MATH U113 · Probability & Statistics · Tutorial companion

Tutorial sheets 1–3: hints, then solutions

Every question from the 5, 12 and 19 Aug tutorial sheets, each mapped to the module section that teaches it, with a hint that names the method and a collapsed full solution. Attempt → hint → solution, in that order.

Start here

The course guide calls these sheets the exam's rough draft — they're the only instructor-written problems you have. Working rule: attempt each question for 5 minutes before touching the hint, and open the solution only to check, never to read first. A solution you read is forgotten; a solution you checked your own attempt against sticks. Every number below has been computed independently, not copied.

Coverage so far: T1 = counting + axioms §2.1–2.3, T2 = conditional probability, Bayes, independence + the first pmf §2.4–2.5, §3.2, T3 = pmf, cdf, expectation §3.2–3.3. Nothing on the three sheets goes beyond Modules 1–2.

Tutorial 1 · 16 Q Tutorial 2 · 7 Q Tutorial 3 · 4 Q Previous-year quizzes T2 Q5–Q6 from zero What the sheets tell you

Tutorial 1 · 05 Aug · counting and the axioms

Prerequisite pages: Module 1 lesson (§3 for the beyond-Devore counting patterns) and notes. Q1–4 and 14 are the patterns the counting drill randomises.

Q1 · Nine flags§2.3 · identical copies

9 flags in a line: 4 white, 3 red, 2 blue, same-colour flags identical. How many signals?

HintCount as if all 9 were distinct, then divide out the reorderings you can't see.

Solution

9! / (4!·3!·2!) = 362880 / 288 = 1260. Equivalent view: choose the 4 white positions, then 3 red from the rest: C(9,4)·C(5,3) = 126·10.

Q2 · Ten officers, three groups§2.3 · labelled groups

Divide 10 officers into patrol (5), station (2), reserve (3).

HintThe groups have different jobs, so they are labelled — it's the same arithmetic as Q1 with people instead of flags.

Solution

10! / (5!·2!·3!) = C(10,5)·C(5,2)·C(3,3) = 252·10·1 = 2520.

Q3 · n-digit strings§2.3 · product rule

Digits 0–9, n positions (leading zeros allowed — "numbers" here means digit strings). (a) No two consecutive digits equal. (b) 0 appears exactly i times.

Hint(a) Fill positions left to right: how many choices for the first, and then for each later one given its neighbour? (b) First decide which positions hold the zeros.

Solution

(a) First digit 10 ways; each subsequent digit must avoid its predecessor: 9 ways. 10·9ⁿ⁻¹.

(b) Choose the i zero positions in C(n,i) ways; the other n−i positions take any non-zero digit: C(n,i)·9ⁿ⁻ⁱ. Sanity: summing over i gives (9+1)ⁿ = 10ⁿ by the binomial theorem.

Q4 · Nationalities together§2.3 · glue trick

4 Americans, 3 French, 3 British in a row, each nationality adjacent.

HintGlue each nationality into one block. Arrange the blocks, then arrange inside each block.

Solution

3! · 4! · 3! · 3! = 6·24·6·6 = 5184.

Q5 · 53 Sundays in a leap year§2.2 · equally likely outcomes

Hint366 = 52 full weeks + 2 leftover days. The leftover pair is one of 7 consecutive-day pairs.

Solution

52 weeks guarantee 52 Sundays; a 53rd needs Sunday among the 2 extra days. The pair is (Sun,Mon), (Mon,Tue), …, (Sat,Sun): 7 equally likely, 2 contain Sunday. 2/7. (Non-leap year: 1 extra day → 1/7.)

Q6 · Ring or necklace§2.2 · complement + addition rule

60% wear neither; 20% ring; 30% necklace.

Hint"Neither" is the complement of "or". Then the addition rule gives "and" from "or".

Solution

(a) P(R ∪ N) = 1 − 0.60 = 0.40. (b) P(R ∩ N) = P(R)+P(N)−P(R∪N) = 0.2+0.3−0.4 = 0.10.

Q7 · Two digits without replacement§2.1 · listing a sample space

From 1–5 choose one, then another from the remaining four. List outcomes with an odd digit (a) first, (b) second, (c) both times.

HintOutcomes are ordered pairs of distinct digits — 5×4 of them. Odd digits are 1, 3, 5. Count before you list, so you know when the list is complete.

Solution

Sample space: 20 ordered pairs (a,b), a ≠ b.

(a) first odd — 3·4 = 12: (1,2)(1,3)(1,4)(1,5)(3,1)(3,2)(3,4)(3,5)(5,1)(5,2)(5,3)(5,4).

(b) second odd — 12: (2,1)(3,1)(4,1)(5,1)(1,3)(2,3)(4,3)(5,3)(1,5)(2,5)(3,5)(4,5).

Q8 · AmEx or VISA§2.2 · addition rule

HintStraight addition rule; "both" is the overlap you subtract.

Solution

24 + 61 − 11 = 74%.

P(A ∪ B) = 0.24 + 0.61 − 0.11 = 0.74 — the 0.11 overlap is inside both circles, so plain addition would count it twice.

Q9 · Two laptops among six computers§2.3 · equally likely combinations

6 computers, 2 laptops; 2 chosen at random (15 outcomes).

HintCount favourable pairs with C(·,·) over the 15 total. For "at least one", go through the complement.

Solution

(a) both laptops: C(2,2)/15 = 1/15. (b) both desktops: C(4,2)/15 = 6/15 = 0.4. (c) at least one desktop = 1 − (a) = 14/15. (d) one of each: 2·4/15 = 8/15. Check: 1/15 + 6/15 + 8/15 = 1.

Q10 · Five dryer purchases§2.2 · complements

Hint"At least 2" is the complement of "at most 1". "At least one of each" is the complement of "all the same".

Solution

(a) 1 − 0.526 = 0.474. (b) 1 − 0.239 − 0.03 = 0.731 (the two "all same" events are disjoint, so they add).

Q11 · Two launches§2.5 · independence → quadratic

Independent, P(A) > P(B), P(A∪B)=0.626, P(A∩B)=0.144. Find both.

HintYou can get the sum P(A)+P(B) from the addition rule and the product from independence. Two numbers with known sum and product are the roots of a quadratic.

Solution

Sum: P(A)+P(B) = P(A∪B)+P(A∩B) = 0.770. Product: P(A)P(B) = 0.144. So they solve t² − 0.77t + 0.144 = 0: discriminant 0.5929 − 0.576 = 0.0169 = 0.13², roots (0.77 ± 0.13)/2. P(A) = 0.45, P(B) = 0.32.

Q12 · Ten forms, six handed over§2.3 · hypergeometric counting

6 withdrawal + 4 substitution forms; 6 given away at random. P(only one type remains on the desk)?

HintFour forms remain. "Only one type remains" = all 4 withdrawal, or all 4 substitution. Count the ways to give away 6 that leave each.

Solution

Leave 4 W: give 2 W + 4 S → C(6,2)·C(4,4) = 15. Leave 4 S: give 6 W → C(6,6) = 1. Total C(10,6) = 210. 16/210 = 8/105 ≈ 0.076.

Q13 · Cricket and football§2.2 · addition rule

P(A)=0.5, P(B)=0.4, P(A∩B)=0.25.

HintDraw the two-circle Venn diagram and fill the overlap first; everything reads off it.

Solution

(a) "one of the two sports" — read as at least one: 0.5+0.4−0.25 = 0.65. (If your tutor meant exactly one: 0.65 − 0.25 = 0.40 — ask which; both are one line once the Venn is filled.) (b) none: 0.35. (c) both: 0.25.

Q14 · REGULATIONS and MISSISSIPPI§2.3 · positions; glue + identical copies

Hint(a) All 11 letters are distinct. Count the position-pairs for R and E that are exactly 5 apart, times 2 for order, times the free arrangement of the other 9. (b) Glue the four S's into one block, then divide by the remaining repeats.

Solution

(a) Position pairs (i, i+5) for i = 1…6: 6 pairs × 2 orders × 9! for the rest, over 11!: 12·9!/11! = 12/110 = 6/55 ≈ 0.109.

(b) Total arrangements 11!/(4!·4!·2!) = 34650. With SSSS glued: 8 items (block, I×4, P×2, M) → 8!/(4!·2!) = 840. 840/34650 = 4/165 ≈ 0.024.

Q15 · P(A ∪ B^c)§2.2 · De Morgan + addition rule

P(A)=P(B)=½, P(A^c∩B^c)=⅓.

HintA^c∩B^c is the complement of A∪B (De Morgan) — that unlocks P(A∩B). Then P(A∩B^c) = P(A) − P(A∩B).

Solution

P(A∪B) = 1 − ⅓ = ⅔, so P(A∩B) = ½+½−⅔ = ⅓. Then P(A∩B^c) = ½ − ⅓ = ⅙, and P(A∪B^c) = P(A)+P(B^c)−P(A∩B^c) = ½+½−⅙ = 5/6.

Q16 · Committee of four from ten§2.3 · constrained committees

3 production, 4 purchase, 2 sales, 1 CA. Committee of 4.

HintDenominator is always C(10,4) = 210. (a) multiply one choice per category; (b) complement of "no purchase officer"; (c) fix the CA, choose the other 3 freely.

Solution

(a) 3·4·2·1/210 = 24/210 = 4/35. (b) 1 − C(6,4)/210 = 1 − 15/210 = 13/14. (c) C(9,3)/210 = 84/210 = 2/5.

Tutorial 2 · 12 Aug · conditional probability, Bayes, first pmf

Prerequisite pages: Module 1 lesson (conditional probability, the two-stage tree, Bayes) and Module 2 lesson for Q7.

Q1 · Three can lines§2.4 · table → conditional probability

HintThe table gives percentages within each line, i.e. conditional probabilities. Convert to counts (line total × %) — then every question is a ratio of counts.

Solution

(a) P(line 1) = 500/1500 = 1/3. Crack counts: 0.50·500 + 0.44·400 + 0.40·600 = 250+176+240 = 666, so P(crack) = 0.444.

(b) P(blemish | line 1) = 0.15 — read straight from the table.

(c) Surface-defect counts: 50 + 32 + 90 = 172; P(line 1 | surface) = 50/172 = 0.291. This is Bayes done with counts — no formula needed.

Total probability as counts: each line contributes (line total × crack rate); the crack total 666 out of 1500 is the denominator-free answer.

Q2 · Fastener recrimping§2.4 · tree diagram

HintDraw the tree: pass (0.95) / fail (0.05) → scrapped (0.2) / recrimp (0.8) → discarded (0.4) / pass (0.6). Multiply along branches.

Solution

(a) P(pass) = 0.95 + 0.05·0.8·0.6 = 0.95 + 0.024 = 0.974. (b) P(initial pass | pass) = 0.95/0.974 = 0.975.

Q3 · Two components§2.4 · definition of conditional probability

P(B)=0.9, P(A∪B)=0.96, P(A∩B)=0.75. Find P(B|A).

HintYou need P(A) first — the addition rule gives it.

Solution

P(A) = 0.96 + 0.75 − 0.90 = 0.81; P(B|A) = 0.75/0.81 = 0.926.

Q4 · Three defect types (eight parts)§2.2, §2.4 · three-circle Venn

HintFirst extract the three pairwise intersections from the three unions (addition rule each time). Then fill the 8 regions of a three-circle Venn diagram from the centre outward. Every part becomes a region lookup.

Solution

Pairwise: P(A₁∩A₂) = 0.12+0.07−0.13 = 0.06, P(A₁∩A₃) = 0.03, P(A₂∩A₃) = 0.02; centre 0.01.

Venn regions: only-1 = 0.12−0.06−0.03+0.01 = 0.04; only-2 = 0.07−0.06−0.02+0.01 = 0; only-3 = 0.05−0.03−0.02+0.01 = 0.01; 1&2 only 0.05; 1&3 only 0.02; 2&3 only 0.01; all three 0.01. Union = 0.14.

(a) 0.88 (b) 0.06 (c) 0.05 (d) 1 − P(all three) = 0.99 (e) 0.06/0.12 = 0.5 (f) 0.01/0.12 = 0.083 (g) exactly one = 0.04+0+0.01 = 0.05, over the union: 0.05/0.14 = 0.357 (h) 0.05/0.06 = 0.833.

Q5 · Impurity, two detections in three§2.5 · Bayes with binomial likelihoods

HintThe "data" is the event "exactly 2 of 3 independent tests detect". Compute its probability under each hypothesis by listing the sequences DDN, DND, NDD (present: detection 0.8, miss 0.2; absent: the sheet's 0.90 is the miss rate, so false-detection is 0.1), then apply Bayes with the priors 0.4 / 0.6.

Stuck? This question from zero — the prior/likelihood/posterior vocabulary, the eight-sequence figure, the tree and the area picture, and what each possible number of detections would have told the engineer.

Solution

Present: C(3,2)·0.8²·0.2 = 0.384. Absent: C(3,2)·0.1²·0.9 = 0.027.

P(present | data) = 0.4·0.3840.4·0.384 + 0.6·0.027 = 0.15360.1698 = 0.905

Posterior = highlighted leaf over the sum of the two highlighted leaves: 0.1536 / (0.1536 + 0.0162) = 0.905.

Q6 · Birthday problem§2.3, §2.5 · product rule for "all different"

Hint"All different" is a falling product over 365^k. "At least two share" is its complement. For (b), compute the running product until the complement crosses 0.5.

Stuck? This question from zero — why the complement, the two views of "all different" (counting and the chain of fractions), the running-product table, a slider-and-simulation widget for part (b), and the "shares my birthday" trap.

Solution

(a) P(all different) = (365·364·…·356)/365¹⁰ = 0.883; at least two share: 0.117.

(b) Running the product: k = 22 gives 0.476, k = 23 gives 0.507. k = 23.

Q7 · Forms, p(y) ∝ y§3.2 · normalising a pmf

Hint"Proportional to y" means p(y) = ky. The probabilities must sum to 1 — that fixes k.

Solution

(a) k(1+2+3+4+5) = 1 ⇒ k = 1/15, so p(y) = y/15, y = 1…5. (b) P(Y ≤ 3) = 6/15 = 0.4. (c) P(2 ≤ Y ≤ 4) = 9/15 = 0.6.

(d) With p(y) = ky²: 1+4+9+16+25 = 55, so p(y) = y²/55; P(Y ≤ 3) = 14/55 ≈ 0.255; P(2 ≤ Y ≤ 4) = 29/55 ≈ 0.527.

Tutorial 3 · 19 Aug · pmf, cdf, expectation

Prerequisite pages: Module 2 lesson and notes (§3.2–3.3).

Q1 · Overbooked flight§3.2 · reading a pmf table

HintTranslate each sentence into an inequality on Y, then add table entries. Standby position k gets on only if at least k seats are empty.

Solution

(a) P(Y ≤ 50) = 0.05+0.10+0.12+0.14+0.25+0.17 = 0.83. (b) P(Y ≥ 51) = 0.17. (c) First standby needs Y ≤ 49: 0.66. Third standby needs Y ≤ 47: 0.27.

The pmf as a probability histogram (the sheet gives 45–50 individually; 51–55 shown lumped, total .17).

The cdf staircase — every part of the question is one horizontal read-off, e.g. (a) F(50) = 0.83.

Q2 · Sum of three cards from 1–7§3.2–3.3 · building a pmf by enumeration; μ, σ²

Hint35 equally likely triples. Tally sums from 6 to 18 — the counts are symmetric about 12, so you only need to tally half. For μ use symmetry; for σ² compute E(W²) − μ².

Solution

Counts (over 35): w = 6:1, 7:1, 8:2, 9:3, 10:4, 11:4, 12:5, 13:4, 14:4, 15:3, 16:2, 17:1, 18:1. So p(w) = count/35.

μ = 12 (symmetry; also 3 × average card 4). E(W²) = Σ w²·count/35 = 5320/35 = 152, so σ² = 152 − 144 = 8.

Q3 · Maximum of two dice§3.2 · pmf and cdf

HintGo through the cdf first: M ≤ m means both dice ≤ m. Then the pmf is the difference of consecutive cdf values. Completely lost? The lesson now walks this exact problem from zero — the 6×6 grid, the L-shaped band, both routes.

Solution

(b) F(m) = P(both ≤ m) = m²/36 for m = 1…6 (a step function: 0 below 1, 1 at and above 6).

(a) p(m) = F(m) − F(m−1) = (2m−1)/36: 1/36, 3/36, 5/36, 7/36, 9/36, 11/36. Sums to 36/36.

Q4 · Order three or four magazines?§3.3 · expectation of a function, E[h(X)]

HintWith n copies ordered, sales = min(X, n) and profit = 4·min(X,n) − 2n. Compute the expected profit for n = 3 and 4 — it's E[h(X)] = Σ h(x) p(x), not h(E[X]).

Solution

n = 3: E[min(X,3)] = (1·1 + 2·2 + 3·12)/15 = 41/15; profit 4·41/15 − 6 = $4.93.

n = 4: E[min(X,4)] = (1 + 4 + 9 + 4·9)/15 = 50/15; profit 4·50/15 − 8 = $5.33. Order four.

Previous-year quizzes

Papers from earlier years surface through seniors — as new ones land, they get added here. Read them chiefly for format: what a 20-minute, answer-only paper feels like is the one thing the tutorial sheets can't teach. But a different year can mean a different paper-setter and different emphasis — treat each entry as a style guide, not a contract.

Quiz · 2026–27 (09 Sep 2026) · the paper she sat

Four variants and the key are filed; every question is rebuilt from zero on the quiz post-mortem page — Bayes with n unknown (urns), a conditional with a union in the condition, and the shuffled-PINs matching problem (derangements). Same format as below: 30 min, 20 marks, five decimals, no partial credit.

Quiz 1 · 2025–26 (05 Feb 2026) · conditional probability + expectation

Want these rebuilt from first principles instead of compressed? The two quiz questions, from zero — every step shown, with figures and all four variants. Two questions, 10 marks each, with about four numeric variants of the same paper across the room (same structure, shuffled numbers — so memorised answers are worthless, methods aren't). For reference while you prepare: this year's Quiz 1 is most probably 09 Sep — class word, unconfirmed; check LMS.

Duration	Marks	Type	Answer format	Partial credit
20 min	20	Closed book	Fill in the blank — final answer only, rounded to five decimal places	None — rough work is not evaluated, and overwriting voids the answer

What this format rewards

A correct method with an arithmetic slip scores zero here — so the skill under test is fluency plus accuracy under time. Three habits worth drilling: sanity-check that every probability lands in [0, 1] before writing it; know the rounding convention cold (variant B's Bayes answer is 0.471803… → write 0.47180, trailing zero included — five decimals means five digits); and do the arithmetic in rough space, check it once, then write in the blank a single time.

Q1 · Two coins, five tosses (variant A)§2.5, §3.4 · binomial tail + Bayes with binomial likelihood

A box holds coins C₁ and C₂; P(choose C₁) = 0.4. C₁ shows heads with probability 0.4 per toss, C₂ with 0.7. One coin is chosen at random and tossed 5 times. (i) P(at least 3 heads | C₁ chosen)? (ii) P(the coin was C₂ | exactly 3 heads observed)?

Hint(i) is a pure binomial tail — the coin is given, so P(C₁) plays no part. (ii) is the T2 Q5 pattern: Bayes where each branch's likelihood is itself a binomial probability.

Solution

(i) Given C₁, heads-count ~ binomial(n = 5, p = 0.4). Add the top three terms, using C(5,3) = 10, C(5,4) = 5, C(5,5) = 1:

P(≥3) = 10(0.4)³(0.6)² + 5(0.4)⁴(0.6) + (0.4)⁵ = 0.2304 + 0.0768 + 0.01024 = 0.31744.

(ii) Two ways the evidence "exactly 3 heads" can happen — through C₁ or through C₂ — and Bayes is the C₂ branch over their sum. Each branch = P(that coin) × P(3 heads with it):

C₂ branch: 0.6 × 10(0.7)³(0.3)² = 0.6 × 0.30870 = 0.18522. C₁ branch: 0.4 × 10(0.4)³(0.6)² = 0.4 × 0.23040 = 0.09216.

P(C₂ | 3H) = 0.18522 / (0.18522 + 0.09216) = 0.18522/0.27738 = 0.66775. Same shape as T2 Q5 — the companion flagged that pattern as one to master, and here it is worth 5 marks.

Q2 · Functions of a two-toss count (variant A)§3.2–3.3 · E[h(X)] term by term; variance of a function

X = number of heads in two fair tosses. (i) E[(X + 1)²]? (ii) If V[aX² + 1] = 2.25, find the positive a.

HintBuild the pmf of X first (three values), then everything is term-by-term sums. For (ii): adding a constant never changes variance, and pulling a scalar out squares it — so you need V[X²], which needs E[X⁴].

Solution

The pmf, by counting the four equally likely toss-pairs (the same build-from-outcomes routine as the dice-grid walkthrough): p(0) = ¼, p(1) = ½, p(2) = ¼.

(i) Term by term: E[(X+1)²] = ¼·1² + ½·2² + ¼·3² = 0.25 + 2 + 2.25 = 4.5. (Cross-check by expanding: E[X²] + 2E[X] + 1 = 1.5 + 2 + 1 = 4.5 ✓.)

(ii) V[aX² + 1] = a²V[X²] (the +1 shifts, doesn't spread; the a comes out squared). Now V[X²] = E[X⁴] − (E[X²])² with E[X²] = ¼·0 + ½·1 + ¼·4 = 1.5 and E[X⁴] = ¼·0 + ½·1 + ¼·16 = 4.5, so V[X²] = 4.5 − 2.25 = 2.25. Then a² · 2.25 = 2.25 ⇒ a = 1.

The one new tool: E[X⁴] — a fourth moment. Nothing deeper than E[X²]: the same term-by-term sum with fourth powers. It's the only place this quiz stepped past what T3 drilled, and one deliberate rep closes it.

All four variants, answered (recomputed and checked against the circulated answer key). Note the pattern: p₂ = 0.7 in every set, and Q1(i) depends only on p₁ — the "P(choose C₁)" number is a decoy for part (i).

Variant (P(C₁), p₁)	Q1(i)	Q1(ii)	Q2(i) asks	Q2(i)	Q2(ii) target	a
(0.4, 0.4)	0.31744	0.66775	E[(X+1)²]	4.5	2.25	1
(0.6, 0.4)	0.31744	0.47180	E[(X−1)²]	0.5	9	2
(0.4, 0.6)	0.68256	0.57262	E[(X+2)²]	9.5	20.25	3
(0.6, 0.6)	0.68256	0.37323	E[(X−2)²]	1.5	36	4

Where each question maps onto this year's sheets:

Quiz question	This year's tutorial pattern
Q1(i) binomial tail, coin given	T2's binomial setups (§3.4 machinery)
Q1(ii) Bayes, binomial likelihoods	T2 Q5 — the exact pattern, third of Bayes's three forms
Q2 E and V of functions of X	T3 Q4 E[h(X)] term by term + the pmf-from-outcomes build

The takeaway: last year's quiz was this year's tutorial patterns in fill-in-the-blank clothing. So prepare with the same material — but timed and to five decimals: the practice drill in short, clock-on bursts is the closest rehearsal available.

What the three sheets tell you

The pattern

T1 is mostly Ross-style counting, not Devore. Five of sixteen questions (1, 2, 4, 12, 14) use identical-copies / labelled-groups / glue patterns that Devore §2.3 never names. They're in the Module 1 lesson §3 and the drill — that's the one place the sheets run ahead of the textbook.

T2 is pure Devore §2.4–2.5. The method that appears three times (Q1c, Q2b, Q5) is Bayes — as counts, as a tree, and with a binomial likelihood. Master all three forms.

T3 is short and entirely §3.2–3.3 — but Q4 is the exam-style trap: expected profit needs E[h(X)] term by term. Q2 signals enumerate-a-pmf questions will be asked even when tedious; symmetry shortcuts are your friend.

Classic trap · seen on these sheets

T1 Q13(a), "one of the two" — ambiguous between "at least one" and "exactly one". Write which reading you took. T2 Q4 — guessing P(A₁∩A₂) instead of deriving it from the union. T3 Q4 — plugging E[X] = 3.67 into the profit formula instead of computing E[profit] term by term.

Ask an AI well

Paste a question and say: "Don't solve it. Tell me which Devore section it maps to and name the first step." Then: "I got ___ — check my reasoning, not just the number." Avoid "solve Tutorial 2 Q5" — you'll read a solution and retain nothing.