Mathematics Advanced • Year 12 • Module 5 • Lesson 2
Probability Rules
Apply the addition, multiplication and complement rules to real-world contexts: insurance, sport, manufacturing, marketing and IT security.
Problem 1 — Insurance risk
An insurer assesses a coastal property. Their data shows P(flood claim in a year) = 0.06, P(fire claim in a year) = 0.03 and P(both in same year) = 0.005.
Set up: What are we solving for?
(i) Find P(at least one claim in a given year) using the addition rule. 2 marks
(ii) Find P(no claim in a given year) two ways: (1) using the complement of part (i); (2) using P(flood′ ∩ fire′) directly. Confirm they agree. 3 marks
(iii) A naive analyst computes P(at least one claim) by adding 0.06 + 0.03 = 0.09. State the percentage error this causes and explain in one sentence why ignoring the overlap inflates the answer. 2 marks
Stuck? Revisit lesson § Real-World Anchor — Insurance Risk.Problem 2 — Basketball free throws
A basketball player makes 80% of her free throws independently. She takes 6 free throws in a game.
Set up: What are we solving for?
(i) Find P(she makes all 6). 1 mark
(ii) Find P(she misses at least one) using the complement. 2 marks
(iii) Across an 82-game season she takes 6 free throws per game. Find the expected number of games in which she makes all 6. State your assumption about independence between games. 3 marks
Problem 3 — Two-machine production line
Machine A produces 60% of all items with a 2% defect rate. Machine B produces 40% with a 5% defect rate. An item is drawn at random and inspected.
Set up: What are we solving for?
(i) Draw a fully labelled tree diagram showing P(A), P(B), and the conditional defect / non-defect probabilities for each branch. 2 marks
(ii) Find P(defective) and P(non-defective). Verify they sum to 1. 3 marks
(iii) Three items are inspected (with replacement, so draws are independent). Find P(all three non-defective). 2 marks
Stuck on (i)? Use 2 first-level branches (A, B) then 2 second-level branches each (def, def′).Problem 4 — Marketing campaign
A retailer mails 1 000 customers a discount code. Long-run data shows P(customer opens email) = 0.40 and P(customer redeems code | opens email) = 0.18. Customers who never open the email never redeem.
Set up: What are we solving for?
(i) Find P(a randomly chosen customer redeems the code). 2 marks
(ii) Expected number of redemptions from the 1 000 mailings. 2 marks
(iii) The retailer wants ≥ 100 redemptions. They cannot change the open-rate, but they can change the redemption rate per opener via a more attractive discount. What is the minimum value of P(redeems | opens) needed to hit 100 redemptions on average from 1 000 mailings? 2 marks
Problem 5 — IT password security
A 4-digit PIN code is randomly assigned (digits 0–9, each independent). An attacker tries random PINs without replacement until they succeed.
Set up: What are we solving for?
(i) Find n(S), the size of the sample space of 4-digit PINs. 1 mark
(ii) Find P(at least two digits are the same) using the complement (i.e. P(all four digits distinct)). 3 marks
(iii) An attacker gets 3 attempts before the account is locked. They try three distinct random PINs. Find P(they unlock the account). 3 marks
Stuck on (iii)? Without replacement of guesses, the three attempts are 3 distinct PINs out of 10 000.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Insurance risk
Set up. We are computing the probability of at least one claim and its complement, using the addition rule on two events with given overlap.
(i) P(at least one claim) = 0.06 + 0.03 − 0.005 = 0.085.
(ii) Method 1: P(no claim) = 1 − 0.085 = 0.915. Method 2: P(flood′) and P(fire′) aren't independent in general, but using P(flood′ ∩ fire′) = 1 − P(flood ∪ fire) = 1 − 0.085 = 0.915 ✓ (same value via De Morgan).
(iii) Naive value = 0.09; true value = 0.085. Percentage error = (0.09 − 0.085)/0.085 ≈ 5.9% overestimate. The overlap P(flood ∩ fire) was added in twice, so adding 0.06 + 0.03 double-counts properties exposed to both hazards.
Problem 2 — Basketball free throws
Set up. Independent Bernoulli trials with p = 0.8; we need joint and complement probabilities, then a season-level expectation.
(i) P(makes all 6) = (0.8)⁶ = 0.2621 (4 d.p.).
(ii) P(misses at least one) = 1 − P(makes all 6) = 1 − 0.2621 = 0.7379.
(iii) Assume games are independent and the shooter maintains 80%. Expected perfect-from-the-line games = 82 × 0.2621 ≈ 21.5 games. Assumption: shot independence within and across games (no fatigue, no clutch effect).
Problem 3 — Two-machine line
Set up. Use a tree diagram with machine source as the first stage and defect status (conditional on machine) as the second.
(i) Branches: A (0.6) → def (0.02) and def′ (0.98); B (0.4) → def (0.05) and def′ (0.95). Each branch labelled with its conditional probability.
(ii) P(defective) = 0.6 × 0.02 + 0.4 × 0.05 = 0.012 + 0.020 = 0.032. P(non-defective) = 0.6 × 0.98 + 0.4 × 0.95 = 0.588 + 0.380 = 0.968. Sum = 0.032 + 0.968 = 1 ✓.
(iii) P(all three non-defective) = (0.968)³ ≈ 0.907.
Problem 4 — Marketing campaign
Set up. Tree with stages "open email" then "redeem | opens". Redeeming requires opening, so the other branches contribute zero.
(i) P(redeems) = P(opens) × P(redeems | opens) = 0.40 × 0.18 = 0.072.
(ii) Expected redemptions = 1 000 × 0.072 = 72 redemptions.
(iii) Need 1 000 × 0.40 × p ≥ 100 ⇒ p ≥ 100/(1000 × 0.40) = 100/400 = 0.25. So P(redeems | opens) must rise from 18% to at least 25%.
Problem 5 — PIN security
Set up. Sample space is all 4-digit codes; we need the complement of "all distinct" and the probability of cracking with multiple guesses.
(i) n(S) = 10⁴ = 10 000.
(ii) P(all four distinct) = (10 × 9 × 8 × 7) / 10 000 = 5040/10 000 = 0.504. So P(at least two equal) = 1 − 0.504 = 0.496.
(iii) With three distinct guesses out of 10 000 possible PINs, P(unlock) = 3/10 000 = 0.0003. (Equivalently, 1 − P(all three guesses miss) = 1 − (9999/10000)(9998/9999)(9997/9998) = 1 − 9997/10000 = 3/10 000 ✓.)