Mathematics Advanced • Year 12 • Module 5 • Lesson 4
Independence and Mutual Exclusivity
Apply the independence and mutual-exclusivity tests to drug trials, sports, surveys, networks and lottery scenarios.
Problem 1 — Drug-trial side effects
A new medication produces two recorded side effects: nausea (N) with P(N) = 0.20 and headache (H) with P(H) = 0.25. Trial data shows P(N ∩ H) = 0.05.
Set up: What are we solving for?
(i) Test whether N and H are independent. 2 marks
(ii) Find P(at least one side effect) and P(neither side effect). 2 marks
(iii) A second drug records survival (S) and death (D) for trial patients. Explain in 1-2 sentences why S and D are mutually exclusive but cannot be independent in a meaningful trial. 2 marks
Stuck? Revisit lesson § Real-World Anchor (drug trials).Problem 2 — Tennis serves
A tennis player's first-serve success rate is P(1st in) = 0.6. If the first serve fails, the second-serve success rate is P(2nd in | 1st out) = 0.85. Successive points are independent.
Set up: What are we solving for?
(i) Find P(double fault) — both serves out. 2 marks
(ii) Are "1st serve in" and "1st serve out" independent? Justify in one sentence. 1 mark
(iii) Across a 12-point service game, find P(at least one double fault). Assume points are independent and each point follows the structure above. 3 marks
Problem 3 — Subject and gender (independence from a table)
A school surveys 400 students on whether they take Economics (E). The data is broken down by gender:
| Takes Econ (E) | Does not take Econ | Total | |
|---|---|---|---|
| Female (F) | 96 | 144 | 240 |
| Male (M) | 64 | 96 | 160 |
| Total | 160 | 240 | 400 |
Set up: What are we solving for?
(i) Find P(F), P(E) and P(F ∩ E). 2 marks
(ii) Test whether "is female" and "takes Economics" are independent. 2 marks
(iii) Suppose the school added 40 more female students, all of whom take Economics. Recompute the table and test independence again. Comment on whether independence is preserved under arbitrary data changes. 3 marks
Stuck on (iii)? Recompute P(F), P(E) and P(F ∩ E) with the new totals.Problem 4 — Server-network reliability
A small data centre has three identical servers. Each server independently has a probability 0.02 of failing on a given day.
Set up: What are we solving for?
(i) Find P(all three servers fail on the same day). State the independence assumption you used. 2 marks
(ii) Find P(at least one server fails on a given day). 2 marks
(iii) In reality, servers in the same room share a cooling system, so failures are positively correlated (not independent). Without recomputing, state whether your answer to (i) is an over- or under-estimate of the true joint failure probability, and explain in one sentence. 2 marks
Problem 5 — Weekly lottery results
A weekly lottery draws 6 numbered balls from a pool of 45. Let A = "ball number 7 is drawn this week" and B = "ball number 7 is drawn next week".
Set up: What are we solving for?
(i) Find P(A) and P(B). 2 marks
(ii) Argue carefully that A and B are independent, and hence find P(A ∩ B). 2 marks
(iii) A "gambler's fallacy" punter claims: "Ball 7 was drawn last week, so it's less likely this week — the lottery balances out." Identify the error in one sentence and state the correct P(A | "drawn last week"). 2 marks
Stuck? Each week's draw is from a fresh full pool — past draws do not influence future draws.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Drug trial
Set up. Test independence of N and H from given probabilities; then critique a mutually exclusive vs independent claim for survival/death.
(i) P(N) × P(H) = 0.20 × 0.25 = 0.05 = P(N ∩ H) ✓ → N and H are independent (they occur together exactly as often as chance would predict).
(ii) P(at least one) = P(N ∪ H) = 0.20 + 0.25 − 0.05 = 0.40. P(neither) = 1 − 0.40 = 0.60.
(iii) S and D are mutually exclusive (a patient cannot both survive and die). With P(S), P(D) > 0, the lesson theorem says they cannot also be independent — knowing the patient survived gives total information that they did not die.
Problem 2 — Tennis serves
Set up. Conditional structure on second serve; then independent points across a game.
(i) P(double fault) = P(1st out) × P(2nd out | 1st out) = (1 − 0.6) × (1 − 0.85) = 0.4 × 0.15 = 0.06.
(ii) "1st serve in" and "1st serve out" are not independent — they are mutually exclusive (one is the complement of the other), so knowing one determines the other; mutually exclusive events with positive probability are dependent.
(iii) P(no double fault on a single point) = 1 − 0.06 = 0.94. Across 12 independent points: P(no DF in game) = (0.94)¹² ≈ 0.476. So P(at least one DF) = 1 − 0.476 = 0.524.
Problem 3 — Gender and Economics
Set up. Use the table to compute marginals and the joint, then test independence; re-run the test after a data change.
(i) P(F) = 240/400 = 0.6; P(E) = 160/400 = 0.4; P(F ∩ E) = 96/400 = 0.24.
(ii) P(F) × P(E) = 0.6 × 0.4 = 0.24 = P(F ∩ E) ✓ → independent. Gender does not predict Economics enrolment in this sample.
(iii) New totals: F = 280, M = 160; E = 200, not E = 240; F ∩ E = 136. Total = 440. P(F) = 280/440 ≈ 0.636; P(E) = 200/440 ≈ 0.455; P(F)·P(E) ≈ 0.289; P(F ∩ E) = 136/440 ≈ 0.309. These differ — no longer independent. Independence is sensitive to data changes: adding 40 economics-taking females tips the balance so that "female" now predicts higher chance of Economics.
Problem 4 — Server reliability
Set up. Independent Bernoulli trials, then a critique of the independence assumption.
(i) Assuming failures are independent across servers: P(all three fail) = (0.02)³ = 0.000008 = 8 × 10⁻⁶.
(ii) P(at least one fails) = 1 − P(none fail) = 1 − (0.98)³ ≈ 1 − 0.9412 = 0.0588.
(iii) The independence assumption under-estimates the true joint failure probability. Positive correlation (shared cooling) means failures cluster: a cooling problem can drop all three servers at once, so the true P(all three fail) is higher than the 8 × 10⁻⁶ predicted by the independence model.
Problem 5 — Lottery
Set up. Each week's draw is from a fresh pool of 45 balls; we need single-event and joint probabilities, then a gambler's-fallacy critique.
(i) P(A) = 6/45 = 2/15 (chance that ball 7 is one of the 6 drawn). Same for next week: P(B) = 2/15.
(ii) Different weeks use independent draws — last week's outcome is physically irrelevant to next week's. So P(A ∩ B) = P(A) × P(B) = (2/15)² = 4/225 ≈ 0.0178.
(iii) The error is the gambler's fallacy — assuming past random draws affect future ones. By independence, P(A | "ball 7 drawn last week") = P(A) = 2/15 ≈ 0.133, exactly the same as the unconditional probability.