcomes in A, B, A ∪ B, A ∩ B, Ac, Bc, and Ac ∩ Bc.
- A student is selected at random from a school. Let A = "the student takes calculus" and B = "the student takes physics." Describe A ∩ B, A ∪ B, Ac, and (A ∪ B)c in words.
- Use De Morgan's first law to rewrite (A ∪ B)c for the events in Problem 1. Verify both forms give the same set of outcomes.
- A single card is drawn. Let A = "the card is red" and B = "the card is a spade." What is A ∩ B? What does this tell you about these two events?
Section 2.2
Probability Axioms
A probability is not just any assignment of numbers to events. For the assignment to qualify as a probability measure, it must satisfy three conditions known as the Kolmogorov axioms, named after the Russian mathematician Andrey Kolmogorov who formalized probability theory in 1933.
The Kolmogorov Axioms
Axiom 1 (Non-negativity). For every event A: P(A) ≥ 0.
Axiom 2 (Normalization). The probability of the entire sample space is one: P(S) = 1. Something always happens.
Axiom 3 (Countable Additivity). If A₁, A₂, A₃, … is a countable collection of mutually exclusive events (Aᵢ ∩ Aⱼ = ∅ for all i ≠ j), then P(A₁ ∪ A₂ ∪ ···) = P(A₁) + P(A₂) + ···
Why Three Axioms Are Enough
Everything else in probability theory — every rule, formula, and theorem — follows logically from these three axioms. The axioms do not say what numbers to assign to events; they say only what properties any valid probability assignment must have. This is the strength of the axiomatic approach: it applies to any probability model, no matter how it was constructed.
Example 2.2.1
Verifying the axioms for a coin-toss experiment
- SetupToss a fair coin twice. Sample space: S = {HH, HT, TH, TT}. Each outcome has probability 1/4.
- Axiom 2
P(S) = 1/4 + 1/4 + 1/4 + 1/4 = 1 ✓
- Axiom 3Let A = {HH, HT} (first toss heads) and B = {TT} (both tails). These are mutually exclusive.
P(A ∪ B) = P(A) + P(B) = 2/4 + 1/4 = 3/4 ✓
- Prove, directly from the axioms, that P(∅) = 0. (Hint: ∅ and S are mutually exclusive and their union is S.)
- Prove that P(A) ≤ 1 for all events A using only the axioms and the complement rule.
- A probability model assigns P({1}) = 0.2, P({2}) = 0.3, P({3}) = 0.15, P({4}) = 0.35. Verify that this is a valid probability measure on S = {1, 2, 3, 4}.
- Someone claims P(A) = 0.6, P(B) = 0.5, P(A ∪ B) = 1.2 for mutually exclusive events A and B. Identify which axiom is violated.
Section 2.3
Complements
Theorem 2.3.1
Complement Rule
For any event A: P(Ac) = 1 − P(A).
Proof. The events A and Ac are mutually exclusive by definition, and A ∪ Ac = S. By Axiom 3 and Axiom 2: P(A) + P(Ac) = P(S) = 1. Rearranging gives P(Ac) = 1 − P(A). ∎
Using Complements Strategically
The complement rule is more than a formula — it is a problem-solving strategy. When an event is difficult to compute directly, its complement is often easier. The most common trigger phrase is "at least one." The complement of "at least one" is "none." In many settings, "none" is far easier to compute.
Example 2.3.1
At least one correct — using the complement
- ProblemA student takes a five-question true-false quiz by guessing randomly. What is the probability of getting at least one question correct?
- StrategyLet A = "at least one correct." Then Ac = "none correct" — all five guesses are wrong.
- SolveThe probability of one wrong guess is 1/2. Since guesses are independent (see Section 2.8):
P(Ac) = (1/2)⁵ = 1/32
P(A) = 1 − 1/32 = 31/32 ≈ 0.969
- InterpretThe probability of getting at least one question correct by random guessing is about 96.9%. Guessing on all five rarely produces a completely blank result.
Answer: 31/32 ≈ 0.969
Example 2.3.2
At least one system failure
- ProblemA system has three independent components. Each fails with probability 0.1. What is the probability of at least one failure?
- ComplementAc = "no failures" — all three components work.
P(Ac) = (0.9)³ = 0.729
P(at least one failure) = 1 − 0.729 = 0.271
Answer: 0.271
- A fair die is rolled four times. Find the probability of rolling at least one six. Use the complement.
- A quality inspector tests five independently produced parts. Each part has a 4% chance of being defective. Find the probability that at least one is defective.
- P(A) = 0.35. Find P(Ac), P((Ac)c), and verify the result makes sense.
- Explain why P(Ac) = 1 − P(A) cannot be used to derive P(Ac) < 0 for any valid event A.
Sections 2.4 – 2.6
Unions, Addition Rules, and Mutually Exclusive Events
General Addition Rule · Inclusion-Exclusion · Mutually Exclusive vs. Independent
Section 2.4
Unions and Intersections
Theorem 2.4.1
General Addition Rule
For any two events A and B:
P(A ∪ B) = P(A) + P(B) − P(A ∩ B).
Why subtract? When you add P(A) and P(B), the outcomes in A ∩ B are counted twice — once in each. Subtracting P(A ∩ B) once removes the double-count, so each outcome is counted exactly once.
Example 2.4.1
Students in math and physics
- SetupIn a group of 100 students, 60 take mathematics, 45 take physics, and 30 take both. A student is chosen at random.
- GivenP(M) = 0.60, P(P) = 0.45, P(M ∩ P) = 0.30
- Solve
P(M ∪ P) = 0.60 + 0.45 − 0.30 = 0.75
- InterpretThe probability that a randomly chosen student takes mathematics or physics (or both) is 0.75.
Answer: 0.75
Extension to Three Events
The inclusion-exclusion principle extends to 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)
The pattern alternates signs at each level: add individual probabilities, subtract pairwise intersections, add back the triple intersection. The formula continues for more events.
- A card is drawn from a standard deck. Find: (a) P(diamond or ace), (b) P(red or queen), (c) P(club or face card).
- In a group of 300 students, 120 own a car, 80 own a bicycle, and 40 own both. Find P(car or bicycle) and P(neither).
- Let P(A) = 0.5, P(B) = 0.4, P(A ∪ B) = 0.7. Find P(A ∩ B).
- Prove the three-event inclusion-exclusion formula using the two-event formula applied twice.
Section 2.5
Addition Rules
| Situation | Rule | Formula |
|---|
| Mutually exclusive events (A ∩ B = ∅) | Simple addition | P(A ∪ B) = P(A) + P(B) |
| General events (any A and B) | Inclusion-exclusion | P(A ∪ B) = P(A) + P(B) − P(A ∩ B) |
| Partition (exhaustive and mutually exclusive) | Partition sum | P(A₁) + P(A₂) + ··· + P(Aₙ) = 1 |
Student Warning — Check Mutual Exclusivity First
Always ask whether A ∩ B = ∅ before summing P(A) + P(B). Applying simple addition when events overlap double-counts the intersection and gives a probability that may even exceed 1. If there is any chance the events can co-occur, use the general addition rule.
Example 2.5.1
Choosing the right addition rule
- Case AA fair die is rolled. Let A = "roll a 3" and B = "roll a 6." These cannot both occur: A ∩ B = ∅.
P(A ∪ B) = 1/6 + 1/6 = 2/6 = 1/3
- Case BLet A = "roll an even number" = {2,4,6} and B = "roll a number > 4" = {5,6}. Then A ∩ B = {6}.
P(A ∪ B) = 3/6 + 2/6 − 1/6 = 4/6 = 2/3
- P(A) = 0.3, P(B) = 0.5, mutually exclusive. Find P(A ∪ B).
- P(A) = 0.4, P(B) = 0.6, P(A ∩ B) = 0.2. Find P(A ∪ B).
- Die roll: P(prime) and P(even). Are these mutually exclusive? Find P(prime ∪ even).
- Events A, B, C partition S. P(A) = 0.35, P(B) = 0.45. Find P(C).
Section 2.6
Mutually Exclusive Events
Definition
Events A and B are mutually exclusive (also called disjoint) if A ∩ B = ∅, meaning they cannot both occur in the same trial. Equivalently, P(A ∩ B) = 0.
Critical Distinction — Mutually Exclusive Is NOT Independent
Confusing mutual exclusivity with independence is one of the most common errors in probability. If A and B are mutually exclusive with P(A) > 0 and P(B) > 0, then knowing A occurred tells you B definitely did not occur. That is extreme dependence — the opposite of independence. Events with positive probability cannot be both mutually exclusive and independent.
Example 2.6.1
Manufacturing quality classification
- SetupA part is classified as excellent, acceptable, or defective. P(excellent) = 0.65, P(acceptable) = 0.28, P(defective) = 0.07.
- VerifyThese categories are mutually exclusive and exhaustive.
0.65 + 0.28 + 0.07 = 1.00 ✓
- UseP(not defective) = P(excellent) + P(acceptable) = 0.65 + 0.28 = 0.93. Or by complement: 1 − 0.07 = 0.93. ✓
Answer: P(not defective) = 0.93
- For each pair, state whether the events are mutually exclusive and explain why: (a) Drawing a heart and drawing a spade; (b) Drawing a heart and drawing a face card; (c) Rolling a 4 and rolling an even number.
- If A and B are mutually exclusive with P(A) = 0.3 and P(B) = 0.4, find P(Ac ∩ Bc).
- True or False: If P(A ∩ B) = 0, then A and B are independent. Justify your answer.
Sections 2.7 – 2.9
Conditional Probability, Independence, and the Multiplication Rule
P(A|B) · Conditioning Changes the Sample Space · Independence · Chain Rule
Section 2.7
Conditional Probability
Suppose you are told that an event B has already occurred. How does this information change the probability that another event A also occurred? This is the question that conditional probability answers — and it is one of the most important ideas in all of probability theory.
When B is known to have occurred, the relevant sample space is no longer all of S. It is reduced to B. Within that reduced sample space, A can only occur in the part where A and B overlap: A ∩ B. Conditional probability rescales the probability of A ∩ B so the total probability of the new sample space (B) equals one.
Definition 2.7.1 — Conditional Probability
The conditional probability of A given B, written P(A | B), is defined for P(B) > 0 as:
P(A | B) = P(A ∩ B) / P(B)
The vertical bar "|" is read "given." So P(A | B) is read "the probability of A given B."
Conditioning Changes the Sample Space
This is the central insight of conditional probability: it does not merely add information. It fundamentally changes what the sample space is. After conditioning on B, you act as if B is the entire world. All the probability axioms hold within this new sample space — complements, unions, intersections, all rules — but within B, not within S.
Example 2.7.1
Conditional probability on a die roll
- ProblemA fair die is rolled. Let A = "result is 2" and B = "result is even" = {2, 4, 6}. Find P(A | B).
- Values
P(A) = 1/6, P(B) = 3/6 = 1/2, P(A ∩ B) = P({2}) = 1/6
- Compute
P(A | B) = P(A ∩ B) / P(B) = (1/6) / (3/6) = 1/3
- InterpretKnowing the result is even restricts possibilities to {2, 4, 6}. Within those three equally likely outcomes, the probability the result is 2 is 1/3.
Answer: P(A | B) = 1/3, compared to P(A) = 1/6 unconditionally.
Example 2.7.2
Conditional probability from a table
- SetupA survey of 200 employees records job type and wellness program participation.
- Table
| Participates | Does Not Participate | Total |
|---|
| Office | 60 | 80 | 140 |
| Field | 30 | 30 | 60 |
| Total | 90 | 110 | 200 |
- Find P(P|F)Among the 60 field employees, 30 participate.
P(P | F) = (30/200) / (60/200) = 30/60 = 0.50
- Find P(P|Fc)Among 140 office employees, 60 participate.
P(P | Fc) = 60/140 ≈ 0.429
- CompareField employees participate at a higher rate (50%) than office employees (42.9%) in this sample.
- A fair die is rolled. Let A = "result is even" and B = "result is at least 5." Find P(A | B) and P(B | A). Explain why these two values differ.
- Two cards are drawn without replacement. Find: (a) P(second is an ace | first is an ace); (b) P(both are aces).
- Let P(A) = 0.4, P(B) = 0.5, P(A ∩ B) = 0.2. Find P(A | B), P(B | A), P(A | Bc), P(Ac | Bc).
- Show algebraically that P(A | B) + P(Ac | B) = 1. What does this mean in words?
Section 2.8
Independence
Definition 2.8.1 — Independence
Events A and B are independent if and only if
P(A ∩ B) = P(A) · P(B).
Equivalently (when P(B) > 0): P(A | B) = P(A) — learning B occurred does not change the probability of A.
Equivalently (when P(A) > 0): P(B | A) = P(B) — learning A occurred does not change the probability of B.
Independence Must Be Verified
Independence is not something you assume because two events "feel" unrelated. It must be verified from the definition. Many events that seem unrelated are not independent. The most reliable source of true independence: repeated independent trials — tossing a fair coin twice, rolling two dice, drawing with replacement.
Example 2.8.1
Testing independence on a die
- SetupRoll a fair die. Let A = "even" = {2,4,6} and B = "at most 3" = {1,2,3}.
- Values
P(A) = 1/2, P(B) = 1/2, A ∩ B = {2}, P(A ∩ B) = 1/6
- Check
P(A) · P(B) = 1/2 · 1/2 = 1/4 ≠ 1/6 = P(A ∩ B)
- ConcludeA and B are not independent. Knowing the result is at most 3 changes P(A) from 1/2 to P(A | B) = (1/6)/(1/2) = 1/3.
Not independent — the definition fails.
Example 2.8.2
Two independent coin tosses
- SetupA fair coin is tossed twice. Let A = "first toss is heads" and B = "second toss is heads."
- Check
P(A) = 1/2, P(B) = 1/2, P(A ∩ B) = P(HH) = 1/4
P(A) · P(B) = 1/2 · 1/2 = 1/4 = P(A ∩ B) ✓
- ConcludeA and B are independent. The result of one coin toss does not affect the result of the other.
Mutual Independence of Multiple Events
Three events A, B, C are mutually independent if all four conditions hold simultaneously:
P(A ∩ B) = P(A)P(B)
P(A ∩ C) = P(A)P(C)
P(B ∩ C) = P(B)P(C)
P(A ∩ B ∩ C) = P(A)P(B)P(C)
All four conditions are required. Three events can satisfy the pairwise conditions without satisfying the triple-intersection condition — that situation is called pairwise independence but not mutual independence.
Independence and Complements
Result
If A and B are independent, then A and Bc are also independent (and so are Ac and B, and Ac and Bc).
Proof: P(A ∩ Bc) = P(A) − P(A ∩ B) = P(A) − P(A)P(B) = P(A)(1 − P(B)) = P(A) · P(Bc). ∎
- A fair die is rolled. Let A = "even" and B = "result is 1, 2, or 3." Are A and B independent? Show all work using the definition.
- Let P(A) = 0.3, P(B) = 0.4, and A and B are independent. Find: (a) P(A ∩ B); (b) P(A ∪ B); (c) P(Ac ∩ Bc); (d) P(A | B).
- True or False (justify each): (a) If A and B are mutually exclusive, they are independent. (b) If P(A|B) = P(A), then A and B are independent. (c) If A and B are independent and B and C are independent, then A and C are independent.
- A system has three independent components. Each works with probability 0.95. The system works only if all three work. Find P(system works).
Section 2.9
Multiplication Rules
Theorem 2.9.1 — General Multiplication Rule
For any events A and B:
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 ∩ B)
For n events (the chain rule):
P(A₁ ∩ ··· ∩ Aₙ) = P(A₁) · P(A₂ | A₁) · P(A₃ | A₁ ∩ A₂) ··· P(Aₙ | A₁ ∩ ··· ∩ Aₙ₋₁)
When A and B are independent, P(B | A) = P(B), so this simplifies to: P(A ∩ B) = P(A) · P(B)
Example 2.9.1
Drawing without replacement
- ProblemA box contains 5 red and 3 blue marbles. Two are drawn without replacement. Find P(both red).
- SolveP(first red) = 5/8. Given the first is red, 4 red and 3 blue remain (7 total).
P(second red | first red) = 4/7
P(both red) = P(first red) · P(second red | first red) = 5/8 · 4/7 = 20/56 = 5/14
- NoteDrawing with replacement would give P = 5/8 · 5/8 = 25/64. The distinction between with and without replacement is fundamental — without replacement creates dependence.
Answer: 5/14 ≈ 0.357
Example 2.9.2
Independent trials — password characters
- ProblemA password has three characters chosen randomly from digits 0–9 and letters A–Z (36 total, repetition allowed). Find P(all three are digits).
- SolveCharacters are chosen independently, so:
P(all digits) = (10/36)³ = (5/18)³ ≈ 0.043
Answer: (10/36)³ ≈ 0.043
- An urn contains 3 red, 4 white, and 5 blue balls. Three balls are drawn without replacement. Find: (a) P(all three blue); (b) P(first red, second white, third blue).
- A fair coin is tossed five times. Find P(all five heads).
- A component fails with probability 0.05. A system requires all four independent components to work. Find P(system works).
- Explain the difference between P(A ∩ B) = P(A)·P(B|A) and P(A ∩ B) = P(A)·P(B). When is each formula the right one to use?
Sections 2.10 – 2.12
Total Probability, Bayes' Theorem, and Tree Diagrams
Partitioning the Sample Space · Prior and Posterior · Tree Diagram Method
Section 2.10
Law of Total Probability
Theorem 2.10.1
Law of Total Probability
If {B₁, B₂, …, Bₖ} is a partition of S (mutually exclusive, exhaustive, each with P(Bᵢ) > 0), then for any event A:
P(A) = Σᵢ P(A | Bᵢ) · P(Bᵢ)
Proof. Because {B₁,…,Bₖ} partitions S, the events A ∩ B₁, …, A ∩ Bₖ are mutually exclusive with union A. By Axiom 3 and the multiplication rule: P(A) = Σᵢ P(A ∩ Bᵢ) = Σᵢ P(A | Bᵢ) · P(Bᵢ). ∎
Strategy — Divide and Conquer
The law of total probability lets you compute P(A) by breaking the problem into cases. Each case corresponds to one element of the partition. Compute P(A) within each case (the conditional probability), weight by the probability of that case, and sum. This is especially useful when A is hard to compute directly but easy conditioned on each Bᵢ.
Example 2.10.1
Factory defect rate
- SetupMachine 1 produces 60% of all parts with a 2% defect rate. Machine 2 produces 40% with a 5% defect rate.
- PartitionB₁ = "produced by Machine 1," B₂ = "produced by Machine 2," D = "defective."
- Solve
P(D) = P(D | B₁)·P(B₁) + P(D | B₂)·P(B₂)
= (0.02)(0.60) + (0.05)(0.40)
= 0.012 + 0.020
= 0.032
- InterpretThe overall defect rate is 3.2%. Machine 2 has a higher defect rate (5%) but produces fewer parts, so it contributes proportionally less to the total.
Answer: P(D) = 0.032 (3.2%)
Example 2.10.2
Disease testing — setting up for Bayes' theorem
- SetupA disease affects 1% of a population. A diagnostic test has 95% sensitivity (P(T⁺|D) = 0.95) and 90% specificity (P(T⁻|Dᶜ) = 0.90, so P(T⁺|Dᶜ) = 0.10).
- Find P(T⁺)
P(T⁺) = P(T⁺|D)·P(D) + P(T⁺|Dᶜ)·P(Dᶜ)
= (0.95)(0.01) + (0.10)(0.99)
= 0.0095 + 0.0990
= 0.1085
- InterpretAbout 10.85% of people test positive. The dominant contributor is false positives from the disease-free majority — a crucial insight for Section 2.11.
Answer: P(T⁺) = 0.1085
- Three factories produce components: Factory A (50%, defect rate 2%), Factory B (30%, defect rate 4%), Factory C (20%, defect rate 6%). Find the overall defect rate.
- A bag contains two types of coins: type I (P(heads) = 0.6, 70% of bag) and type II (P(heads) = 0.3, 30% of bag). A coin is chosen at random and flipped. Find P(heads).
- In a city, 40% of voters are Party A (60% turnout), 35% are Party B (50% turnout), and 25% are Independent (40% turnout). Find the overall voter turnout rate.
Section 2.11
Bayes' Theorem
The law of total probability computes P(A) by conditioning on cases Bᵢ. Bayes' theorem inverts this: given that A has occurred, what is the probability that a particular case Bⱼ was the cause? This is the problem of inference: given an observation (data, a test result, an outcome), what can we conclude about the underlying cause?
Theorem 2.11.1
Bayes' Theorem
If {B₁, …, Bₖ} is a partition of S with P(Bᵢ) > 0 for all i, and A is any event with P(A) > 0, then for each j:
P(Bⱼ | A) = P(A | Bⱼ) · P(Bⱼ) / Σᵢ P(A | Bᵢ) · P(Bᵢ)
Proof. P(Bⱼ | A) = P(A ∩ Bⱼ)/P(A) = P(A | Bⱼ)·P(Bⱼ)/P(A). Substituting the law of total probability for P(A) in the denominator gives the result. ∎
Vocabulary: Prior, Likelihood, and Posterior
| Term | Symbol | Meaning |
|---|
| Prior probability | P(Bⱼ) | Probability of Bⱼ before observing A |
| Likelihood | P(A | Bⱼ) | Probability of observing A if Bⱼ is true |
| Posterior probability | P(Bⱼ | A) | Updated probability of Bⱼ after observing A |
Example 2.11.1
The disease diagnosis problem — a surprising answer
- ProblemUsing the setup from Example 2.10.2: a person tests positive. What is the probability they actually have the disease?
- Apply Bayes
P(D | T⁺) = P(T⁺ | D) · P(D) / P(T⁺)
= (0.95)(0.01) / 0.1085
= 0.0095 / 0.1085
≈ 0.0876
- InterpretGiven a positive test, there is only about an 8.76% chance the person has the disease. This surprises most people who expect the answer to be close to 95%.
Answer: P(D | T⁺) ≈ 0.088 — about 8.8%.
The Base Rate Fallacy
A test with 95% sensitivity and 90% specificity produces mostly false positives when the disease prevalence is only 1%. The prior probability of disease is so low that even strong positive evidence does not bring the posterior probability very high. This phenomenon — the base rate fallacy — has major practical consequences in medical screening, security screening, and machine learning. Always incorporate the prior probability P(B) when applying Bayes' theorem.
Example 2.11.2
Factory defect source — three-case Bayes' theorem
- SetupLine 1 produces 50% of output (1% defect rate), Line 2 produces 30% (2% defect rate), Line 3 produces 20% (4% defect rate). A defective part is found. What is P(Line 3 | defective)?
- P(D) first
P(D) = (0.01)(0.50) + (0.02)(0.30) + (0.04)(0.20)
= 0.005 + 0.006 + 0.008 = 0.019
- Apply Bayes
P(L₃ | D) = P(D | L₃) · P(L₃) / P(D)
= (0.04)(0.20) / 0.019
= 0.008 / 0.019
≈ 0.421
- InterpretLine 3 produces only 20% of output but accounts for about 42.1% of defective parts, because of its higher defect rate.
Answer: P(L₃ | defective) ≈ 0.421
Sequential Updating
One of the most powerful features of Bayesian reasoning is that it is sequential: the posterior from one update becomes the prior for the next. If new evidence arrives, compute the new posterior using the previous posterior as the new prior. Each piece of evidence refines the probability estimate. This sequential, updatable structure makes Bayesian reasoning central to machine learning, filtering algorithms, and statistical inference (Chapter 10).
- A spam filter correctly identifies 92% of spam and 98% of legitimate messages. Suppose 30% of all email is spam. (a) Find P(classified as spam). (b) Find P(actually spam | classified as spam). (c) Find P(actually spam | classified as legitimate).
- A bag contains 70% type I coins (P(heads) = 0.6) and 30% type II coins (P(heads) = 0.3). A coin is selected and flipped. It comes up heads. (a) Find P(type I | heads). (b) The coin is flipped again and comes up heads a second time. Update the probability the coin is type I.
- Construct your own Bayes' theorem problem in a context of your choosing. Define the partition, state the prior and likelihood values, and compute the posterior.
Section 2.12
Tree Diagrams and Partitioned Sample Spaces
A tree diagram is a visual tool that organizes probability calculations for experiments with multiple stages. Each stage corresponds to a level of branches. Each branch is labeled with a conditional probability. The probability of any path is the product of all branch probabilities along that path.
Rules for Tree Diagrams
1. Each set of branches from a single node represents a partition — their probabilities must sum to 1.
2. Branch probabilities at later stages are conditional on everything that came before.
3. The probability of a complete path = product of all branch probabilities along that path.
4. The sum of all complete path probabilities = 1. Use this as a check.
Example 2.12.1
Tree diagram — disease testing
- Stage 1Disease status: D (prob 0.01) or Dc (prob 0.99).
- Stage 2Test result (conditional on Stage 1):
From D: T⁺ (0.95), T⁻ (0.05)
From Dc: T⁺ (0.10), T⁻ (0.90)
- Path table
| Path | Calculation | Probability |
|---|
| D ∩ T⁺ | 0.01 × 0.95 | 0.0095 |
| D ∩ T⁻ | 0.01 × 0.05 | 0.0005 |
| Dc ∩ T⁺ | 0.99 × 0.10 | 0.0990 |
| Dc ∩ T⁻ | 0.99 × 0.90 | 0.8910 |
| Total | | 1.0000 ✓ |
- Read offP(T⁺) = 0.0095 + 0.0990 = 0.1085. P(D | T⁺) = 0.0095 / 0.1085 ≈ 0.0876. These match our earlier calculations.
Example 2.12.2
Tree diagram — sampling without replacement
- SetupAn urn has 4 red and 6 blue balls. Two balls are drawn without replacement.
- Stage 1First draw: R₁ (4/10 = 0.4) or B₁ (6/10 = 0.6).
- Stage 2Second draw (9 balls remain):
From R₁: P(R₂|R₁) = 3/9, P(B₂|R₁) = 6/9
From B₁: P(R₂|B₁) = 4/9, P(B₂|B₁) = 5/9
- Paths
| Outcome | Calculation | Probability |
|---|
| R₁ ∩ R₂ | 4/10 · 3/9 | 12/90 = 2/15 |
| R₁ ∩ B₂ | 4/10 · 6/9 | 24/90 = 4/15 |
| B₁ ∩ R₂ | 6/10 · 4/9 | 24/90 = 4/15 |
| B₁ ∩ B₂ | 6/10 · 5/9 | 30/90 = 5/15 |
| Total | | 15/15 = 1 ✓ |
- A bag contains 5 red chips and 7 blue chips. Two chips are drawn without replacement. Draw a tree diagram showing all outcomes and their probabilities. Find P(both same color) and P(second is blue | first is red).
- Redraw the factory defect problem from Example 2.11.2 as a tree diagram. Verify all path probabilities sum to 1 and use the tree to compute P(L₂ | defective).
- A biased coin with P(H) = 0.6 is tossed twice. Given the result was not two tails, find P(two heads). Use a tree diagram to organize your work.
Sections 2.13 – 2.14
Common Mistakes and the Road to Random Variables
Eight Recurring Errors · Transposition Fallacy · Connections to Chapters 3–14
Section 2.13
Common Probability Rule Mistakes
Understanding where probability reasoning goes wrong is as important as understanding where it goes right. This section catalogs the eight most frequent errors students make with probability rules and offers precise corrections.
Mistake 1 — Adding Probabilities Without Checking Mutual Exclusivity
P(heart or face card) = 13/52 + 12/52 = 25/52
Error: The events overlap — three cards are both hearts and face cards. Correct: P(heart or face card) = 13/52 + 12/52 − 3/52 = 22/52. Always check whether A ∩ B = ∅ before summing probabilities.
Mistake 2 — Confusing Mutually Exclusive with Independent
"They can't both happen, so they must be independent."
Error: If A and B are mutually exclusive with positive probabilities, then P(A ∩ B) = 0 ≠ P(A)·P(B). They fail the independence definition. In fact, mutual exclusivity with positive probability is the opposite of independence — knowing A occurred tells you B definitely did not.
Mistake 3 — Reversing Conditional Probability (The Transposition Fallacy)
P(A | B) = P(B | A) — treating the two as equal
Error: P(disease | positive test) ≈ 0.088, not 0.95. The probability of a positive test given disease is entirely different from the probability of disease given a positive test. This error appears in legal reasoning (the prosecutor's fallacy) and medical interpretation. Use Bayes' theorem to compute P(B | A) from P(A | B).
Mistake 4 — Ignoring the Base Rate
"The test is 95% accurate and it's positive, so there's a 95% chance I have the disease."
Error: Base rate fallacy. Even a highly accurate test produces mostly false positives when the condition is rare, because the disease-free population vastly outnumbers the diseased population. Always incorporate P(B) when applying Bayes' theorem.
Mistake 5 — Applying the Independent Multiplication Rule Without Justification
Drawing two cards: P(both aces) = 4/52 · 4/52 (treating as independent)
Error: Sampling without replacement creates dependence. The correct computation: P(both aces) = 4/52 · 3/51. Only use P(A ∩ B) = P(A)·P(B) when independence has been established or stated.
Mistake 6 — Using a Conditional Probability as an Unconditional One
"80% of lung cancer patients smoked, so the probability of cancer if you smoke is 80%."
Error: Confusing P(smoked | cancer) with P(cancer | smoked). These are entirely different quantities. P(smoked | cancer) tells you about smoking among cancer patients. P(cancer | smoked) tells you about cancer risk among smokers. Use Bayes' theorem or direct data to compute the one you need.
Mistake 7 — Conditioning on an Impossible Event
Writing P(A | B) when P(B) = 0
Error: P(A | B) is only defined when P(B) > 0. Conditioning on an impossible event is undefined — the formula P(A ∩ B)/P(B) has a zero denominator. Before writing P(A | B), verify that P(B) > 0.
Mistake 8 — Assuming Independence from Physical Separation
"They come from different processes, so they must be independent."
Error: Independence is a mathematical property that must follow from the probability model, not from physical intuition alone. In some problems, outcomes from different processes share a common cause and are not independent. Always verify through the definition P(A ∩ B) = P(A)·P(B) or through the structure of the model.
Each problem contains one or more errors. Identify the error, explain why it is wrong, and provide a corrected solution.
- A student writes: "The probability the first card is an ace is 4/52. The probability the second card is an ace is also 4/52. Therefore P(both aces) = 4/52 · 4/52." What is the error?
- "Events A and B can't both happen at the same time, so they must be independent."
- "A disease test has a false positive rate of only 1%. I tested positive, so there's a 99% chance I have the disease."
- "Since P(A | B) = 0.7 and P(A) = 0.7, the events are mutually exclusive."
- A student computes P(at least one six in four rolls) = 4 × 1/6 = 4/6 = 2/3. What is wrong? Find the correct answer using the complement.
- A student writes P(A ∪ B) = P(A) + P(B) − P(A)·P(B). (a) Under what conditions is this formula correct? (b) Give an example where this gives the wrong answer.
Section 2.14
Preparing for Random Variables
In Chapters 1 and 2, you have worked with events — sets of outcomes — and computed their probabilities using a systematic set of rules. In Chapter 3, the perspective shifts. Instead of studying events directly, you will assign numbers to outcomes and study those numbers.
A random variable is a function from the sample space to the real numbers. It turns uncertain outcomes into uncertain numerical quantities. Instead of studying the event "two heads in a row," you define a random variable X = "number of heads in two coin tosses" and study the probability that X = 0, X = 1, or X = 2. This numerical framing is far more flexible and powerful than working with events alone — it allows you to compute averages, measure spread, and use the full machinery of calculus.
None of what you have learned in this chapter is left behind. Every rule carries directly forward.
Connections to Later Chapters
| This Chapter | Later Connection |
|---|
| Conditional Probability (§2.7) | Conditional Distributions (Ch. 5), Bayesian Inference (Ch. 10) |
| Independence (§2.8) | Independence of Random Variables (Ch. 5) |
| Multiplication Rule (§2.9) | Joint Distributions (Ch. 5), Maximum Likelihood Estimation (Ch. 10) |
| Law of Total Probability (§2.10) | Law of Total Expectation (Ch. 6), Sampling Distributions (Ch. 7) |
| Bayes' Theorem (§2.11) | Bayesian Estimation and Posterior Distributions (Ch. 10) |
| Partition and Tree Diagrams (§2.12) | Decomposition Methods in Distribution Theory (Ch. 6, 7) |
- A fair coin is tossed three times. (a) List all outcomes. (b) Let X = number of heads. List the values X can take. (c) Find P(X = 2). This is the kind of question Chapter 3 will develop systematically.
- Explain in your own words how conditional probability in Section 2.7 will become a conditional distribution in Chapter 5.
- Explain how the independence condition P(A ∩ B) = P(A)·P(B) will be extended to define independence of random variables.
Chapter 2 — Back Matter
Chapter Summary
- Sec 2.1Events are subsets of the sample space. Standard set notation — Ac, A ∪ B, A ∩ B — describes how events combine. De Morgan's laws: (A ∪ B)c = Ac ∩ Bc and (A ∩ B)c = Ac ∪ Bc.
- Sec 2.2The Kolmogorov axioms define a valid probability measure: non-negativity, normalization (P(S) = 1), and countable additivity. All other probability rules follow logically from these three axioms.
- Sec 2.3Complement rule: P(Ac) = 1 − P(A). The complement strategy — compute "none" first, then subtract from one — is especially powerful for "at least one" problems.
- Sec 2.4General addition rule: P(A ∪ B) = P(A) + P(B) − P(A ∩ B). Subtracting the intersection corrects for double-counting. Extends to three or more events via inclusion-exclusion.
- Sec 2.5For mutually exclusive events: P(A ∪ B) = P(A) + P(B). For general events: use the general addition rule. Always check mutual exclusivity before summing probabilities.
- Sec 2.6Mutually exclusive events share no outcomes: A ∩ B = ∅. Mutually exclusive events with positive probability are dependent, not independent — knowing one occurred tells you the other definitely did not.
- Sec 2.7Conditional probability: P(A | B) = P(A ∩ B)/P(B) for P(B) > 0. Conditioning changes the sample space to B. All probability rules apply within the conditioned sample space.
- Sec 2.8Events A and B are independent if P(A ∩ B) = P(A)·P(B), equivalently P(A | B) = P(A). Independence must be verified, not assumed. Mutual independence of n events requires the product rule for every subset.
- Sec 2.9General multiplication rule: P(A ∩ B) = P(A)·P(B | A). Chain rule extends to n events. For independent events: P(A ∩ B) = P(A)·P(B). Sampling without replacement creates dependence; use the general rule.
- Sec 2.10Law of total probability: P(A) = Σᵢ P(A | Bᵢ)·P(Bᵢ) when {Bᵢ} partitions S. Computes an overall probability by averaging conditional probabilities over the partition — a divide-and-conquer strategy.
- Sec 2.11Bayes' theorem: P(Bⱼ | A) = P(A | Bⱼ)·P(Bⱼ) / Σᵢ P(A | Bᵢ)·P(Bᵢ). Inverts conditional probability to find the probability of a cause given an observation. Vocabulary: prior, likelihood, posterior.
- Sec 2.12Tree diagrams organize multi-stage probability calculations. Each branch is labeled with a conditional probability. Path probabilities are products of branch probabilities and must sum to one — use this as a check.
- Sec 2.13Eight key errors: adding without checking exclusivity, confusing exclusivity with independence, reversing conditional probability, ignoring base rates, applying the independent multiplication rule without justification, using conditionals as unconditionals, conditioning on P(B) = 0, and assuming independence from physical separation.
- Sec 2.14Chapter 3 extends these rules to random variables — functions that assign numbers to outcomes. All rules from Chapter 2 carry forward. The fundamental shift: from events as sets to probabilities as functions of numbers.
Key Terms Review
axiom
A foundational statement accepted without proof from which other results are logically derived.
Bayes' theorem
A formula for computing P(Bⱼ | A) from the likelihoods P(A | Bⱼ) and prior probabilities P(Bⱼ). It inverts conditional probability.
base rate
The prior probability of an event in the population before conditioning on any observed evidence.
base rate fallacy
The error of ignoring the prior probability when updating beliefs based on evidence.
complement
Ac: all outcomes in S not in A. P(Ac) = 1 − P(A).
conditional probability
P(A | B) = P(A ∩ B) / P(B). The probability of A given that B has occurred. Conditioning changes the sample space.
De Morgan's laws
(A ∪ B)c = Ac ∩ Bc and (A ∩ B)c = Ac ∪ Bc. Complement distributes across union and intersection by flipping the operation.
disjoint
Another word for mutually exclusive. Two events are disjoint if A ∩ B = ∅.
event
A subset of the sample space. A collection of outcomes sharing some property of interest.
exhaustive
A collection of events is exhaustive if their union equals the full sample space S.
false positive
A test result indicating the presence of a condition that is actually absent.
inclusion-exclusion
The general addition rule P(A ∪ B) = P(A) + P(B) − P(A ∩ B), extended to three or more events by alternating signs at each level.
independence
Events A and B are independent if P(A ∩ B) = P(A)·P(B). Equivalently, P(A | B) = P(A): knowing B does not change the probability of A.
intersection
A ∩ B: the event that both A and B occur.
Kolmogorov axioms
The three axioms — non-negativity, normalization, countable additivity — that define a valid probability measure.
law of total probability
P(A) = Σᵢ P(A | Bᵢ) P(Bᵢ) when {Bᵢ} partitions S. Computes overall probability by averaging over a partition.
likelihood
In Bayes' theorem, P(A | Bⱼ): the probability of observing event A if Bⱼ is the true state.
multiplication rule
P(A ∩ B) = P(A) · P(B | A). For independent events: P(A ∩ B) = P(A) · P(B).
mutually exclusive
A ∩ B = ∅: events A and B cannot both occur in the same trial. With positive probabilities, they are dependent, not independent.
mutually independent
Events A₁, …, Aₙ are mutually independent if the product rule P(Aᵢ₁ ∩ ··· ∩ Aᵢₖ) = P(Aᵢ₁) ··· P(Aᵢₖ) holds for every subset.
partition
A collection of events that are mutually exclusive, exhaustive, and each have positive probability.
posterior probability
P(Bⱼ | A): the updated probability of Bⱼ after observing event A.
prior probability
P(Bⱼ): the probability of Bⱼ before observing any evidence.
probability measure
A function P from events to [0,1] satisfying the three Kolmogorov axioms.
sensitivity
In testing, P(positive | condition present): the probability the test correctly identifies a true case.
specificity
In testing, P(negative | condition absent): the probability the test correctly identifies a true negative.
transposition fallacy
Incorrectly treating P(A | B) as equal to P(B | A). Also called the prosecutor's fallacy.
tree diagram
A branching diagram organizing the probabilities of a multi-stage experiment. Path probabilities = products of branch probabilities; all paths sum to 1.
union
A ∪ B: the event that A or B or both occur — "at least one."
Probability Rule Practice
- A fair die is rolled. Let A = "result is odd" and B = "result is a multiple of 3." (a) List outcomes in A, B, A ∪ B, A ∩ B, Ac, Bc. (b) Compute P(A ∪ B) using the general addition rule. (c) Verify using the complement of Ac ∩ Bc.
- A component fails in its first year with probability 0.15. Find: (a) P(does not fail); (b) P(at least one of three independent components fails). Use the complement for (b).
- Prove from the axioms that 0 ≤ P(A) ≤ 1 for all events A.
- Events A, B, C partition S with P(A) = 0.35, P(B) = 0.45. Find P(C) and P(A ∪ B).
- A survey of 500 people records handedness and dominant eye. Right-handed/right-eyed: 300; right-handed/left-eyed: 100; left-handed/right-eyed: 60; left-handed/left-eyed: 40. Find: (a) P(right-handed); (b) P(right-eyed | right-handed); (c) P(left-handed | left-eyed); (d) P(right-eyed | left-handed).
- A biased coin has P(H) = 0.6. Tossed twice. Given the result was not two tails, find P(two heads).
- Let P(A) = 0.4, P(B) = 0.5, P(A ∩ B) = 0.2. Find P(A | B), P(B | A), P(A | Bc), P(Ac | Bc).
- A card is drawn from a standard deck. Let A = "the card is a club" and B = "the card is a face card." Are A and B independent? Show using the definition.
- Events A, B, C are mutually independent with P(A) = 0.5, P(B) = 0.4, P(C) = 0.3. Find: (a) P(A ∩ B ∩ C); (b) P(A ∪ B ∪ C) using inclusion-exclusion; (c) verify (b) using the complement.
- A system has two subsystems in parallel (works if at least one subsystem works). Each subsystem has three components in series (works only if all components work). Each component works with probability 0.9 independently. Find P(system works).
- A medical test has sensitivity 0.98 and specificity 0.95. The condition affects 0.5% of the population. (a) Find P(positive test). (b) Find P(condition | positive test). (c) Find P(no condition | negative test). (d) Interpret your results.
- Three factories: A (50%, defect rate 2%), B (30%, defect rate 4%), C (20%, defect rate 6%). (a) Find overall defect rate. (b) Given a defective part, find P(came from each factory). (c) Which factory is most likely source of a randomly chosen defective part?
- A bag contains two coin types: 70% type I (P(heads) = 0.6) and 30% type II (P(heads) = 0.3). A coin is selected and flipped. (a) P(heads). (b) Given heads, P(type I). (c) The coin is flipped again and comes up heads a second time. Update P(type I).
Applications
- Genetics. Each parent (genotype Dr) independently passes one allele to an offspring. (a) List all possible offspring genotypes. (b) Find P(at least one dominant allele). (c) Find P(genotype DD | shows dominant phenotype, i.e., has at least one D).
- Communications. A binary channel transmits 0s and 1s, flipping each bit with probability 0.05 independently. (a) P(single bit received correctly). (b) P(entire 8-bit message received without error). (c) P(at least 7 of 8 bits correct).
- Finance. Three stocks rise on any given day with probabilities 0.6, 0.55, 0.5 independently. (a) P(all three rise). (b) P(none falls). (c) P(at least two rise).
- Epidemiology. A disease is transmitted with probability 0.3 per contact. A healthy person contacts 3 independent infected people. (a) P(not infected by any contact). (b) P(person becomes infected).
- Law. The probability of observing evidence as extreme as presented, if the defendant is innocent, is 0.01. Suppose 0.1% of suspects for this type of crime are actually guilty. Apply Bayes' theorem to find P(guilty | evidence). Discuss the base rate problem in this context.
Error Analysis
Each problem shows a student's work. Identify the error and write the correct solution.
Problem 1
"Both aces" without replacement: P = 4/52 · 4/52 = 16/2704
What is the error? What is the correct answer for P(both cards are aces) when drawing without replacement?
Problem 2
"A and B can't both happen, therefore A and B are independent."
Identify the conceptual error. Explain why mutually exclusive events with positive probability cannot be independent.
Problem 3
P(at least one six in 4 rolls) = 4 × 1/6 = 4/6 = 2/3
What is wrong with this calculation? Find the correct probability using the complement.
Problem 4
"Test sensitivity is 95%; I tested positive; therefore I have a 95% chance of having the disease."
Name the fallacy. Explain what information is missing. Describe how Bayes' theorem gives the correct answer.
Problem 5
P(A ∪ B) = P(A) + P(B) − P(A) · P(B) = 0.4 + 0.6 − 0.24 = 0.76 [where P(A ∩ B) = 0.3, not 0.24]
Under what conditions is P(A ∪ B) = P(A) + P(B) − P(A)·P(B) correct? What was wrong here, and what is the correct answer given P(A ∩ B) = 0.3?
Chapter 2 Checkpoint
Chapter 2 Checkpoint
Complete without assistance — assesses all key skills and concepts from Chapter 2
Section 1 — Vocabulary and Concepts
- State the three Kolmogorov axioms. Explain in one sentence why each one is necessary.
- Distinguish mutually exclusive from independent. Give an example of each and explain why one is not the other.
- Explain in your own words what it means to say that conditioning changes the sample space. Give a concrete example.
- State the law of total probability and Bayes' theorem. Describe one real-world scenario where each applies.
Section 2 — Skills
- A fair die is rolled. Let A = "result is prime" = {2,3,5} and B = "result is odd" = {1,3,5}. Find P(A ∪ B), P(A ∩ B), P(Ac ∩ B). Are A and B mutually exclusive? Are they independent? Find P(A | B) and P(B | A).
- A bag contains 5 red and 7 blue chips. Two chips are drawn without replacement. Draw a tree diagram. Find P(both same color) and P(second is blue | first is red).
- Let P(A) = 0.5, P(B) = 0.4, P(A ∪ B) = 0.7. Find P(A ∩ B), P(A | B), P(B | A). Are A and B independent?
- A fair coin is tossed five times. Find P(at least one tail) using the complement. Show your work.
Section 3 — Total Probability and Bayes' Theorem
- Supplier X provides 60% of components (3% defect rate). Supplier Y provides 40% (6% defect rate). (a) Find P(component is defective). (b) A defective component is found — find P(came from Supplier Y).
- In a city, 40% of voters are Party A (60% turnout), 35% Party B (50% turnout), 25% Independent (40% turnout). (a) Find overall voter turnout. (b) A voter is selected at random from those who voted — find P(registered with Party A).
Section 4 — Applications
- Three mutually independent events A, B, C have P(A) = 0.3, P(B) = 0.4, P(C) = 0.5. Find P(A ∩ B ∩ C), P(A ∪ B ∪ C) by inclusion-exclusion, and verify using the complement P(Ac ∩ Bc ∩ Cc).
- A communications channel transmits each bit correctly with probability 0.97 (errors are independent). A message of 10 bits is sent. Find P(no errors) and P(at least one error).
Section 5 — Error Analysis
- A student computes P(heart or face card) = 13/52 + 12/52 = 25/52. What is the error? What is the correct answer?
- A student says P(A | B) = P(A)/P(B) and argues this is the formula for conditional probability. Identify the error and write the correct formula.
- Write a brief explanation (3–5 sentences) of why the base rate matters when interpreting a positive test result. Use a numerical example to support your explanation.
Looking Ahead
Chapter 3: Discrete Random Variables
Chapter 2 gave you the logical structure of probability: how to combine events, condition on information, verify independence, and apply Bayes' theorem. Chapter 3 introduces a fundamentally new object: the random variable. Rather than labeling outcomes as sets, a random variable assigns a real number to each outcome. This allows you to use the full machinery of algebra and eventually calculus — computing averages as weighted sums, measuring spread through variance, and studying named distributions such as the Bernoulli, binomial, geometric, and Poisson. All the probability rules from this chapter carry forward directly, now operating on the probabilities that numbers take specific values. The fundamental shift in Chapter 3: from events as sets to probabilities as functions of numbers — and everything that follows in this course builds on that foundation.
MGU Library Connection
- Probability Rules Reference — Appendix E of this textbook
- Set and Event Notation Reference — Appendix D of this textbook
- Chapter 1: What Probability and Statistics Study — foundational sample space and event definitions
- Chapter 10: Likelihood, Maximum Likelihood, and Bayesian Foundations — develops Bayes' theorem into a full statistical framework
- MGU Dictionary entries: conditional probability, independence, Bayes' theorem, partition, mutually exclusive