Lesson 15: Probability

Probability is the mathematics of uncertainty — quantifying 'how likely?' with the same rigour as algebra. A-level probability builds on GCSE foundations into conditional probability, Bayes-like reasoning, and the systematic use of Venn diagrams and tree diagrams for multi-stage problems. This lesson develops the rules from first principles, emphasises the independence/exclusivity distinction that examiners probe relentlessly, and applies probability to medical testing, risk, and everyday decisions.

The rules — from axioms to tools

Probability satisfies: 0 ≤ P(A) ≤ 1; P(S) = 1 for the sample space; P(A') = 1 − P(A). The addition rule: P(A ∪ B) = P(A) + P(B) − P(A ∩ B) — subtract the double-counted overlap (proved by counting or by Venn areas). For mutually exclusive events (A ∩ B = ∅), P(A ∪ B) = P(A) + P(B). The multiplication rule: P(A ∩ B) = P(A) × P(B|A) — always true; it simplifies to P(A)P(B) only when independent.

Venn diagrams organise two- or three-set problems. Example: in a sixth form of 200, 120 study maths (M), 90 study physics (P), 50 study both. Then P(M ∪ P) = (120 + 90 − 50)/200 = 160/200 = 0.8; P(neither) = 0.2; P(M only) = 70/200 = 0.35. For three sets, fill from the centre outward: all-three region first, then pairwise-only, then singles-only — systematic filling prevents double-counting, the classic Venn error.

Two-way tables are Venn diagrams in disguise: rows/columns for categories, with totals as the check. Always verify totals sum correctly — a table that does not balance has an error. Examiners give tables with a missing entry to find via the totals; the totals are your equation.

Conditional probability

P(A|B) = P(A ∩ B)/P(B): the probability of A given B occurred — restrict the sample space to B. Example: a fair die is rolled; given the result is even, what is P(> 3)? Restrict to {2, 4, 6}: two of three exceed 3, so P = 2/3. Via formula: P(> 3 ∩ even) = P({4, 6}) = 1/3; P(even) = 1/2; ratio = 2/3 ✓. The formula and the restriction are the same idea — use whichever is faster.

Worked example (conditional from a table). In a survey of 200 adults, 90 hold a driving licence, 130 own a car, and 70 both hold a licence and own a car. Find P(owns a car | holds a licence). Restrict to the 90 licence holders: 70 of them own a car, so P = 70/90 = 7/9 ≈ 0.78. Note P(owns a car) overall is 130/200 = 0.65 — conditioning on the licence raised the probability, exactly what 'given that' means.

Worked: 5% of emails are spam; the filter flags 98% of spam and 2% of genuine mail. P(flagged|spam) = 0.98, P(flagged|genuine) = 0.02. If 1,000 emails arrive: 50 spam (49 flagged), 950 genuine (19 flagged). So P(spam|flagged) = 49/(49 + 19) ≈ 0.72 — nearly a third of flagged mail is genuine! This is Bayes' theorem in action: P(spam|flagged) = P(flagged|spam)P(spam)/P(flagged). Base rates dominate: rare events stay unlikely even after positive tests.

The prosecutor's fallacy: confusing P(evidence|innocent) with P(innocent|evidence). A DNA match with 1-in-a-million random-match probability does not mean 1-in-a-million innocence — in a city of 8 million, ~8 innocent matches are expected. Courts have convicted on this confusion. Conditional probability is not academic: it is justice.

Independence vs exclusivity

Independent: P(A|B) = P(A) — B's occurrence does not affect A; equivalently P(A ∩ B) = P(A)P(B). Mutually exclusive: P(A ∩ B) = 0 — they cannot co-occur. These are different ideas, and confusing them is the top A-level probability error. Note: mutually exclusive events with positive probability are NEVER independent (if A occurred, B certainly did not — maximal dependence!). Example: rolling a 6 and rolling an odd number are mutually exclusive but dependent; rolling a 6 on two separate dice are independent but not exclusive.

Testing independence: check P(A ∩ B) = P(A)P(B). Example: cards — A = 'red', B = 'king'. P(A ∩ B) = 2/52 (red kings); P(A)P(B) = (1/2)(4/52) = 2/52 ✓ — independent (colour and rank do not interact). But A = 'red', C = 'heart': P(A ∩ C) = 13/52 = 1/4; P(A)P(C) = (1/2)(1/4) = 1/8 ✗ — dependent (hearts are always red). In context: rain today and rain tomorrow are dependent (weather persists); two roulette spins are independent (the wheel has no memory — the gambler's fallacy).

Sampling with vs without replacement: drawing two cards without replacement — dependent (the deck changes); with replacement — independent. A box of 10 fuses, 3 faulty: P(both faulty, no replacement) = (3/10)(2/9) = 1/15; with replacement = (3/10)² = 9/100. Quality-control sampling is without replacement — adjust the fractions at each draw.

Tree diagrams — multi-stage problems

Trees handle sequential probability: branches with conditional probabilities, path probability = product along the path, event probability = sum of paths. Example: a factory has two machines: A makes 60% of bolts, 2% defective; B makes 40%, 5% defective. P(defective) = 0.6(0.02) + 0.4(0.05) = 0.012 + 0.02 = 0.032. Given a defective bolt, P(from A) = 0.012/0.032 = 0.375 — Bayes via tree: (branch)/(sum of branches).

Worked: medical test — 1% of a population has a condition; test is 99% sensitive (P(pos|sick) = 0.99) and 95% specific (P(neg|healthy) = 0.95). For 10,000 people: 100 sick (99 positive), 9,900 healthy (495 positive). P(sick|positive) = 99/(99 + 495) ≈ 0.167 — only 17%! Most positives are healthy — the base-rate lesson again. This is why screening rare conditions needs confirmatory testing, and why '99% accurate' headlines mislead.

Trees also solve 'at least one' problems via complements: P(at least one six in 4 rolls) = 1 − P(no sixes) = 1 − (5/6)⁴ ≈ 0.518. And expected counts: in 1,000 bolts, expect 32 defective (1000 × 0.032). Draw trees for every multi-stage problem — the diagram is the solution's skeleton, and examiners award method marks for correct branch probabilities even if the final sum slips.

Risk, expected value, and exam technique

Expected value E = Σ x·P(x): the long-run average. A £2 lottery ticket with 1-in-1000 chance of £500: E(winnings) = 500(1/1000) = £0.50 — expected loss £1.50 per ticket. Insurance: a £300 premium for a 1-in-500 chance of a £100,000 payout — insurer's expected profit per policy: 300 − 100000/500 = £100. Expected value prices risk; the loading (£100 here) covers costs and profit. Fair price = expected payout — real prices add margins.

Decision trees combine probability and value: a company considers a £50,000 trial with 70% chance of leading to a £200,000 contract. Expected gain: 0.7(200000) − 50000 = £90,000 > 0 — proceed (risk-neutral). But expected value ignores risk appetite: a 1% chance of £10M has E = £100,000, yet most people would not pay £100,000 for it — utility, not expectation, governs real choices (St Petersburg paradox). Mention this nuance in 'discuss' questions.

Exam technique: (1) define events with letters and state what they mean; (2) write the rule before numbers — P(A ∪ B) = P(A) + P(B) − P(A ∩ B) with substitution; (3) for conditional, write P(A|B) = ... explicitly; (4) never confuse 'given that' direction — P(A|B) ≠ P(B|A) in general; (5) check answers: probabilities ≤ 1, and sanity-check against intuition; (6) in 'show independent', verify the product rule numerically — assertion without verification earns nothing.

The Monty Hall problem

Three doors: a car behind one, goats behind two. You pick door 1; the host (who knows) opens door 3 revealing a goat, and offers the switch to door 2. Switch or stay? Conditional probability says switch: initially P(car at 1) = 1/3. The host's action concentrates the remaining 2/3 probability onto door 2 — because the host deliberately avoids the car. P(win|switch) = 2/3, P(win|stay) = 1/3. Simulations and mathematicians alike confirm it; intuition rebels anyway.

Tree-diagram proof: enumerate the car's true door (1, 2, 3, each 1/3). If car at 1 (prob 1/3): host opens 2 or 3; switching loses. If car at 2 (1/3): host must open 3; switching wins. If car at 3 (1/3): host must open 2; switching wins. Switching wins in 2 of 3 equally likely cases. The key: the host's choice is CONDITIONAL on your pick and the car's location — naive 'two doors, fifty-fifty' reasoning ignores this conditioning.

Common mistakes

• Confusing independent with mutually exclusive. Exclusive events (with positive probability) are maximally dependent: P(A ∩ B) = 0 ≠ P(A)P(B). The two words are not synonyms — examiners probe this relentlessly.
• Reversing the conditional. P(positive|sick) = 0.99 does not make P(sick|positive) = 0.99. The base rate sits between them; ignoring it is the prosecutor's fallacy.
• Adding probabilities down a tree branch. Along a path you multiply; across paths you add. Swapping the two operations is the commonest tree error.
• Forgetting the sample space changes without replacement. After removing one red ball from 5 red and 4 blue, the second draw is out of 8, with 4 red. Unadjusted fractions signal replacement thinking.
• Skipping the complement for 'at least one'. P(at least one) = 1 − P(none) is one line; summing four cases is four chances to slip.

Exam Experience

Questions in the style of the A-level statistics paper. Attempt them before reading the mark scheme.

1. A bag contains 5 red counters and 4 blue counters. Two counters are drawn at random without replacement. (a) Find the probability that both counters are red. [2] (b) Find the probability that the two counters are the same colour. [3] (c) Given that the first counter drawn is red, find the probability that the second counter is also red. [1]

2. Events A and B satisfy P(A) = 0.5, P(B) = 0.4 and P(A ∩ B) = 0.2. Show that A and B are independent. [2]

Mark scheme.

• 1(a) (5/9) × (4/8) (M1) for the product with the second fraction adjusted for no replacement. = 20/72 = 5/18 ≈ 0.278 (A1).
• 1(b) P(both blue) = (4/9) × (3/8) = 12/72 = 1/6 (M1). P(same colour) = 5/18 + 1/6 = 8/18 = 4/9 ≈ 0.444 (A1) — adding the two mutually exclusive cases (A1).
• 1(c) After a red is removed, 4 red remain from 8 counters: P = 4/8 = 1/2 (A1).
• 2. P(A) × P(B) = 0.5 × 0.4 = 0.2 (M1) for computing the product. This equals P(A ∩ B) = 0.2, so A and B are independent (A1) — the numerical verification is the proof; the statement alone earns nothing.