STOR 435 Midterm 1 Study Guide

Introduction to Probability · UNC Fall 2026 · Section 002 (Bhamidi) · Covers Lectures 1–4 and Homework 1–4

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0. Exam logistics and how to use this guide

What is in scope

The information sheet says the exam covers "material from Lectures 1–4, together with the corresponding homework, worksheets, and prerequisite ideas used in those lectures." You should "be able to recognize which tools apply, carry out the calculation, and interpret the answer." Homework 1 is explicitly part of the syllabus, "especially Problems 1–9 and 14," because they cover prerequisite material that does not appear on the slides.

How to study with this guide

• Each lecture chapter follows the slide deck in order, expands the terse slides, works every board-work problem the slides posed, and ends with "Check yourself" questions with hidden answers.
• Boxes: blue "Key idea" boxes hold definitions and formulas quoted from the slides (memorise these: there is no formula sheet). Orange "Watch out" boxes flag the mistakes that cost points on the homework and the practice bank. Green "Exam tip" boxes say how a multiple-choice question on that topic tends to be phrased.
• Homework problems are reworked inside the relevant chapter, including the two that were marked wrong on the graded copies (HW2 Q9 and HW3 Q9). Those are exactly the kind of slip a multiple-choice exam punishes.
• The practice exam at the end is a mixed set in the style of the practice bank: multiple-choice, all four lectures shuffled, with the "which correction is appropriate" audit style the instructor likes.
• The cram sheet at the very end is one screen of formulas and traps. Read it last.

HW11. Prerequisites the slides assume

Homework 1 is on the exam syllabus "especially Problems 1–9 and 14" because Lectures 1–4 use these tools without stopping to teach them. Each problem below is reworked with the reasoning the course wants, and the "why this matters" line from the homework is kept because it tells you where the tool reappears.

1.1 Set difference

HW1 Q2. \(A=\{1,2,3,4\}\), \(B=\{3,4,5\}\). Then \(A\setminus B=\{1,2\}\): remove from \(A\) whatever also appears in \(B\). Note the 5 in \(B\) is irrelevant. Lecture 2 uses \(E\setminus F\) to translate phrases like "the second ball is black but the third is not."

1.2 Summation notation and finite sums

HW1 Q3. \(\sum_{k=2}^{4}(2k-1)\) expands term by term: \(k=2\) gives 3, \(k=3\) gives 5, \(k=4\) gives 7, so it equals \(3+5+7\). The index runs from the lower limit to the upper limit inclusive.

HW1 Q4. \(\sum_{j=1}^{10}(3j-2)=3\sum_{j=1}^{10}j-\sum_{j=1}^{10}2=3\cdot 55-20=145\), using \(\sum_{j=1}^{n}j=n(n+1)/2\) and the fact that a constant summed \(n\) times is \(n\) times the constant.

HW1 Q5 (double sum). \(\sum_{i=1}^{3}\sum_{j=0}^{2}(i+2j)\). Do the inner sum for fixed \(i\): \(\sum_{j=0}^{2}(i+2j)=3i+2(0+1+2)=3i+6\). Then the outer sum: \(\sum_{i=1}^{3}(3i+6)=3\cdot 6+18=36\). Lecture 3 sums point masses over the outcomes in an event; later lectures sum over one coordinate to get a marginal.

1.3 Binomial coefficient as an algebraic coefficient

HW1 Q6. Coefficient of \(x^3y^4\) in \((x+y)^7\) is \(\binom{7}{3}=\binom{7}{4}=35\).

1.4 Geometric series

HW1 Q7. \(\sum_{k=0}^{5}3\left(\tfrac12\right)^k=3\cdot\dfrac{1-(1/2)^6}{1-1/2}=3\cdot 2\cdot\dfrac{63}{64}=\dfrac{378}{64}=5.90625\).

HW1 Q8. \(\sum_{k=0}^{\infty}4\left(\tfrac13\right)^k\) converges because \(|1/3|\lt 1\), to \(\dfrac{4}{1-1/3}=\dfrac{4}{2/3}=6\). "Infinitely many positive terms" does not by itself force divergence. Lecture 6 will use this to check that infinitely many probabilities add to 1.

1.5 Exponential series and limit

HW1 Q9. \(\sum_{k=0}^{\infty}\dfrac{(-2)^k}{k!}=e^{-2}\approx 0.1353\).

HW1 Q10. \(\lim_{n\to\infty}(1+5/n)^n=e^{5}\). (Q10 is not in the "especially" list, but it is one line.)

1.6 Normalising a density

HW1 Q14. For which \(c\) does \(f(x)=cx(1-x)\) on \([0,1]\) (and 0 elsewhere) integrate to 1? Compute \(\int_0^1 x(1-x)\,dx=\int_0^1 (x-x^2)\,dx=\tfrac12-\tfrac13=\tfrac16\), so \(c\cdot\tfrac16=1\) and \(c=6\). The logic is identical to the discrete case in Lecture 3: point masses (or a density) must add (integrate) to exactly 1 before they define a model. HW3 Q16 is the discrete twin: \(\P(G)=c,\P(Y)=2c,\P(O)=3c,\P(R)=4c\) forces \(10c=1\).

Extra context beyond the homework: Q11–13 (definite integral \(=12\), Riemann sum \(=\int_0^1x^2\,dx\), double integral \(=0.5\)) and Q15 (committee with Ana and Bo forced in: \(\binom{6}{3}=20\)) are not in the "especially" list; Q15 is really a Lecture 1 problem and reappears in §2.

1.7 Check yourself

L012. Counting and combinatorics

Lecture 1 spends its first half on motivation (mammograms, Sally Clark, the O. J. Simpson trial, the course roadmap) and its second half on the counting toolkit. The motivation is context: the exam-relevant content starts at the "Basic principle of counting" slide. The slides' own summary of the goals: "describe what one outcome looks like and decide whether order or repetition matters; use the multiplication principle and distinguish permutations from combinations; count the outcomes in a finite set needed for a probability calculation; check a counting answer by working out a small case."

2.1 Why the motivation slides still matter

Three stories from the deck recur in later lectures and in the instructor's phrasing. Know what each one is about, not the numbers.

• Mammogram (Gigerenzer). Base rate 0.8%, sensitivity 90%, false-positive rate 7%. Doctors guessed 90% or 50–80% for \(\P(\text{cancer}\mid+)\); the correct value is about 9%. Lesson: "Probability separates what a test reports from what we want to know." The calculation itself is Bayes and is not on this exam.
• Sally Clark. A paediatrician multiplied \(1/8500\times 1/8500\) for two SIDS deaths. The deck's question: "Why should the two deaths be independent of each other?" Lesson: multiplying probabilities assumes independence, which must be justified.
• O. J. Simpson / Dershowitz. "Fewer than 1 homicide per 2500 incidents of abuse" was presented as evidence that battery is not a predictor of murder. The deck notes that a "more representative" calculation (conditioning on the woman having been murdered) gives about 88%. Lesson: "A numerical argument can look precise while relying on the wrong event."

Extra context beyond the slides: all three are really conditional-probability stories, which is why they are worth remembering when you reach Lecture 4's "check the direction of the bar."

2.2 The basic principle of counting

The condition "for every Stage 1 outcome" is the whole point. The number of second-stage choices must not depend on which first-stage choice was made, only on how many were made. That is what lets you multiply.

Bridge example (slide 29). Roll a die twice: \(6\cdot 6=36\) ordered outcomes. Exactly six have sum 7: \((1,6),(2,5),(3,4),(4,3),(5,2),(6,1)\). If all 36 are equally likely, \(\P(\text{sum }7)=6/36=1/6\). Next lecture: the 36 pairs are a sample space, the six pairs an event.

2.3 The board-work examples, worked

2.4 Permutations

Arranging \(r\) of \(n\) distinct objects in order (no repetition) is \(P(n,r)=n(n-1)\cdots(n-r+1)=\dfrac{n!}{(n-r)!}\). The practice bank uses the notation \(P(8,2)=56\), \(P(12,3)=1320\). Convention: \(0!=1\).

Extra context beyond the slides: your calculator has \(n!\) and usually nPr / nCr keys; make sure you know where they are, since you may not use a phone.

2.5 Combinations

Derivation the board work gives: ordered selections of \(r\) from \(n\) number \(n(n-1)\cdots(n-r+1)\), and each unordered group has been counted \(r!\) times (once per ordering), so \[\binom{n}{r}=\frac{n(n-1)\cdots(n-r+1)}{r!}=\frac{n!}{r!\,(n-r)!}.\] Useful identities: \(\binom{n}{r}=\binom{n}{n-r}\); \(\binom{n}{1}=n\); \(\sum_{r=0}^{n}\binom{n}{r}=2^n\) (put \(x=y=1\) in the binomial theorem, and compare with the subsets count).

2.6 The binomial theorem

Why it is true: expanding \((x+y)^n\) means picking \(x\) or \(y\) from each of the \(n\) factors; the term \(x^ry^{n-r}\) arises once for each choice of which \(r\) factors supply the \(x\). See §1.3 for the scalar trap.

2.7 Multinomial coefficients

The slides give two interpretations. You must recognise both from wording.

Interpretation 1: labelled groups

"Number of ways to divide \(n\) distinct objects into \(r\) distinct groups of sizes \(n_1,\dots,n_r\)." Objects \(a,b,c,d\) into groups of sizes 1, 1, 2: \(\{a\}\{b\}\{c,d\}\), \(\{b\}\{a\}\{c,d\}\), … 12 in all. The groups are labelled (Folder A vs Folder B), so \(\{a\}\{b\}\) and \(\{b\}\{a\}\) are different. Derivation: \(\binom{n}{n_1}\binom{n-n_1}{n_2}\cdots\), which telescopes to the formula.

Interpretation 2: arrangements with repeated objects

"Number of ordered arrangements of \(n\) objects of which \(n_1\) are alike, \(n_2\) are alike, …, \(n_r\) are alike." \(A,A,A,B,C,C\): \(\binom{6}{3,1,2}=60\) arrangements. Derivation: \(6!\) arrangements of six labelled letters, divided by \(3!\) for the interchangeable A's and \(2!\) for the C's.

Worked (HW2 Q10, bank Q8). BANANAS has 7 letters: A×3, N×2, B, S. Arrangements \(=\frac{7!}{3!\,2!\,1!\,1!}=\frac{5040}{12}=420\). BALLOON: L×2, O×2: \(\frac{7!}{2!2!}=1260\).

Worked (HW2 Q12, bank Q7). Nine distinct samples into labelled freezers A (4), B (3), C (2), no order within a freezer: \(\binom{9}{4,3,2}=\frac{9!}{4!3!2!}=\frac{362880}{288}=1260\). Same number as the folders problem \(\binom{9}{2,3,4}\), since the denominator is symmetric.

2.8 The counting decision guide

Diagnose three counting errors (slide 50).

1. "Three different letters are arranged in order from the alphabet: \(26^3\)." Wrong because \(26^3\) allows repeats. One outcome is an ordered triple of distinct letters: \(26\cdot 25\cdot 24=15{,}600\).
2. "Three students are arranged in order from a class of 10: \(10!\)." Wrong because \(10!\) arranges all ten. Only three positions are filled: \(10\cdot 9\cdot 8=720\).
3. "A three-student committee from a class of 10: \(10\cdot 9\cdot 8\)." Wrong because a committee is unordered; each was counted \(3!\) times: \(\binom{10}{3}=120\).

Constraint techniques.

• Complement. "At least one librarian" among four chosen from 5 librarians + 6 archivists (bank Q5): all panels minus no-librarian panels, \(\binom{11}{4}-\binom{6}{4}=330-15=315\). HW4 Q8 is the probability version: \(1-\binom{7}{3}/\binom{11}{3}=1-35/165=26/33\).
• Cases. Pizza with exactly 4 of 8 toppings, never Pepperoni and Pineapple together (bank Q55): neither \(\binom{6}{4}=15\), exactly one \(2\binom{6}{3}=40\); total 55. Party of 5 from 8 friends, two feuding (bank Q53): \(\binom{6}{5}+2\binom{6}{4}=6+30=36\).
• Forced members. Ana and Bo must both serve on a 5-person committee from 8 (HW1 Q15): choose the other 3 from the remaining 6, \(\binom{6}{3}=20\).
• Two-stage selection. Pick a 4-person team from 11, then a lead from the team (bank Q4): \(\binom{11}{4}\cdot 4=1320\). Contrast with 4 ordered roles: \(11\cdot 10\cdot 9\cdot 8=7920\).
• Block trick for adjacency. Tiles R,R,R,B,B,G,Y with the B's not adjacent (bank Q24): all \(7!/(3!2!)=420\), adjacent (glue BB into one object) \(6!/3!=120\), so \(300\).
• Disjoint patterns add. Lineup FFSSS or FFFSS from 8 faculty and 12 students (bank Q22): \(P(8,2)P(12,3)+P(8,3)P(12,2)=73920+44352=118272\).
• Subsets with a forbidden pair. Subsets of 10 students not containing both Brad and John (bank Q50): \(2^{10}-2^{8}=1024-256=768\).

2.9 Counting becomes probability

Capstone (slides 51–52). A committee of five is selected uniformly from twelve students: seven juniors, five seniors. \(\P(\text{exactly two seniors})\)? "Describe the equally likely outcomes, count all outcomes, count the favorable outcomes, and take their ratio."

HW2 Q13. Three-person panel from 5 statistics + 4 CS majors; \(A\) = both majors present. \(|S|=\binom{9}{3}=84\). Complement: all-stats \(\binom{5}{3}=10\) plus all-CS \(\binom{4}{3}=4\), so \(|A|=84-14=70\) and \(\P(A)=70/84=0.8333\).

Bank Q10. Four from 5 biology + 4 physics, exactly 2 biology: \(\binom{5}{2}\binom{4}{2}/\binom{9}{4}=60/126=10/21\).

2.10 Extra reading (not examined)

The deck ends with token sequences (\(V^T\) strings from a vocabulary of \(V\) over \(T\) positions, e.g. \(5^4=625\)) and CRISPR guide sequences (\(4^{20}\approx 1.1\times 10^{12}\)). Both are the product rule again; the deck's point is that "possible does not mean equally likely." The information sheet excludes Extra Reading, so this is background only.

2.11 Check yourself

L023. Sample spaces and events

Lecture 2's goal statement: "choose a sample space at the right level of detail, express events as subsets, translate words into event algebra, and check whether a model can answer the question asked." It was taught over two meetings: outcomes, granularity and events first; then event operations, laws, translation and model audits.

3.1 Intuition can mislead: three opening lessons

• Conjunction fallacy (Linda). "Linda is a bank teller and active in the feminist movement" cannot be more likely than "Linda is a bank teller."
  Check · Event inclusion"For any events \(E\) and \(F\), \(E\cap F\subseteq E\). A plausible narrative cannot reverse this inclusion."
• Random-looking patterns. For six fair tosses, \(HTHTTH\) and \(HHHTTT\) are two individual outcomes of the same sample space; "neither becomes less likely merely because it looks more clustered." The modelling question is whether you are comparing individual sequences or larger events.
• Coloured-die patterns. A die with four green and two red faces is rolled 20 times; you win if your pattern appears consecutively: (1) \(RGRRR\), (2) \(GRGRRR\), (3) \(GRRRRR\). Debrief: if \(E\) = "\(GRGRRR\) appears at a given position" and \(F\) = "\(RGRRR\) appears one roll later," then \(E\subset F\), because the last five symbols of the longer pattern are the shorter pattern. So pattern 1 is the best choice and "a pattern may look more representative of randomness without being more likely. Define the outcomes and compare the corresponding events before trusting appearance."

HW3 Q12 is the inclusion lesson in clinic clothing: adding a second condition cannot make \(E\cap F\) contain visits outside \(E\); the correct audit is "\(E\cap F\subseteq E\) and \(E\cap F\subseteq F\); every visit in the intersection satisfies both conditions." HW3 Q13: the single sequence \(HTHTHT\) is one outcome; "exactly three H's" contains \(\binom{6}{3}=20\) outcomes, so \(B\) is 20 times larger than \(A\).

3.2 The five-step modelling template

Two dice (slides 10–11). \(S_{\text{pair}}=\{(i,j):i,j\in\{1,\dots,6\}\}\) versus \(S_{\text{sum}}=\{2,\dots,12\}\). Can each represent (1) the sum is 8, (2) blue exceeds white, (3) the dice match? \(S_{\text{pair}}\) "can represent sums, comparisons, and matches." \(S_{\text{sum}}\) "can represent sum questions but loses which die produced which value." The map \((i,j)\mapsto i+j\) "combines many detailed outcomes into one summary. That compression is useful only when no later question needs the discarded information."

Audit a flawed model (slide 25). With \(S=\{2,\dots,12\}\), the event "first die exceeds the second" is not a subset of \(S\): outcomes \((6,2)\) and \((2,6)\) both have sum 8 but disagree about the event. Repair with the smallest natural change: record ordered pairs.

3.3 Three sizes of sample space

Board work (slide 12): specify \(S\) and classify it.

HW3 Q3 and bank Q12 test the "countably infinite" recognition: "number of login attempts until the first success" and "day number of the first alert, with no stated end" are \(\{1,2,3,\dots\}\). A lifetime is continuous; three coin tosses or a finishing order is finite.

Stopping rules (bank Q21). A client records S or F per attempt and stops after the first success or the third attempt. The complete records are \(\{S, FS, FFS, FFF\}\). Strings like \(F\) or \(FF\) are unfinished, \(SSS\) continues past a success, \(FFFF\) exceeds the limit. "These records need not be equally likely."

3.4 Events are subsets

Two ordered tosses, \(S=\{HH,HT,TH,TT\}\): \(E=\{\text{both agree}\}=\{HH,TT\}\), \(F=\{\text{at least one head}\}=\{HH,HT,TH\}\). HW3 Q2 (sensor RR/RB/BR/BB): "first is red" \(\cap\) "reports agree" \(=\{RR\}\).

3.5 Union, intersection, complement, difference

Union \(E\cup F\)

"outcomes in \(E\), in \(F\), or in both." Means "at least one occurs." "The word or is inclusive unless a problem explicitly says otherwise." With \(E=\{\text{first toss }H\}=\{HH,HT\}\) and \(F=\{\text{tosses differ}\}=\{HT,TH\}\): \(E\cup F=\{HH,HT,TH\}\).

Intersection \(E\cap F\)

"outcomes in both \(E\) and \(F\)." Here \(E\cap F=\{HT\}\). "\(E\) and \(F\)" always means intersection.

Disjoint (mutually exclusive)

\(E\cap F=\emptyset\). "Disjointness says the two events cannot occur together; it does not say that either event must occur."

Complement \(E^c\)

\(E^c=S\setminus E\), the outcomes not in \(E\). For \(E=\{\text{first toss }H\}\), \(E^c=\{TH,TT\}\). Also \(S^c=\emptyset\), \(\emptyset^c=S\).

Difference \(E\setminus F\)

\(E\cap F^c\): \(E\) occurs and \(F\) does not.

Inclusion \(E\subseteq F\)

Every outcome of \(E\) is in \(F\), so "\(E\) occurs ⟹ \(F\) occurs." \(E=F\) exactly when \(E\subseteq F\) and \(F\subseteq E\).

Countable operations (slide 20). \(\bigcup_{n\ge 1}E_n\) = "outcomes in at least one \(E_n\)"; \(\bigcap_{n\ge 1}E_n\) = "outcomes in every \(E_n\)." Translation cue: "Union corresponds to 'at least one index'; intersection corresponds to 'every index.'"

3.6 Three families of laws, and De Morgan

Superheroes (slide 23). \(E\) = Superman saves Metropolis, \(F\) = Batman saves Gotham. \(E\cup F\) = "at least one city is saved"; \((E\cup F)^c\) = "neither city is saved" = "Superman fails and Batman fails" \(=E^c\cap F^c\). Likewise \((E\cap F)^c\) = "not both saved" = "at least one of them fails" \(=E^c\cup F^c\).

Nested De Morgan (bank Q17). \([A\cup(B\cap C)]^c=A^c\cap(B\cap C)^c=A^c\cap(B^c\cup C^c)\). Apply the law to the outer operation first, then to the inner one.

Three events (bank Q54). \((A\cup B\cup C)^c=A^c\cap B^c\cap C^c\). The bank's correct option was written as a union of three copies of that same triple intersection with different bracketing; associativity makes them identical, so their union is still \(A^c\cap B^c\cap C^c\). Option "\(A^c\cup B^c\cup C^c\)" is the complement of the intersection, not the union.

3.7 Words ⟷ symbols

Board work (slide 24), with the standard answers. These are the phrasings that appear on every homework and in the bank.

3.8 Synthesis: coloured balls without replacement

Slide 26: five black and two red balls; draw three without replacement, recording only the ordered colour sequence.

1. Template. Experiment: three ordered draws. \(S\) = colour strings of length 3 with at most two R's: \(\{BBB,BBR,BRB,RBB,BRR,RBR,RRB\}\), seven outcomes. Key assumption: without replacement, so \(RRR\) is impossible (only two reds exist). The seven outcomes are not equally likely.
2. Events. \(E\) = second ball black \(=\{BBB,BBR,RBB,RBR\}\). \(F\) = third black, so \(F^c\) = third red \(=\{BBR,BRR,RBR\}\). \(E\cap F^c=\{BBR,RBR\}\).
3. Words. \(E\cap F^c\) = "second ball black and third ball red." Check each listed outcome: \(BBR\) (second B, third R ✓), \(RBR\) (second B, third R ✓).

3.9 Simplify an event expression

Slide 28: simplify \((A\cup B)\cap(A^c\cup B)\cap(A\cup B^c)\).

By laws. Distributivity in the form \((X\cup Z)\cap(Y\cup Z)=(X\cap Y)\cup Z\) gives \((A\cup B)\cap(A^c\cup B)=(A\cap A^c)\cup B=\emptyset\cup B=B\). Then \(B\cap(A\cup B^c)=(B\cap A)\cup(B\cap B^c)=(A\cap B)\cup\emptyset=A\cap B\).

By membership. An outcome in \(A\cap B\) is in all three brackets. An outcome in \(A\setminus B\) fails \(A^c\cup B\). One in \(B\setminus A\) fails \(A\cup B^c\). One in neither fails \(A\cup B\). So the expression is exactly \(A\cap B\).

Bank Q19 is the same technique: \((R\cap S)\cup S^c=(R\cup S^c)\cap(S\cup S^c)=(R\cup S^c)\cap S=R\cup S^c\) (with \(S\) here standing for the sample space in the last step). Bank Q49: \([(A\cup B)\cap(A\cup C)\cap(B^c\cap C^c)]\): on the region \(B^c\cap C^c\), both \(A\cup B\) and \(A\cup C\) reduce to \(A\), so the event is \(A\cap B^c\cap C^c\).

3.10 Verify the model

Extra reading in this deck (threat circuitry, clusters and John Snow, AI-benchmark coordinates, extreme-heat profiles) is excluded from the exam. Its single transferable lesson is the granularity principle again: a daily-maximum record can represent "max \(\ge u\)" but not "three consecutive hours \(\ge u\)."

3.11 Check yourself

L034. Axioms of probability and their consequences

Lecture 3's goals: "audit a probability assignment against the axioms, derive useful probability identities by splitting events into disjoint pieces, and combine exact calculation with independent checks." The opening question: for \(S=\{H,T\}\), can we assign \(\P(H)=0.60\) and \(\P(T)=0.60\)? "Each number looks plausible by itself. The problem appears only when the assignments are required to describe one coherent model."

4.1 Frequency is evidence, not the definition

Agreement between the two supports the model; disagreement says to examine the assumptions, the data, or both. Some questions (a future stock price, one person's accident risk) need a model even though the experiment cannot be repeated. HW3 Q15's wrong option "cannot be audited until observed login frequencies are available" is exactly this confusion: validity against the axioms is checked before any data.

4.2 The three axioms

Kolmogorov's 1933 monograph gave these axioms. A probability space is \((S,\text{events},\P)\): a sample space, an allowed collection of events, and an assignment satisfying the axioms. "For finite models in this course, every subset of \(S\) may be treated as an event." Three roles: "Assumptions choose \(S\), the events, and the point masses ⟶ Axioms constrain the assignment ⟶ Theorems derive probabilities of new events."

4.3 First consequences

The symbol \(\dot\cup\) means "union of disjoint sets" and is the instructor's cue that additivity applies.

4.4 Auditing an assignment

Coin audit (slide 8). \(S=\{H,T\}\), \(0\lt p\lt 1\).

1. \(\P(H)=p,\ \P(T)=1-p\): nonnegative, sums to 1. Valid.
2. \(\P(H)=p,\ \P(T)=1-p^2\): nonnegative, but the sum is \(1+p-p^2\gt 1\) for \(0\lt p\lt 1\). Fails normalisation (via finite additivity). Invalid.
3. \(\P(H)=1.10,\ \P(T)=-0.10\): fails nonnegativity at \(\P(T)\). Invalid. (The sum happens to be 1; that does not rescue it.)

HW3 Q15. \(\P(N)=0.52,\P(R)=0.31,\P(L)=0.22\): "invalid because the three disjoint singleton probabilities sum to 1.05, contradicting normalization and finite additivity." Bank Q26 is the same with four states summing to 1.05. The distractor "a valid model must assign the same probability to all outcomes" is false: equal likelihood is an assumption, not an axiom.

4.5 Finite models are built from point masses

Unfair die (slide 10). \(\P(1)=\P(2)=\P(3)=\tfrac14\), \(\P(4)=\P(5)=\P(6)=\tfrac1{12}\). (1) Valid: all nonnegative and \(3\cdot\tfrac14+3\cdot\tfrac1{12}=\tfrac34+\tfrac14=1\). (2) \(\P(\text{even})=\P(2)+\P(4)+\P(6)=\tfrac14+\tfrac1{12}+\tfrac1{12}=\tfrac{5}{12}\). (3) Quick bound: it must lie between \(\P(2)=\tfrac14\) and \(1-\P(1)-\P(3)=\tfrac12\). ✓

HW3 Q16. \(\P(G)=c,\P(Y)=2c,\P(O)=3c,\P(R)=4c\). Normalisation: \(10c=1\), \(c=0.1\). \(\P(\{Y,R\})=0.2+0.4=0.6\). Bank Q27: \(\P(\{D,O\})=0.28+0.25=0.53\), because singletons are disjoint.

The simulation slide's workflow: "Predict using your board calculation → Simulate repeated rolls → Compare the empirical proportion with your exact value → Explain ordinary variation or a model mismatch." A finite empirical proportion need not equal the exact value.

4.6 Equally likely outcomes

4.7 Sampling two people: three models that agree (or nearly)

900 adults: 600 Democrats, 300 Republicans. Two names selected uniformly. Event: different party labels.

• Unordered, without replacement (slide 13). \(S\) = all pairs of distinct adults, \(|S|=\binom{900}{2}=404{,}550\), each equally likely. Mixed pairs: \(600\cdot 300=180{,}000\). \(\P=180000/404550\approx 0.4449\).
• Ordered, without replacement (slide 14). \(|S|=900\cdot 899\), each ordered pair probability \(1/(900\cdot 899)\). Mixed: \(600\cdot 300+300\cdot 600=360{,}000\). \(\P=360000/809100\approx 0.4449\). It must agree because each unordered pair corresponds to exactly two ordered pairs, in both numerator and denominator.
• With replacement (slide 15). \(S_{\text{with}}=\{(u,v)\}\), \(\P((u,v))=1/900^2\); repeated names now possible. Mixed: \(360000/810000=0.4444\). Slightly smaller: the first draw no longer removes a person of its own party.

HW4 Q4 (route cards 1–8, two draws). Model W (with replacement): \(|S|=64\), \(\P(\text{match})=8/64=1/8\). Model N (without): \(|S|=56\), \(\P(\text{match})=0\). HW4 Q3: 5 high-capacity of 12 packs, exactly one high in two draws: without replacement \(\binom51\binom71/\binom{12}{2}=35/66=0.5303\); with replacement \(2\cdot\tfrac{5}{12}\cdot\tfrac{7}{12}=0.4861\); difference \(0.0442\). Bank Q29: 3 blue, 2 gold; one of each: without \(\tfrac35\cdot\tfrac24+\tfrac25\cdot\tfrac34=\tfrac35\); with \(2\cdot\tfrac35\cdot\tfrac25=\tfrac{12}{25}\).

4.8 One decomposition drives every identity

Bank Q30. \(A\subseteq B\), \(\P(A)=0.57\), \(\P(B^c)=0.18\). Then \(\P(B)=0.82\) and \(\P(B\cap A^c)=0.82-0.57=0.25\).

4.9 Two-event inclusion–exclusion

Campus services (slide 20). 80% used Recreation or a Union program; 60% Recreation, 25% Union. Overlap \(=0.60+0.25-0.80=0.05\). Upper bound before calculating: overlap \(\le\min(0.60,0.25)=0.25\). Union check: \(0.80\) lies between \(\max(0.60,0.25)=0.60\) and \(\min(1,0.85)=0.85\). ✓

Bank Q32. \(\P(R)=0.72,\P(U)=0.41\): \(\P(R\cup U)\in[0.72,1]\), so among the offered values only 0.88 is feasible (overlap 0.25).

"Exactly one" of two events. The union is the disjoint union of the exactly-one region and the overlap, so \[\P(\text{exactly one})=\P(E\cup F)-\P(E\cap F)=\P(E)+\P(F)-2\P(E\cap F).\] HW4 Q5: \(\P(B)=0.62,\P(S)=0.47,\P(B\cup S)=0.81\) → overlap \(0.28\), exactly one \(0.81-0.28=0.53\). Bank Q31: \(0.64+0.46-0.82=0.28\), exactly one \(0.54\).

HW3 Q9 (dice-event union), corrected. Blue and white dice, \(|S|=36\). \(A\) = blue odd: \(|A|=18\). \(B\) = sum \(\ge 9\): sums 9, 10, 11, 12 have 4, 3, 2, 1 outcomes, \(|B|=10\). \(A\cap B\): blue 3 with white 6; blue 5 with white 4, 5, 6: \(|A\cap B|=4\). \(|A\cup B|=18+10-4=24\), \(\P=24/36=0.6667\). The graded copy entered 0.5556 (\(=20/36\)) and lost the point; recount \(|A\cap B|\) by listing.

4.10 Three-event inclusion–exclusion

Audit by membership pattern (slide 23). An outcome in exactly one event is counted \(1\) time. In exactly two: \(2-1=1\). In all three: \(3-3+1=1\). In none: \(0\). So every outcome of the union is counted exactly once.

Course enrollment (slides 24–25). 100 students; \(|P|=65,|L|=45,|C|=40\), \(|P\cap L|=30,|P\cap C|=25,|L\cap C|=20\), \(|P\cap L\cap C|=10\).

• Union: \(65+45+40-30-25-20+10=85\). None: \(1-0.85=0.15\).
• Exactly one: \(P\) only \(=65-30-25+10=20\) (the triple was subtracted twice, add it back once); \(L\) only \(=45-30-20+10=5\); \(C\) only \(=40-25-20+10=5\). Total 30, probability \(0.30\).

HW4 Q6: none of three flags \(=1-(0.46+0.39+0.31-0.18-0.14-0.12+0.05)=1-0.77=0.23\). HW4 Q7: pair-only counts \(55-25=30\), \(45-25=20\), \(40-25=15\); exactly two \(=65/240=0.2708\). HW4 Q9: \(A\) only \(=95-40-30+15=40\), \(B\) only \(=80-40-25+15=30\), \(C\) only \(=70-30-25+15=30\); exactly one \(=100/200=0.500\). Bank Q33: none \(=1-0.78=0.22\). Bank Q46 (Olympics): at least one \(=0.28+0.29+0.19-0.14-0.12-0.10+0.08=0.48\), none \(=0.52\in(0.5,0.6]\).

4.11 Choosing groups: unordered versus sequential

Three-student group (slide 29). Six first-years, four sophomores; group of three uniform over \(\binom{10}{3}=120\). One first-year and two sophomores: \(\binom61\binom42=6\cdot 6=36\); \(\P=36/120=0.3\).

Sequential check (slide 30). Ordered draws: \(10\cdot 9\cdot 8=720\) sequences. Favourable: choose which draw is the first-year (3 ways), fill it (6), fill the other two with sophomores (\(4\cdot 3\)): \(3\cdot 6\cdot 12=216\). \(216/720=0.3\). Same answer, as it must be.

Tablet giveaway (slide 31). 120 people, six names drawn without replacement. \(\P(\text{you win})=\binom{119}{5}/\binom{120}{6}=6/120=1/20\). General: \(\binom{n-1}{k-1}/\binom{n}{k}=k/n\). HW4 Q10 (Maya, 4 of 18): by symmetry \(4/18\); by counting \(\binom{17}{3}/\binom{18}{4}=2/9\). The option "\(1-(17/18)^4\)" is wrong because draws are without replacement, not repeated independent selections.

Deer (bank Q47). 300 deer, 100 tracked; sample 3 without replacement; \(\P(\le 1\text{ tracked})=\dfrac{\binom{200}{3}+\binom{100}{1}\binom{200}{2}}{\binom{300}{3}}\). "At most one" = zero or exactly one: two disjoint counts added.

4.12 Birthday collisions

Assume birthdays independent and uniform over 365 days (no February 29). For \(n\) students, \(\P(\text{at least two share})\)? "Count the complement event in which all birthdays are different."

Denominator: \(365^n\) ordered assignments, equally likely. Numerator: ordered assignments with all distinct days, \(P(365,n)\). HW4 Q11: the product \(365\cdot 364\cdots 350/365^{16}\) is \(\P(\text{all 16 different})\), so the shared-birthday probability is \(1-\) that. The distractor "denominator should be \(\binom{365}{16}\)" mixes ordered numerator with unordered denominator. Bank Q34: 12 files, 20 suffixes: \(\P(\text{collision})=1-\dfrac{20\cdot 19\cdots 9}{20^{12}}\). Bank Q52: 5 people, \(\P(\text{all distinct})=\dfrac{365\cdot364\cdot363\cdot362\cdot361}{365^5}\approx 0.9729\).

Bulgarian lottery (slide 34). The same six numbers came up on September 6 and 10, 2009. A specified later draw matches a fixed draw with probability \(1/\binom{49}{6}=1/13{,}983{,}816\); but across many drawings there are many pairs that could match, and about 4,404 drawings make a repeat more likely than not. "Repeat events need the right comparison set." V-1 impacts: visible clustering "does not prove targeting"; random spatial models also produce clusters.

4.13 Poker hands

Two pairs (bank Q35). \(\dfrac{\binom{13}{2}\binom42^2\cdot 11\cdot 4}{\binom{52}{5}}=\dfrac{123552}{2598960}\approx 0.0475\). The two pair ranks are chosen together with \(\binom{13}{2}\) (unordered), the fifth card from the 11 remaining ranks.

4.14 Aside: Maxwell–Boltzmann and Bose–Einstein

Marked "context, not syllabus" on the slides. \(r\) labelled particles in \(n\) cells: \(n^r\) ordered configurations (Maxwell–Boltzmann). Recording only occupation numbers \((r_1,\dots,r_n)\) with \(\sum r_i=r\): stars and bars gives \(\binom{r+n-1}{n-1}\) vectors (Bose–Einstein). Which outcomes are equally likely is a modelling assumption, not an axiom. The extra reading (hash collisions, XKCD multiple testing, union bound \(\P(\bigcup F_i)\le\sum\P(F_i)\), Bonferroni \(\alpha/k\)) is excluded.

4.15 Check yourself

L045. Conditional probability

Lecture 4's goals: "identify the observed information, restrict the sample space, compute a conditional probability, use multiplication along a sequence, and read a probability tree." The two question columns on the slide: Interpret (What information was observed? Which outcomes remain possible?) and Calculate and check (What is the new denominator? Did we reverse the conditional?).

5.1 Update before using a formula

Three red balls (slide 3). A friend uses one of two boxes: \(M_1\): 2 red, 8 white; \(M_2\): 8 red, 2 white. You draw three without replacement and all three are red. Which model remains possible? \(M_1\) has only two red balls, so three reds is impossible under \(M_1\). Under the stated assumptions you can be certain it is \(M_2\). The lesson (Bayes and Price, Laplace): "Observed information changes the probabilities that matter." No formula was needed, only restriction of the possible models.

5.2 Restrict the table when information arrives

The modelled class of 100 students:

Before conditioning, \(\P(\text{In state})=67/100\) and \(\P(\text{STAN})=70/100\); "every one of the 100 students is still in the reference population."

1. Given STAN, \(\P(\text{In state}\mid\text{STAN})\): target = in state, observed = STAN, remaining students = the 70 in the STAN row, new denominator 70. Answer \(52/70\approx 0.743\).
2. Given non-STAN: remaining = 30, answer \(15/30=0.5\).

5.3 The definition: intersect, then renormalise

In an equally likely model this is just \(\P(E\mid F)=|E\cap F|/|F|\): count the survivors in \(F\), then count how many of them are also in \(E\).

5.4 List the restricted universe

At least one girl (slide 10). Two children, ordered older/younger, \(S=\{BB,BG,GB,GG\}\) equally likely. \(A\) = at least one girl \(=\{BG,GB,GG\}\), \(B\) = both girls \(=\{GG\}\). \(\P(B\mid A)=\P(B\cap A)/\P(A)=(1/4)/(3/4)=1/3\). Three outcomes survive conditioning; one of them is \(GG\). (Contrast: given the older child is a girl, survivors are \(\{GB,GG\}\) and the answer is \(1/2\).)

Two hearts given at least one heart (slide 11). Two cards without replacement. \(B\) = both hearts, \(A\) = at least one heart, so \(B\subseteq A\) and \(B\cap A=B\). Numerator \(|B|=\binom{13}{2}=78\). Restricted denominator by complement: \(|A|=\binom{52}{2}-\binom{39}{2}=1326-741=585\). \(\P(B\mid A)=78/585=2/15\approx 0.133\).

Bank Q39. Two spinners \(\{1,2,3\}\), nine equally likely pairs. Given at least one 3, survivors are \((3,1),(3,2),(3,3),(1,3),(2,3)\): five. Sum 4: \((3,1),(1,3)\). \(\P=2/5\). Bank Q43: committee of 3 from 4 stats + 3 OR; given at least one OR, exactly one OR: \(\binom31\binom42\big/\bigl[\binom73-\binom43\bigr]=18/31\).

5.5 The multiplication rule

Actuarial exams (slide 15). Pass June with 0.9; given a June pass, pass July with 0.8. "Passes both" \(=J\cap L\), \(\P(J\cap L)=\P(J)\P(L\mid J)=0.9\cdot 0.8=0.72\). The conditional factor is the July pass. Bank Q56: John attends 0.65, Joe attends given John 0.8: both \(=0.52\). HW4 Q14: \(\P(A\cap D)=0.37\cdot 0.68=0.2516\). Bank Q37: \(0.65\cdot 0.40=0.26\).

Reversing with the definition (HW4 Q15, HW4 Q13, bank Q45). You are given \(\P(E)\), \(\P(B\mid E)\) and \(\P(B)\) and asked for \(\P(E\mid B)\). Two steps: multiplication rule for the intersection, then the definition with \(B\) as the new universe.

• HW4 Q15: \(\P(E\cap B)=\tfrac{29}{50}\cdot\tfrac{13}{20}=\tfrac{377}{1000}\); \(\P(E\mid B)=\tfrac{377/1000}{49/100}=\tfrac{377}{490}\approx 0.7694\). (The graded copy entered the decimal; the question asked for the exact fraction \(377/490\), already in lowest terms.)
• HW4 Q13: \(\P(R\cap S)=0.82\cdot 0.70=0.574\); \(\P(R\mid S)=0.574/0.61=0.9410\).
• Bank Q45: \(\P(A\cap B)=0.24\cdot 0.75=0.18\); \(\P(A\mid B)=0.18/0.45=0.40\); \(\P(A^c\mid B)=1-0.40=0.60\). The complement rule works inside the conditioned universe.

Extra context beyond the slides: this two-step manoeuvre is Bayes' rule in disguise, but the exam excludes Bayes as a named formula. Using multiplication then the definition is Lecture 4 material and is exactly what HW4 Q15 asked for.

5.6 The chain rule

With versus without replacement (slide 14). 50 students, 20 prefer probability. Both selected prefer probability: without replacement \(\tfrac{20}{50}\cdot\tfrac{19}{49}=\tfrac{380}{2450}\approx 0.1551\); with replacement \(\tfrac{20}{50}\cdot\tfrac{20}{50}=0.16\). The second factor is what changes.

Bank Q42. 5 blue, 3 red, 2 gold; blue then red then gold without replacement: \(\tfrac{5}{10}\cdot\tfrac{3}{9}\cdot\tfrac{2}{8}=\tfrac{30}{720}=\tfrac1{24}\). Bank Q40: 4 green, 2 gold; \(\P(\text{2nd gold}\mid\text{1st gold})\) is \(1/5\) without replacement and \(2/6=1/3\) with. Bank Q44: the student wrote \(\P(R_1\cap B_2)=\tfrac35\cdot\tfrac25\); the second factor must be \(\P(B_2\mid R_1)=\tfrac24\), giving \(\tfrac{3}{10}\).

5.7 Probability trees: multiply along, add across

The response problem (slides 16–18). Shankar picks one of two STOR 435 sections uniformly, then emails a uniformly chosen student from that section. Section 01 is 30% actuarial, Section 02 is 50%. An actuarial student replies with probability 0.30 in Section 01 (midterm week) and 0.80 in Section 02. A non-actuarial student never replies.

\(\P(\text{Section 01 and reply})=0.045\). Overall \(\P(\text{reply})=0.045+0.200=0.245\). The two routes are disjoint because a student is selected from only one section. (Lecture 5 names this "total probability"; on this exam you just need the tree.)

"Exactly one" on a tree (bank Q41, Q57, HW-style). Reading done with 0.70; office hours given reading 0.90, given no reading 0.40. Exactly one of the two events: reading-and-no-OH \(0.70\cdot 0.10=0.07\) plus no-reading-and-OH \(0.30\cdot 0.40=0.12\): total \(0.19\). Bank Q57: John 0.65, Joe given John 0.8, Joe given no John 0.45: exactly one \(=0.65\cdot 0.2+0.35\cdot 0.45=0.13+0.1575=0.2875\).

5.8 Conditioning checklist

The extra reading (the brain as a betting engine, \(\P(\text{cause}\mid\text{cue})\propto\P(\text{cue}\mid\text{cause})\P(\text{cause})\), fear and high stakes, prediction error) is excluded from the exam. Lecture 5 (total probability, Bayes, independence) is excluded too.

5.9 Check yourself

6. Mixed practice exam

Thirty-four multiple-choice questions in the style of the practice bank and the previous practice midterms: all four lectures shuffled, some "which correction is appropriate" audits, and a few interval-style answers. Work them closed-book with a calculator in about 75 minutes. The tag on each question says which lecture it comes from; the real exam will not tell you.

7. One-screen cram sheet

There is no formula sheet in the exam. Everything below must be in your head.