How to start a probability problem

• The routine
• Word cues → model
• Phrases → symbols
• Shooters
• Two coins
• 30 units
• Particles
• Chips
• Servers
• Battery
• Coin test
• In the exam room

1 · The routine: four moves before any thinking

When you read a question and freeze, the freeze is usually a search for "the idea". The routine replaces that search with four writing jobs. Each one is small, each one is always possible, and after the fourth the problem has turned into a formula with blanks to fill.

Questions come in two kinds, and the routine has a version for each. Event questions talk about things happening or not (a shot hits, a chip is defective, a coin is fair). Random-variable questions talk about a number that varies (the number of defectives, a battery's life in hours, how many heads).

For event questions

1. LettersGive every event in the story a capital letter and write what it means in words. A = "X hits the target". If the story has two sources (two machines, two coins), each source gets a letter.
2. GivensTurn every number in the question into a P( ) statement. "5% of M₁'s chips are defective" is P(D | M₁) = 0.05 — a percentage "of those that …" is a conditional probability. Words carry givens too: "independently" means P(A ∩ B) = P(A)P(B); "at random, equal probability" means the sources are ½ each.
3. TargetTurn the question sentence into P(… | …). What comes after "given that" goes on the right of the bar; what you're asked the probability of goes on the left. If a phrase has no letter yet ("exactly one hit"), give it one now and write what it is in terms of your letters.
4. FormulaWrite the definition or rule for your target, with the letters in: P(B | E) = P(B ∩ E) / P(E), or Bayes, or the complement rule. Now every piece is something you can read off a table or a tree.

For random-variable questions

1. Name the rvIn words, starting "X = the number of …" or "X = the time until …". Include the unit and the "of what": "X = number of the 4 assigned units that need a speed test".
2. Name the distribution, with parameters"X ~ Bin(10, 0.07)", "X ~ Poisson(2)", "X ~ Exp(mean 20 h)". Add a one-line reason from the word cues in the table below ("drawn without replacement from a fixed pool of 30 → hypergeometric"). This line is the "specify all distributions properly" mark.
3. Target with the exact inequalityTranslate the words into P(X …) and write the inequality carefully: "at least 3" is X ≥ 3, "less than 700" is X ≤ 699 for a count, "no more than 15 hours" is X ≤ 15.
4. FormulaWrite the pmf, cdf or approximation you'll use, with the parameters in: P(X = x) = C(20, x) C(10, 4 − x) / C(30, 4). If the target is "at least one", write the complement here: 1 − P(X = 0).

Many mid-sem questions stack both kinds: an event question whose likelihood comes from a random variable (the chips and servers questions below). Then you run the routine twice, and the second run's answer feeds the first. The servers opening shows it.

2 · Word cues → which model

Move 2 of the random-variable routine is the one people fear, because it feels like it needs judgement. Mostly it doesn't: the question tells you, through a few set phrases. Look for these.

3 · Phrases → symbols

Move 3 is translation. This dictionary covers almost every phrase on the three past papers. With letters A, B for two events, or a count X:

4 · Opening 1 — the two shooters (slowly)

This is the one that felt totally new. Knowing "it's conditional probability" is true but doesn't tell you what to write, so here is what to write, in order. Nothing below needs an idea; each line is a translation of a sentence in the question.

1. LettersA = "X hits the target", B = "Y hits the target". (Not X and Y themselves: those are people, and in this course capital X, Y usually mean random variables. Fresh letters avoid the tangle.)
2. GivensP(A) = 0.7, P(B) = 0.4. The word "independently" is a given too: P(A ∩ B) = 0.7 × 0.4 = 0.28. From these, every one of the four "who hit" boxes is a product, so draw the 2×2 table (figure). The four boxes add to 1 — your first in-room check.
3. Target(a) "Exactly one shot hit" has no letter yet, so give it one: E = (A ∩ Bᶜ) ∪ (Aᶜ ∩ B). "It was Y's shot" is B. "Given that" puts E on the right: the target is P(B | E).
   (b) "The target is hit" means at least one hit: H = A ∪ B. The target is P(B | H).
4. FormulaThe definition of conditional probability, letters in:

   P(B | E) = P(B ∩ E)P(E)

   Now read the pieces off the table. B ∩ E means "Y hit, and exactly one hit", so X missed: that's the single box Aᶜ ∩ B = 0.12. E is the two shaded boxes, 0.42 + 0.12. For (b), B sits inside H (if Y hit, the target was hit), so B ∩ H = B and the formula becomes P(B) / P(A ∪ B), where P(A ∪ B) = 1 − 0.18 (everything except "both miss").

Stop here. The rest is one division per part: 0.12 / 0.54 and 0.4 / 0.82. Full solution, answers and traps →

Look at what happened: the "idea" you were waiting for (the 2×2 table, the single box) appeared as a result of writing moves 1–3, not before them. That's the point of the routine.

5 · Opening 2 — the gambler's two coins

1. LettersTwo sources: F = "the fair coin was chosen", T = "the two-headed coin was chosen". Evidence: H₁, H₂ = heads on flip 1, flip 2; T₃ = tails on flip 3.
2. Givens"At random (that is, with equal probability)": P(F) = P(T) = ½. What each coin does: P(H | F) = ½, P(H | T) = 1, P(tails | T) = 0. Flips of the same coin are independent once you know which coin, so P(H₁ ∩ H₂ | F) = ¼.
3. Target(a) P(F | H₁). (b) P(F | H₁ ∩ H₂). (c) P(F | H₁ ∩ H₂ ∩ T₃). Same shape each time: only the evidence grows.
4. FormulaBayes with two sources, evidence e:

   P(F | e) = P(e | F) P(F)P(e | F) P(F) + P(e | T) P(T)

   Each part is this formula with a different P(e | ·): for (b) it's (½)² versus 1².

Stop here. The rest is fractions: ¼ / (¼ + ½), then the same with ⅛. For (c), notice what P(tails | T) = 0 does. Full solution →

6 · Opening 3 — thirty units, four assigned

This is a random-variable question: it asks "how many".

1. Name the rvX = the number of the 4 assigned units that need a speed test.
2. DistributionX is hypergeometric with N = 30 units, M = 20 of the "speed test" kind, n = 4 drawn. Reason: "without replacement" from a fixed pool of two kinds (the cue table's first row). Not binomial: each unit taken changes what's left.
3. Target(a) "none" → P(X = 0). (b) "only 2" → P(X = 2). (c) "at least 3" → P(X ≥ 3) = P(X = 3) + P(X = 4) (4 is the most possible).
4. FormulaChoose which speed-test units and which others, over all ways to choose 4:

   P(X = x) = C(20, x) · C(10, 4 − x)C(30, 4)

Stop here. The rest is C(30, 4) = 27 405 and three numerators. Full solution →

7 · Opening 4 — particles on five squares

1. Name the rvsX = number of particles in one given 1 m² region. For (b), a second one: Y = number of the 5 regions that receive no particles.
2. DistributionsX ~ Poisson(λ × area) = Poisson(2 × 1) = Poisson(2) — a rate per square metre, counting how many. Y ~ Bin(5, p₀) with p₀ = P(X = 0): five independent regions (given), each one "empty" or "not empty", same chance each — fixed n, independent, count the successes.
3. Target(a) "at most one" → P(X ≤ 1) = P(X = 0) + P(X = 1). (b) "exactly three of the five receive none" → P(Y = 3).
4. FormulaP(X = k) = e⁻² 2^k / k!, so p₀ = e⁻². Then P(Y = 3) = C(5, 3) p₀³ (1 − p₀)².

Stop here. The rest is 3e⁻² and one binomial term. Full solution →

8 · Opening 5 — chips from two machines (22 marks)

🎧 Watch it done first: the narrated walkthrough, Bayes as areas (about 5 min, sound on) acts out the four moves on this question, and pauses three times for your next move.

Three parts, three different openings. Run the routine once per part.

Part (a) — an event question

1. LettersM₁, M₂ = "the chip came from machine 1 / 2"; D = "the chip is defective".
2. GivensP(M₁) = 0.6, P(M₂) = 0.4. "5% of its chips" is conditional: P(D | M₁) = 0.05, P(D | M₂) = 0.10.
3. Target"Found to be defective" is the given; "produced by M₂" is asked: P(M₂ | D).
4. FormulaBayes: P(M₂ | D) = P(D | M₂)P(M₂) / [P(D | M₁)P(M₁) + P(D | M₂)P(M₂)]. The denominator is P(D), the factory's overall defect rate — keep it, part (c) needs it.

Part (b) — an event question whose likelihood is a random variable

1. Letters and rvSame M₁, M₂ — but now for "which machine was selected". X = number of defective chips among the 5 tested.
2. Givens"A machine is selected randomly" → P(M₁) = P(M₂) = ½. This is a new prior: the 60/40 was about chips, not about picking a machine. Given M₁, the shipment of 100 has 5 defectives, so X | M₁ is hypergeometric (N = 100, M = 5, n = 5) — "without replacement" again. Given M₂: 10 defectives, M = 10.
3. TargetThe evidence is X = 2: P(M₁ | X = 2).
4. FormulaThe same Bayes shape as (a), with likelihoods from the hypergeometric: P(X = 2 | M₁) = C(5, 2) C(95, 3) / C(100, 5) and P(X = 2 | M₂) = C(10, 2) C(90, 3) / C(100, 5).

Part (c) — a random-variable question

1. Name the rvY = number of defective chips among the 10 inspected.
2. DistributionY ~ Bin(10, p) with p = P(D) from part (a): 10 chips, inspected independently, same chance each.
3. Target"At most 2" → P(Y ≤ 2) = P(0) + P(1) + P(2).
4. FormulaP(Y = k) = C(10, k) p^k (1 − p)^{10 − k}.

Stop here. Each part is now arithmetic. Full solution, and the two misreadings that give wrong answers →

9 · Opening 6 — the servers (14 marks, three routines stacked)

This looks like the hardest question on the paper, and it's the one where the routine helps most, because it breaks into three small openings you've already seen. Work from the question sentence backwards.

1. Letters (the outer event question)E₁, E₂ = "the rack is from Enigma-1 / Enigma-2". A = "at least 2 of the 10 servers have failed by 1 500 h". Givens: P(E₁) = 0.8, P(E₂) = 0.2. Target: "posterior probability … Enigma-2" is P(E₂ | A). Formula: Bayes, which needs P(A | E₁) and P(A | E₂). Those are the two gaps; the next moves fill them.
2. Inner rv: one server's lifeT = failure time of one server, hours. Given E₁, T ~ Exp(mean 12 000), rate 1/12 000. Target: p₁ = P(T ≤ 1 500 | E₁), the chance one server has failed by then. Formula: 1 − e^−t/mean. Same for E₂ with mean 5 000.
3. Inner rv: how many of the tenN = number of the 10 servers failed by 1 500 h. Given the maker, N ~ Bin(10, p_i): ten servers, fail independently (given), same chance each. A = {N ≥ 2}, and "at least" → complement: P(A | E_i) = 1 − P(N = 0) − P(N = 1).
4. Formula, assembledPut the two likelihoods from move 3 into the Bayes formula from move 1. Everything is now defined, and the answer is arithmetic.

Stop here. Full solution with every number →

10 · Opening 7 — the battery

1. Name the rvsX = battery life in hours, talk-time use. Y = battery life in hours, internet use.
2. Distributions"Gamma with α = 1" is the exponential (cue table, row 4). "20 hours on an average" is the mean, so X ~ Exp(mean 20), rate 1/20; Y ~ Exp(mean 7), rate 1/7.
3. Target(a) "no more than 15 hours" → P(X ≤ 15). (b) "used for 5 hours and still functioning" is the given, Y > 5; "at least another 3 hours" means lasting to 5 + 3 = 8 hours in total: P(Y ≥ 8 | Y > 5).
4. Formula(a) cdf 1 − e^−x/20. (b) The definition P(Y ≥ 8 ∩ Y > 5) / P(Y > 5) = P(Y ≥ 8)/P(Y > 5), which memorylessness shortens to P(Y ≥ 3). Writing the definition first earns the marks even if you forget the shortcut.

Stop here. Full solution →

11 · Opening 8 — testing the coin 1 000 times

1. Name the rvX = number of heads in the 1 000 tosses.
2. Distribution"If the coin is actually fair": X ~ Bin(1000, 0.5) — 1 000 independent tosses, same chance, count heads. "Approximate" with n this large, and a Φ value printed on the paper → normal approximation with μ = np = 500, σ = √(npq) = √250.
3. TargetA false conclusion for a fair coin is concluding "biased", which the test does when heads is 525 or more: P(X ≥ 525).
4. FormulaContinuity correction: the count 525 is the bar from 524.5 to 525.5, so P(X ≥ 525) ≈ 1 − Φ((524.5 − 500)/√250). That z rounds to 1.55 — exactly the Φ value the paper prints, which confirms you're on the right track.

Stop here. Full solution →

12 · In the exam room

This routine also works on the fear. Blank-page panic comes from waiting for an idea with nothing to do while you wait. With the routine there is always a next line to write, and writing it is progress. By the time the four lines are down, the question has usually turned into one you recognise.

If the fear ever feels bigger than ordinary exam nerves, the campus counselling service is there for exactly that, and plenty of first-years use it.