MATH U113 · Quiz (09 Sep 2026) · post-mortem

The quiz, question by question, from zero

The paper is in the repo (four variants, answer key). This page rebuilds each question from first principles, names the exact step where marks were lost, tabulates all four variants, and ends each question with a twin. Read it for the mid-semester exam on 7 Oct (30%, closed book), which will reuse these three patterns.

Start here · the marks arithmetic, and an honest difficulty read

This quiz is a single 10% component (20 of 200 marks). Whatever happened on it caps the damage at 10% of one course. The mid-sem (30%) and compre (40%) decide the grade, and both allow working room this paper did not.

Difficulty, honestly: Q1 and Q2 were fair — Bayes with a parameter to solve for, and one set-algebra identity plus independence; both are Module 1 material the pages teach. Q3 was not fair to this batch. Half of its 10 marks needed the pmf of "how many of 4 shuffled PINs land correctly", which requires derangement counts (9 of the 24 orders leave nobody correct). Devore never covers derangements; JEE-drilled classmates have them memorised. That is a genuine but tiny gap — three numbers, now in the Module 2 notes — and the other half of Q3 was the same Var(cX + d) pattern as last year, which the quiz-prep page rehearsed.

The format Q1 · urns, find n Q2 · conditional with a union Q3 · shuffled PINs All four variants Next time: the 30-minute plan

The format, and what it rewards

Duration	Marks	Type	Answer format	Partial credit
30 min	20 (5 + 5 + 5 + 5)	Closed book	Fill in the blank, simplified form, five decimal places	None — rough work not evaluated, overwriting voids the answer, no extra sheets

Identical format to last year's paper; four numeric variants in the room. Two of the four answers per variant are integers by design (Q1's n, Q3's a) — a non-integer there is an arithmetic slip, not a surprise.

Q1 · Two black balls: which urn? Then solve for n §2.3–2.4 · 5 marks

Variant 1, as set Each of n urns contains four white and six black balls, while another urn contains five white and five black balls. An urn is chosen at random from the (n + 1) urns, and two balls are randomly drawn from it. Given that the drawn balls are black, the probability that five white and three black balls remain in the chosen urn is 1/7. Find n. (Key: 4.)

Where this lives Notes §2.4 · Bayes, total probability · Notes §2.3 · combinations · Tutorial 2 Q5 from zero (same machine, likelihoods built by counting) · Devore §2.4

What is really being asked

Two decodings before any formula. (1) "Five white and three black remain" after two black balls were removed means the urn started with five white and five black — it is the odd urn out. So the event whose posterior you're told is "the 5W5B urn was chosen". (2) The unknown is not a probability but the number of urns n: set up the posterior with n as a letter, equate to 1/7, solve. Bayes with algebra, not Bayes with numbers.

Step 1Name the states and the priors

Let U = "the 5W5B urn was chosen" and U′ = "one of the n 4W6B urns was chosen". Urn chosen at random from n + 1:

P(U) = 1n + 1, P(U′) = nn + 1

Step 2Build the likelihoods by counting — two black from ten

Let B = "both drawn balls are black". Two balls from ten, order irrelevant: C(10, 2) = 45 equally likely pairs.

P(B | U) = C(5, 2)C(10, 2) = 1045 = 29, P(B | U′) = C(6, 2)C(10, 2) = 1545 = 13

Unpack this step: C(5, 2) = 10 and C(6, 2) = 15, or the sequential route

C(5, 2) = 5·4/2 = 10, C(6, 2) = 6·5/2 = 15, C(10, 2) = 10·9/2 = 45. Or draw in order: (5/10)(4/9) = 2/9 and (6/10)(5/9) = 1/3 — same numbers, no combinations needed.

The tree. The posterior of U is the top leaf divided by the sum of the two marked leaves.

Step 3Bayes, with n as a letter

P(U | B) = P(B | U) P(U)P(B | U) P(U) + P(B | U′) P(U′) = (2/9) · 1/(n + 1)(2/9) · 1/(n + 1) + (1/3) · n/(n + 1)

The factor 1/(n + 1) is in every term — cancel it. Then multiply top and bottom by 9:

P(U | B) = 2/92/9 + n/3 = 22 + 3n

A clean formula, and a check: at n = 0 (only the 5W5B urn exists) it gives 1 ✓; as n grows it shrinks ✓.

Step 4Equate and solve

23n + 2 = 17 ⇒ 3n + 2 = 14 ⇒ n = 4

Integer, as it must be (you cannot have 3.7 urns). That is the in-room check for this question: if n is not a whole number, a likelihood is wrong.

Classic traps

1) Misreading the hypothesis. "Five white and three black remain" is a clue about the starting contents. Students who took it as "the 4W6B urn" got 3n/(3n+2) = 1/7 and a negative n. Negative or fractional n = go back to the decoding. 2) Forgetting there are n urns of the first kind. The prior is n/(n+1), not 1/(n+1) for each type. 3) Likelihood of one black ball, not two. 6/10 and 5/10 are one-draw numbers; the question says two balls. 4) Rounding n. The blank asked for n, an integer — "4.00000" is fine, "0.14286" (the given posterior copied back) is not.

Try the twin: same urns, but both drawn balls are white and the posterior that the 5W5B urn was chosen is 1/4. Find n.

P(WW | U) = C(5,2)/45 = 10/45 = 2/9; P(WW | U′) = C(4,2)/45 = 6/45 = 2/15. Posterior = (2/9) / (2/9 + 2n/15); multiply top and bottom by 45: 10/(10 + 6n) = 5/(3n + 5). Set equal to 1/4: 3n + 5 = 20, n = 5. Integer ✓.

Q2 · A conditional probability with a union in the condition §2.1, 2.4, 2.5 · 5 marks

Variant 1, as set If A and B are two independent events with P(A) = 1/3 and P(B) = 3/4, then P(B^c | A^c ∪ B) is ▢. (Key: 2/11 = 0.18182.)

Where this lives Notes §2.1 · events as sets, De Morgan · Notes §2.4 · conditional probability · Notes §2.5 · independence (complements stay independent) · Devore §2.4–2.5

Step 1Write the definition, then simplify the intersection

P(B^c | A^c ∪ B) = P(B^c ∩ (A^c ∪ B))P(A^c ∪ B)

The top is where marks were won or lost. Distribute the intersection over the union:

B^c ∩ (A^c ∪ B) = (B^c ∩ A^c) ∪ (B^c ∩ B) = (A^c ∩ B^c) ∪ ∅ = A^c ∩ B^c

In words: "not B, and (not A or B)" can only be satisfied by "not A and not B", because the "or B" half is killed by "not B". The Venn diagram says it in one look:

Left: the condition A^c ∪ B (shaded) — everything except "A only". Right: keep only the part outside B: what survives is outside both circles, A^c ∩ B^c.

Step 2Numerator by independence

If A and B are independent, so are their complements (§2.5). P(A^c) = 2/3, P(B^c) = 1/4:

P(A^c ∩ B^c) = 23 · 14 = 16

Step 3Denominator — two routes, same number

Addition rule: P(A^c ∪ B) = P(A^c) + P(B) − P(A^c)P(B) = 2/3 + 3/4 − 1/2 = 8/12 + 9/12 − 6/12 = 11/12.

Complement (faster): the only thing outside A^c ∪ B is "A only" = A ∩ B^c, so P(A^c ∪ B) = 1 − P(A)P(B^c) = 1 − (1/3)(1/4) = 11/12 ✓ (the left Venn panel).

Step 4Divide and round

P(B^c | A^c ∪ B) = 1/611/12 = 16 · 1211 = 211 = 0.18182

Check: it is a probability in [0, 1] ✓, and it is smaller than the unconditional P(B^c) = 1/4 — sensible, since the condition includes "B happened" as one of its ways of being true.

Classic traps

1) Treating the condition as if it were just A^c. The union matters: P(B^c | A^c) = 1/4 by independence, but that is not the question. 2) Numerator = P(B^c) · P(A^c ∪ B). B^c is not independent of A^c ∪ B (they overlap through B). Only A and B (and their complements) are independent — nothing built from both. 3) Forgetting the −P(A^c ∩ B) in the union. 2/3 + 3/4 is more than 1 — alarm bell. 4) Rounding. 2/11 = 0.181818…, so 0.18182 (five places, round half up on the sixth digit 1 → no, 8 rounds up the 1). Write the fraction too; "2/11 = 0.18182" is what the key shows.

Try the twin: A, B independent, P(A) = 1/2, P(B) = 1/3. Find P(Aᶜ | A ∪ Bᶜ).

Numerator: A^c ∩ (A ∪ B^c) = (A^c ∩ A) ∪ (A^c ∩ B^c) = A^c ∩ B^c, probability (1/2)(2/3) = 1/3. Denominator: outside A ∪ B^c is "B only" = A^c ∩ B, so P = 1 − (1/2)(1/3) = 5/6. Ratio (1/3)/(5/6) = 2/5 = 0.40000.

Q3 · Four PINs shuffled: E and V of the number of correct matches §2.3, 3.2–3.3 · 5 + 5 marks

Variant 1, as set Four customers apply for ATM cards at a bank. The bank generates a unique PIN for each customer. Due to a system error, the PINs are accidentally assigned to the four customers in a random order, each customer receiving exactly one PIN. For every person who receives the correct PIN, the bank awards 5 security points. Let X be the number of customers who receive the correct PIN. (i) The expected total security points is ▢. (ii) The value a > 0 such that Var((a² + a)X + 5) = 36 is ▢. (Key: 5 and 2.)

Where this lives Notes §3.3 · E, V, rescaling · Notes §3.3 · worked example 2, the matching problem (new) · Quiz prep · find a from a variance (same pattern as last year) · Devore §3.3

What is really being asked — and what was missing

The variable is X = number of "fixed points" when 4 items are randomly permuted — the classic matching problem (hats, letters in envelopes, PINs). Part (i) needs E(X); part (ii) needs V(X), then the rescaling rule. Both need the pmf of X, and the pmf needs you to count, for each k, how many of the 4! = 24 assignments give exactly k correct. The count for k = 0 is a derangement number, which Devore never names. That is the missing piece, and it is small: D₂ = 1, D₃ = 2, D₄ = 9.

Step 1Build the pmf by counting the 24 orders

Label the customers 1–4 and an assignment as the order the PINs land in; customer i is correct when position i holds PIN i. Count assignments with exactly k correct: choose which k are correct, C(4, k) ways, and then the other 4 − k must all be wrong — a derangement of 4 − k items, D_4−k ways.

k correct	count = C(4, k) · D_4−k	orders	p(k)
0	C(4,0) · D₄ = 1 · 9	9	9/24
1	C(4,1) · D₃ = 4 · 2	8	8/24
2	C(4,2) · D₂ = 6 · 1	6	6/24
3	C(4,3) · D₁ = 4 · 0	0	0
4	C(4,4) · D₀ = 1 · 1	1	1/24

Unpack this step: where D₂ = 1, D₃ = 2, D₄ = 9 come from

A derangement is an order with nobody correct. D₁ = 0 (one item must go to its own place). D₂ = 1 (swap). D₃ = 2: of the 6 orders of 123, only 231 and 312 fix nothing. D₄ = 9: the 24 orders minus those with ≥ 1 correct; or the recursion D_n = (n − 1)(D_n−1 + D_n−2): D₄ = 3(2 + 1) = 9. Check the table: 9 + 8 + 6 + 0 + 1 = 24 ✓ — and "exactly 3 correct" is impossible, because the fourth is then forced correct too.

All 24 assignments (customer order 1 2 3 4; a highlighted digit is in its own position). Counts 9 · 8 · 6 · 0 · 1 — read straight off the picture.

Step 2E(X) and V(X) from the table

E(X) = 0·9 + 1·8 + 2·6 + 3·0 + 4·124 = 2424 = 1

E(X²) = 0·9 + 1·8 + 4·6 + 9·0 + 16·124 = 4824 = 2, V(X) = E(X²) − [E(X)]² = 2 − 1 = 1

Worth memorising as a check: for the matching problem with any n ≥ 2, E(X) = 1 and V(X) = 1. (On average exactly one person gets their own PIN, however many people there are.) In the exam, build the table — it is the only justification available from Module 2's tools — but if your table gives anything other than 1 and 1, recount.

Step 3(i) Expected points: rescale the mean

Points Y = 5X, so E(Y) = 5 E(X) = 5. (Variants: 6, 7, 8 — always the points-per-person, because E(X) = 1.)

Step 4(ii) Find a: rescale the variance

The rule V(cX + d) = c² V(X) with c = a² + a and d = 5 (the shift does nothing):

(a² + a)² · 1 = 36 ⇒ a² + a = 6 (the positive root, since a > 0 makes a² + a > 0) ⇒ a² + a − 6 = 0 ⇒ (a + 3)(a − 2) = 0

a = 2 (reject −3). Same in all four variants — the shift constant 5, 6, 7, 8 is a decoy. The designed check, exactly as last year: a is a small integer; a surd means V(X) was miscounted.

Classic traps

1) Guessing the pmf is binomial. Each customer has probability ¼ of being correct, but the events are not independent (if three are correct, the fourth is forced), so Bin(4, ¼) is wrong — it would give V(X) = 3/4 and a surd for a. 2) Including k = 3. It cannot happen; a table with p(3) > 0 will not sum to 1. 3) Squaring only a. The multiplier is the whole bracket a² + a; the variance picks up its square. 4) Adding the constant into the variance. The +5 shifts the mean and leaves the variance alone. 5) Taking a = −3. The question says a > 0.

Try the twin: three customers, 4 points per correct PIN. Find E(points), and a > 0 with Var((a² − a)X + 4) = 36.

Orders of 3: six of them. Correct counts: 0 correct in D₃ = 2 orders, 1 correct in C(3,1)·D₂ = 3, 2 correct impossible, 3 correct in 1. pmf 2/6, 3/6, 0, 1/6. E(X) = (3 + 3)/6 = 1; E(X²) = (3 + 9)/6 = 2; V(X) = 1 (the memorised check holds). Points: E = 4 · 1 = 4. Variance: (a² − a)² = 36 ⇒ a² − a = 6 ⇒ (a − 3)(a + 2) = 0 ⇒ a = 3.

All four variants, one table

Variant	Q1 given posterior	Q1 n	Q2 P(A), P(B)	Q2 answer	Q3 points each	Q3 (i)	Q3 (ii) a
1	1/7	4	1/3, 3/4	2/11 = 0.18182	5	5	2
2	2/11	3	2/3, 1/4	1/2 = 0.50000	6	6	2
3	2/17	5	1/4, 2/3	3/11 = 0.27273	7	7	2
4	1/4	2	3/4, 1/3	1/3 = 0.33333	8	8	2

Q1: the posterior is always 2/(3n + 2), so n = (2/p − 2)/3 — an integer in every variant. Q2: numerator (1 − P(A))(1 − P(B)), denominator 1 − P(A)(1 − P(B)). Q3: (i) equals the points-per-person, (ii) is 2 regardless of the constant. Every variant re-verified.

Next time: the 30-minute plan for an answer-only paper

Q2 first (5 marks, 5 minutes) — one identity and two multiplications. Then Q3 (10 marks, 12 minutes): write the 5-row table, then E, E(X²), V, then the two answers. Q1 last: the decoding costs thinking time and the algebra is short once decoded.
Integer alarms. Q1's n and Q3's a must be whole numbers. Q3's table must sum to 24/24. Q2's answer must be in [0, 1] and below P(B^c).
Write the fraction and the five-place decimal both — the key does; it protects you from a rounding dispute.
Do not build a decision tree for Q3 from scratch under the clock — the counts 9 · 8 · 6 · 0 · 1 (and 2 · 3 · 0 · 1 for three people) are worth having in memory, along with E = V = 1.

Ask an AI well · about this paper

"Devore §2.4: n urns have 4W6B and one urn has 5W5B; two balls drawn from a random urn are both black; posterior that the odd urn was chosen is 1/7 — walk me to n with the likelihoods built by counting, stop after each step." · "Devore §3.3: X = number of fixed points of a random permutation of 4; check my pmf 9/24, 8/24, 6/24, 0, 1/24 and my V(X) = 1, then Var((a²+a)X + 5) = 36." Stay within §2.1–2.5 and §3.1–3.4.