STOR 435 Midterm 1 Study Guide

Counting · Events · Probability rules · Conditional probability

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0. Before you start

How to use this guide

• Read sections 1 to 5 in order. Each idea builds on the one before. Blue boxes are the formulas you must know. Orange boxes are the mistakes that cost points. Green boxes tell you how a question will probably be phrased.
• Do every "Check yourself" question with pen and calculator before opening the answer. On the exam you need the actual number, not just the setup.
• Section 6 is material taught just before the exam that is officially for Midterm 2. Read it if you have time.
• Section 8 is a full practice exam with worked answers. Section 9 is a one-page summary for the night before.

If you have one day

1. Sections 1 and 2, about 45 minutes. Do the counting questions.
2. Sections 3 and 4, about 45 minutes. Be able to write the three-event inclusion–exclusion formula from memory.
3. Section 5, about 30 minutes. Do the multiplication-rule questions.
4. Take the practice exam in section 8 in 50 minutes, then read every solution.
5. Read the summary in section 9 before bed and again before the exam.

1. Three math facts you need

These are not probability, but they appear on the exam as quick questions and inside harder ones.

1.1 Geometric series

How to use it. Factor out whatever every term has in common until what is left looks like 1 + a + a² + ···, then apply the formula.

• 1 + (1/4)² + (1/4)⁴ + (1/4)⁶ + ··· has ratio a = 1/16, so it equals 1/(1 − 1/16) = 16/15.
• (1/3)²(1/4)² + (1/3)³(1/4)⁴ + (1/3)⁴(1/4)⁶ + ··· : the n-th term is (1/3)(1/3)ⁿ(1/16)ⁿ = (1/3)(1/48)ⁿ. Summing from n = 1 gives (1/3) · (1/48)/(1 − 1/48) = 1/141 ≈ 0.0071.

1.2 The number e

• Σ (−2)ⁿ/n! = e⁻² ≈ 0.1353.   Σ (−3)ⁿ/n! = e⁻³ ≈ 0.0498.
• lim (1 − 3/n)ⁿ = e⁻³ ≈ 0.0498.   lim (1 + 2/n)ⁿ = e².

1.3 Factorials and the calculator

• n! = n × (n − 1) × ··· × 2 × 1, and 0! = 1.
• A partial product like 10 × 9 × 8 equals 10!/7!. In general n(n − 1)···(n − r + 1) = n!/(n − r)!.
• Calculators overflow around 70!. Never compute 365!. Multiply fractions one at a time instead: (364/365)(363/365)···.
• Cancel before multiplying: (3003) = (300 · 299 · 298)/6.
• Find the nCr key on your calculator and practice with it before the exam.

1.4 Check yourself

2. Counting

Why counting comes first. When every outcome is equally likely, the probability of an event is simply (outcomes in the event) ÷ (all outcomes). To use that, you must be able to count outcomes without listing them. This section gives you five counting tools and the patterns for combining them.

2.1 Multiply the stages

The only requirement is that the number of options at each stage does not depend on earlier choices. The options themselves may change (after using the letter A you cannot use it again), as long as the count stays the same.

• Two-letter strings, repeats allowed: 26 · 26 = 676. No repeats: 26 · 25 = 650.
• License plate with 3 letters then 4 digits: 26³ · 10⁴ = 175,760,000. With no repeated letter or digit: (26 · 25 · 24)(10 · 9 · 8 · 7) = 78,624,000.
• A pizza has 4 sizes and any subset of 8 toppings: 4 · 2⁸ = 1024 options. Two pizzas ordered one after the other: 1024².

Example: a forbidden pair. How many subsets of a 10-student class do not contain both Brad and John? All subsets: 2¹⁰ = 1024. Subsets containing both: the other 8 students are free, 2⁸ = 256. Answer 1024 − 256 = 768.

2.2 Permutations: order matters

Example: books in blocks. 12 distinct books: 5 math, 3 sci-fi, 2 cookbooks, 1 environment, 1 teaching. Books of the same type must sit together. Treat each type as one block: 5 blocks in 5! orders. Inside the blocks: 5! · 3! · 2! · 1! · 1!. Total 5! · 5! · 3! · 2! = 172,800.

Example: visiting cities. Visit 4 of the 6 largest US cities. If the order of visits matters: 6 · 5 · 4 · 3 = 360. If it does not: (64) = 15. Same story, two different questions.

2.3 Combinations: order does not matter

Where it comes from. Ordered lists of r objects: n!/(n − r)!. Every group of r objects shows up r! times in that list, once per ordering. Divide by r! to count each group once.

Example: committee with a feud. From 5 juniors and 7 seniors, choose 3 juniors and 2 seniors. Two seniors refuse to serve together.

1. Ignore the feud: (53)(72) = 10 · 21 = 210.
2. Count committees with both feuding seniors: their two seats are taken, choose the 3 juniors: (53) = 10.
3. Subtract: 210 − 10 = 200.

Example: party guests. Invite 5 of 8 friends; two of them will not both attend. (85) − (63) = 56 − 20 = 36 possible groups.

Example: "at least" with cases. Choose 6 of 20 courses: 5 Statistics, 6 Physics, 9 other. Need at least 2 Statistics and at least 3 Physics. List the possible (Stats, Physics, Other) splits: (2,3,1) gives (52)(63)(91) = 1800; (2,4,0) gives (52)(64) = 150; (3,3,0) gives (53)(63) = 200. Total 2150.

Example: no two defectives together. 100 batteries in a line, 80 working and 20 defective, same-type batteries identical. How many orderings have no two defectives adjacent? Lay out the 80 working ones; that creates 81 gaps (before, between, after). Put at most one defective in a gap: choose 20 of 81 gaps, (8120).

2.4 The binomial theorem

Setting x = y = 1 gives Σ_r (nr) = 2ⁿ, the subset count again.

2.5 Multinomial coefficients: several groups, or repeated letters

This one number answers two different questions:

Splitting into labeled groups

The number of ways to divide n distinct objects into groups of sizes n₁, …, n_r when the groups have names or jobs. It equals (nn₁)(n − n₁n₂)··· (fill group 1, then group 2 from what is left, and so on).

Arranging objects with repeats

The number of ways to arrange n objects in a row when n₁ are identical, n₂ are identical, and so on. Start with n! and divide by n_i! for each type, because swapping identical objects changes nothing.

• MISSISSIPPI: 11 letters, I×4, S×4, P×2, M×1: 11!/(4! 4! 2!) = 34,650. SEATTLE: 7!/(2! 2!) = 1260.
• Bridge: 52 cards into four players' hands of 13: 52!/(13!)⁴.
• Chess: 7 games, each a win (1 point), draw (½) or loss (0) for player A; A finishes with 4 points. The number of draws must be even. 0 draws: 4W 3L, 7!/(4! 3!) = 35. 2 draws: 3W 2D 2L, 7!/(3! 2! 2!) = 210. 4 draws: 2W 4D 1L, 105. 6 draws: 1W 6D, 7. Total 357.

Labeled or unlabeled groups?

2.6 Seating problems

8 animals sit in 8 chairs, 4 on each side of a table. Every chair is different, so a seating is an arrangement of 8 animals: 8! = 40,320 in total.

• 6 cats and 2 dogs, dogs not on the same side. Seat dog 1 anywhere (8), dog 2 on the opposite side (4), the cats fill the rest (6!): 8 · 4 · 720 = 23,040. Check by subtraction: same side = 2 sides × 4 · 3 chairs for the dogs × 6! = 17,280, and 40,320 − 17,280 = 23,040.
• 4 dogs and 4 cats, probability all dogs on one side. Favorable: choose the dogs' side (2), arrange the dogs there (4!), arrange the cats opposite (4!). Probability 2(4!)²/8! ≈ 0.0286.

2.7 Which formula?

2.8 Check yourself

3. Sample spaces and events

The idea. Before you can find the probability of something, you must say exactly what that something is. The sample space is the list of everything that could happen. An event is a part of that list. "Or", "and", "not" become union, intersection, complement. Exam questions test whether you can move between words, set notation, and pictures without slipping.

3.1 Sample spaces

The sample space depends on what you write down. Flip a coin twice. Recording each flip gives S = {HH, HT, TH, TT}. Recording only the number of heads gives S = {0, 1, 2}. Same experiment, different sample space. Read carefully what is being recorded.

• Six runners finish with no ties: S is the set of all orderings, and its size is 6! = 720. The sample space is the set; 720 is its size.
• Flip a coin until the first head: S = {H, TH, TTH, TTTH, …}. Infinite, but listable.
• Lifetime of a light bulb in hours: S = [0, ∞). Continuous.
• Draw 2 marbles from {2 red, 1 green, 1 blue}, recording colors in order. With replacement there are 9 outcomes (RR, RG, RB, GR, GG, GB, BR, BG, BB). Without replacement, GG and BB are impossible, leaving 7. RR stays possible because there are two reds.
• Roll a black die and note even/odd, then a white die and note even/odd: S = {(E,E), (E,O), (O,E), (O,O)}. Ordered pairs, black die first. Not the faces, not unordered pairs.
• Pick one of 27 flags on each of 3 days, repeats allowed: S = all triples of flags, 27³ = 19,683 outcomes.

3.2 Events and the three operations

Union E ∪ F

Outcomes in E, in F, or in both. "E or F occurs." The "or" includes both.

Intersection E ∩ F

Outcomes in both. "E and F occur."

Complement E^c

Outcomes not in E. "E does not occur." S^c = ∅.

Empty event ∅

Contains no outcomes; cannot occur. If E ∩ F = ∅, E and F are disjoint (mutually exclusive): they cannot both happen.

Containment E ⊆ F

Every outcome of E is in F. If E happens, F happens. E = F means each contains the other.

For several events, ∪_j E_j means "at least one occurs" and ∩_j E_j means "all occur".

Example. S = {HH, HT, TH, TT}. E = first coin is H = {HH, HT}. F = the coins differ = {HT, TH}. Then E ∪ F = {HH, HT, TH}, E ∩ F = {HT}, E^c = {TH, TT}.

3.3 Rules for combining events

De Morgan works for any number of events: the complement of "at least one E_j" is "no E_j", and the complement of "all E_j" is "at least one fails".

A useful split. E ∪ F = E ∪ (E^c ∩ F). "E or F" is "E" plus "F but not E", and those two pieces do not overlap. This is how the addition rule in section 4 is proved.

3.4 Turning words into events

For three events A, B, C:

Method: list the cases that satisfy the words, write each case as an intersection (what happens ∩ what does not), then join the cases with ∪.

3.5 Simplifying a messy expression

The exam gives a complicated expression and asks which simple event it equals. Two approaches.

Reason in words. (A ∪ B) ∩ (A ∪ C) ∩ (B^c ∩ C^c). The last part says B and C both fail. Then "A or B" forces A, and "A or C" forces A. The whole thing is A ∩ B^c ∩ C^c: only A.

Use the rules. (A ∪ B) ∩ (A^c ∪ B) ∩ (A ∪ B^c). First, (A ∪ B) ∩ (A^c ∪ B) = B: if B occurs both parts hold; if B fails you would need A and A^c at once. Then B ∩ (A ∪ B^c) = (B ∩ A) ∪ (B ∩ B^c) = A ∩ B.

3.6 Worked examples

Colored balls. Draw 3 balls without replacement from 5 black and 2 red, recording colors in order. RRR is impossible, so S = {BBB, BBR, BRB, RBB, BRR, RBR, RRB}. A₂ = second ball black = {BBB, BBR, RBB, RBR}. A₃^c = third ball red = {BBR, BRR, RBR}. A₂ ∩ A₃^c = {BBR, RBR}.

Integers from −7 to 7. A = {N ≤ 3}, B = {2 ≤ N ≤ 10} = {2,…,7} within range, C = {N even and positive} = {2, 4, 6}.

• A ∩ B ∩ C = {2}.
• A^c ∪ (B ∩ C^c) = {4,5,6,7} ∪ {3,5,7} = {3,4,5,6,7}.
• (A ∪ B) ∩ C^c: A ∪ B is everything, so remove {2,4,6}: {−7,…,1,3,5,7}.
• (A ∩ B) ∩ (A ∪ C)^c: A ∩ B = {2,3}; (A ∪ C)^c = {5,7}; no overlap, ∅.

Coin until the first tail. S = {T, HT, HHT, …}. E_k = at least k tosses needed. E₄ ∩ E₈^c = at least 4 but fewer than 8 tosses = {HHHT, HHHHT, HHHHHT, HHHHHHT}.

Elevator. Three people each pick a floor from 1 to 10: S has 10³ = 1000 outcomes. E = they pick three consecutive floors. The floor set is {k, k+1, k+2} for k = 1 to 8, and the three people can be matched to those floors in 3! ways: |E| = 8 · 6 = 48.

3.7 Check yourself

4. The rules of probability

The idea. Probability is not defined as "long-run frequency" (how long is long? and you cannot repeat "Apple stock rises next year"). Instead, three rules say what any probability assignment must satisfy, and everything else follows. You need to: check whether an assignment obeys the rules, use the consequences (complement, inclusion–exclusion), and compute probabilities by counting when outcomes are equally likely.

4.1 The three axioms

Checking an assignment. S = {H, T} with P(H) = p and P(T) = 1 − p, 0 < p < 1. Axiom 1: both numbers are in [0, 1]. Axiom 2: P(S) = 1. Axiom 3: the only disjoint union to check is {H} ∪ {T} = S, and p + (1 − p) = 1. Valid.

Finite sample spaces. Give each outcome x_i a number p_i ≥ 0 with p₁ + ··· + p_N = 1. Then P(E) is the sum of the p_i for outcomes in E. Unfair die with P(1) = P(2) = P(3) = 1/4 and P(4) = P(5) = P(6) = 1/12: P(even) = 1/4 + 1/12 + 1/12 = 5/12.

4.2 What follows from the axioms

Why the addition rule subtracts. Adding P(E) and P(F) counts the overlap twice. Formally, E ∪ F = E ∪ (E^c ∩ F) is a disjoint union, and P(E^c ∩ F) = P(F) − P(E ∩ F).

Example: two overlapping groups. 80% of students have Facebook or Twitter; 60% have Facebook and 25% Twitter. Both? 0.80 = 0.60 + 0.25 − P(both), so P(both) = 0.05.

Example: three courses. 100 students: 65 in Prob, 45 in LA, 40 in COMP; 30 in Prob and LA, 25 in Prob and COMP, 20 in LA and COMP; 10 in all three.

1. In at least one: 65 + 45 + 40 − 30 − 25 − 20 + 10 = 85. In none: 15, so P(none) = 0.15.
2. In exactly one: Prob only = 65 − 30 − 25 + 10 = 20 (the +10 restores the all-three students, who were subtracted twice); LA only = 45 − 30 − 20 + 10 = 5; COMP only = 40 − 25 − 20 + 10 = 5. Total 30, P(exactly one) = 0.30.

Example: three sports. 28% watched track, 29% swimming, 19% gymnastics; 14% track and swimming, 12% swimming and gymnastics, 10% track and gymnastics; 8% all three. P(at least one) = 0.28 + 0.29 + 0.19 − 0.14 − 0.12 − 0.10 + 0.08 = 0.48, so P(none) = 0.52.

Example: bounds. If P(A) = 0.4 and P(B) = 0.7 and nothing else is known: since P(A ∪ B) ≤ 1, the addition rule gives P(A ∩ B) ≥ 0.4 + 0.7 − 1 = 0.1; since A ∩ B ⊆ A, P(A ∩ B) ≤ 0.4.

Example: working backwards. Given P(A^c ∩ B^c) = 0.1, P(A ∩ B) = 0.3, P(B^c) = 0.25. By De Morgan, P(A ∪ B) = 1 − 0.1 = 0.9. P(B) = 0.75. So P(A) = 0.9 − 0.75 + 0.3 = 0.45. Draw the Venn diagram if the algebra feels shaky.

4.3 Equally likely outcomes: probability by counting

Three ways to sample two people from 900 (600 Democrats, 300 Republicans):

• Without replacement, keeping order: 900 · 899 outcomes, each equally likely.
• Without replacement, ignoring order: (9002) outcomes, each equally likely. Fine for any question that only depends on who was picked, since each pair corresponds to exactly two ordered outcomes.
• With replacement: 900² ordered outcomes, each equally likely. Keep the order here: an unordered pair {x, y} with x ≠ y arises two ways but {x, x} only one way, so unordered outcomes would not be equally likely.

Example: ordered and unordered agree. 10 students, 6 first-years and 4 sophomores; pick 3 at random. P(1 first-year and 2 sophomores)?

Example: voters. 40 voters: 20 D, 15 R, 5 I. Pick 3 without replacement. P(no Independents) = (353)/(403) = 6545/9880 ≈ 0.6624. P(one of each) = (20 · 15 · 5)/9880 ≈ 0.1518.

Example: deer. 300 deer, 100 tagged. Pick 3 without replacement. P(at most 1 tagged) = [(2003) + (1001)(2002)] / (3003) ≈ 0.7415. Answers built from (2/3)³ would treat the draws as independent, which is wrong without replacement.

Example: prizes. 120 guests, 6 prizes drawn at random. P(you win) = 6/120. Counting confirms it: (1195)/(1206) = 6/120. With n guests and k prizes: k/n.

Example: difference divisible by 4. Pick m then n with replacement from 5 to 14 (100 ordered outcomes). |m − n| is a multiple of 4 when the difference is 0 (10 outcomes), 4 (12 outcomes), or 8 (4 outcomes). P = 26/100 = 13/50.

4.4 "At least one": use the complement

Birthday problem. n people, 365 equally likely birthdays. P(some two share) = 1 − P(all different), and

P(all different) = 365365 · 364365 · 363365 ··· 365 − n + 1365

• Flags. One of 27 flags is chosen on each of 3 days. P(the Dutch flag appears at least once) = 1 − (26/27)³ ≈ 0.1070.
• Shoes. 10 pairs, choose 8 shoes at random. P(no complete pair) = (108) · 2⁸ / (208) ≈ 0.0915: choose which 8 pairs contribute one shoe, then left or right from each.
• Lottery repeat. A 6-of-42 lottery has N = (426) = 5,245,786 combinations. P(the next draw matches a fixed one) = 1/N. P(some pair of draws matches among m draws) is a birthday problem with N "birthdays", and crosses 50% at about m = 2697 draws. A specific coincidence is rare; some coincidence is not.

4.5 Poker hands

5 cards from 52: (525) = 2,598,960 equally likely hands. A full house is three of one denomination plus two of another. Build it in the order you would describe it: the triple's denomination (13), its suits (43) = 4, the pair's denomination (12), its suits (42) = 6. Favorable = 13 · 4 · 12 · 6 = 3744, P ≈ 0.00144.

4.6 Check yourself

5. Conditional probability

The idea. Learning something changes probabilities. "The probability of E given that F happened" is written P(E | F). Once you know F happened, F becomes your new, smaller sample space. That single picture explains every formula in this section.

5.1 Definition

Example: a table.

Pick a student at random. P(in-state) = 67/100. Told the student is a Statistics major, only that row matters: P(in-state | Statistics) = 52/70. Told "other major": 15/30 = 1/2.

Example: two children. Outcomes {BB, BG, GB, GG}, equally likely. A = at least one girl = {BG, GB, GG}. B = both girls = {GG}. P(B | A) = P(GG)/P(A) = (1/4)/(3/4) = 1/3. Reduced sample space: three outcomes have at least one girl, one of them is GG.

Example: two hearts. Two cards without replacement. B = both hearts, A = at least one heart. Hands with at least one heart: (522) − (392) = 1326 − 741 = 585. Hands with two hearts: (132) = 78. P(B | A) = 78/585 = 2/15.

Example: divisibility. Pick an integer from 1 to 60. A = divisible by 4 or by 9, B = divisible by 3. Given B, the sample space is the 20 multiples of 3. Of those, A means divisible by 12 or by 9: 5 multiples of 12, 6 multiples of 9, 1 multiple of both (36). |A ∩ B| = 5 + 6 − 1 = 10, so P(A | B) = 10/20 = 1/2.

5.2 The multiplication rule: probability of a chain

• Two draws. 50 students, 20 love probability. Pick 2 without replacement. P(both love it) = (20/50)(19/49) ≈ 0.1551. Same as (202)/(502).
• Two exams. P(pass first) = 0.9, P(pass second | passed first) = 0.8. P(pass both) = 0.72.
• Four draws. 8 red and 6 blue candies drawn one at a time without replacement. P(blue, red, red, blue) = (6/14)(8/13)(7/12)(5/11) ≈ 0.0699. Each fraction is what is left in the box at that moment.

5.3 Splitting into cases

Example: emailing a student. Two course sections, each chosen with probability 1/2. Section 1 is 30% actuarial students, Section 2 is 50%. An actuarial student replies with probability 0.30 in Section 1 and 0.80 in Section 2; other students never reply.

1. P(Section 1 and a reply): a reply needs an actuarial student, so multiply along the chain: (1/2)(0.30)(0.30) = 0.045.
2. P(reply): add the two sections. Section 2 gives (1/2)(0.50)(0.80) = 0.20. Total 0.045 + 0.20 = 0.245.

Example: a screening test. 1% of babies have a condition. The test is positive for 90% of affected babies and for 5% of unaffected babies. P(positive) = (0.01)(0.90) + (0.99)(0.05) = 0.0585.

5.4 Conditioning on the wrong thing

5.5 Check yourself

6. Just in case: total probability, Bayes, independence

These were taught in the two classes before the exam but are officially on Midterm 2. They are short extensions of section 5. Learn 6.1 at minimum.

6.1 Law of total probability

• Triage. 15% of patients are critical (50% die), 45% serious (20% die), 40% stable (2% die). P(die) = 0.075 + 0.09 + 0.008 = 0.173. P(survive) = 0.827.
• Voting. 45% independents (30% vote), 30% Democrats (50%), 25% Republicans (60%). P(voted) = 0.135 + 0.15 + 0.15 = 0.435.
• Spam. 85% of email is spam. "Free money" appears in 10% of spam and 1% of the rest. P(phrase) = 0.085 + 0.0015 = 0.0865.
• Mammogram. P(cancer) = 0.008, P(+ | cancer) = 0.9, P(+ | no cancer) = 0.07. P(+) = 0.0072 + 0.0694 = 0.0766.

6.2 Bayes' formula: flipping the condition

• Mammogram. P(cancer | +) = 0.0072/0.0766 ≈ 0.094. Only 9%, because the false positives from the large healthy group outnumber the true positives.
• Screening test. P(condition) = 0.01, positive for 90% of affected and 5% of unaffected. P(condition | +) = 0.009/0.0585 ≈ 0.154.
• Rare disease, near-perfect test. P(disease) = 0.0001, sensitivity 0.999, false-positive rate 0.0001. P(disease | +) = 0.0000999/(0.0000999 + 0.00009999) ≈ 0.50. A positive result is a coin flip.
• Noisy channel. A 0 or 1 is sent with equal probability. A sent 0 flips with probability 0.05, a sent 1 with 0.10. P(sent 1 | received 1) = (0.5)(0.9)/[(0.5)(0.9) + (0.5)(0.05)] = 0.45/0.475 ≈ 0.947.

Chained milestones. P(A₂ | A₁) = 0.8, P(A₂ | A₁^c) = 0.3, and the same for A₃ given A₂; only the most recent milestone matters. Then P(A₃ | A₁) = (0.8)(0.8) + (0.3)(0.2) = 0.70, P(A₃ | A₁^c) = (0.8)(0.3) + (0.3)(0.7) = 0.45, and with P(A₁) = 0.75, P(A₃) = (0.75)(0.70) + (0.25)(0.45) = 0.6375.

6.3 Independence

• Two coin tosses, E = first is H, F = second is T: P(E ∩ F) = 1/4 = (1/2)(1/2). Independent.
• One card, E = club, F = a 6: P(E ∩ F) = 1/52 = (1/4)(1/13). Independent.
• Two dice, F = first die is 4. E₁ = sum is 6: P(E₁ ∩ F) = 1/36 but P(E₁)P(F) = (5/36)(1/6). Not independent. E₂ = sum is 7: P(E₂ ∩ F) = 1/36 = (6/36)(1/6). Independent.
• If E and F are independent, so are E and F^c.
• Three events are independent if every pair multiplies and P(E ∩ F ∩ G) = P(E)P(F)P(G).

Four friends, one flat tire. Each independently names which of four tires went flat. Two pick uniformly (1/4 each). One favors left tires (1/3 each, 1/6 for right); one favors right (the reverse). P(all four name the same tire) = 4 tires × (1/4)(1/4)(1/3)(1/6) = 1/72, since for every tire the two biased friends contribute 1/3 · 1/6 either way.

5 before 7. Roll two dice repeatedly. P(a sum of 5 appears before a sum of 7) = P(5)/(P(5) + P(7)) = (4/36)/(10/36) = 0.4. Rolls that are neither are ignored; only the first decisive roll counts.

7. Exam strategy

7.1 Four questions before you compute

1. What is the sample space, and are outcomes equally likely? If you use |E|/|S|, count both at the same level (ordered with ordered, unordered with unordered).
2. Does order matter? Swapping two items: new outcome or not? Permutation or combination.
3. With or without replacement? With: powers like 27³ or (26/27)³. Without: shrinking fractions like (20/50)(19/49), or binomial coefficients.
4. Is the complement easier? "At least one", "no two the same", "not both": compute the opposite and subtract from 1.

7.2 Multiple-choice tactics

• Interval answers: compute to 4 decimals, then check whether the endpoints are included. (0.8, 0.85) excludes both; (0.1, 0.2] includes 0.2.
• Formula answers: check the pieces. Right denominator? Groups labeled or not? Are all the cases added ("at most 1" = 0 + 1)?
• Sanity limits: probabilities are in [0, 1]; P(E ∩ F) ≤ min(P(E), P(F)); P(E ∪ F) ≥ max(P(E), P(F)); an added condition can only lower a probability.
• Set questions: test concrete outcomes or write the eight-row yes/no table.
• Time: about 3.5 minutes per question. Every question is worth the same; skip and come back.

7.3 The traps

8. Practice exams

Part A is the practice midterm posted for the course, with worked answers. Part B is a new set in the same style. Do each part in 50 minutes with only a calculator, then read every solution, including the ones you got right. The real exam will not just be variants of these, so focus on the reasoning.

8.1 Part A: the posted practice midterm

8.2 Part B: a second exam

9. One-page summary

Read this the night before and again right before the exam.