Mathematics Advanced • Year 12 • Module 5 • Lesson 1
Introduction to Probability
Apply probability rules to multi-step word problems drawn from sport, surveys, lotteries, manufacturing and weather forecasting.
Problem 1 — School subject survey
A Year 12 cohort of 200 students was surveyed about elective subjects. 110 study Economics (E), 90 study Geography (G), and 40 study both.
Set up: What are we solving for? (Write the probability symbols you will use.)
(i) Draw a two-circle Venn diagram and fill in all four regions (E only, G only, both, neither). 2 marks
(ii) A student is chosen at random. Find P(E ∪ G), the probability they study Economics or Geography (or both). 2 marks
(iii) Find P(exactly one of E or G) — that is, Economics only or Geography only, but not both. Show this is equal to P(E ∪ G) − P(E ∩ G), and confirm with your Venn-region counts. 3 marks
Stuck? Revisit lesson § Worked Example (sport survey).Problem 2 — Lottery odds
In Oz Lotto, you choose 7 numbers from 1 to 47. The number of distinct unordered selections is C(47, 7) = 62 891 499.
Set up: What are we solving for?
(i) Find P(winning the division-1 jackpot with one ticket). Give your answer as a decimal in scientific notation to 3 significant figures. 1 mark
(ii) If a syndicate buys 1 000 tickets (all distinct number combinations), find P(at least one winning ticket). Use the complement rule. 3 marks
(iii) Tickets cost $1.30 each, and the syndicate wins on average $25 000 000 when they hit the jackpot (after splitting). In one sentence, explain why expected winnings = (probability) × (payout) suggests this is a poor investment. (No formal calculation of expected value needed — that's Lesson 5.) 1 mark
Problem 3 — Weekend weather forecast
A meteorologist gives Saturday a 30% chance of rain (event R) and Sunday a 40% chance of rain (event S). Long-run records show P(R ∩ S) = 0.15 (both days rain).
Set up: What are we solving for?
(i) Find P(R ∪ S), the probability of rain on at least one of the two days. 2 marks
(ii) Find P(no rain all weekend), i.e. P((R ∪ S)′). Then state the relationship between P(R ∪ S) and P((R ∪ S)′) in one line. 2 marks
(iii) A weather app reports "60% chance of rain this weekend" by simply adding 30% + 40% − 10% (incorrect overlap). Identify the error and state the correct figure. 2 marks
Stuck? Revisit lesson § Complementary Events.Problem 4 — Quality control on a phone production line
A factory inspects 500 randomly selected phones. The results are recorded by two defect types — screen flaw (F) and battery flaw (B):
| Battery flaw (B) | No battery flaw (B′) | Total | |
|---|---|---|---|
| Screen flaw (F) | 12 | 28 | 40 |
| No screen flaw (F′) | 18 | 442 | 460 |
| Total | 30 | 470 | 500 |
Set up: What are we solving for?
(i) Find P(F), P(B), and P(F ∩ B). 2 marks
(ii) Find P(at least one defect) = P(F ∪ B). Show both the addition-rule method and the complement method, and confirm they agree. 3 marks
(iii) The plant manager wants the chance of "any defect" to be below 5%. Is the current process meeting that target? Justify in one sentence. 1 mark
Problem 5 — Card-game probabilities
A single card is drawn at random from a standard 52-card deck. Let H = "card is a heart" and K = "card is a king".
Set up: What are we solving for?
(i) Find P(H), P(K) and P(H ∩ K). 2 marks
(ii) Use the addition rule to find P(H ∪ K), the probability of drawing "a heart or a king". 2 marks
(iii) A poker hand of 5 cards is dealt from a fresh deck. Use the complement to find P(at least one ace). (Hint: P(no ace) = C(48,5)/C(52,5).) Give the answer to 3 decimal places. 3 marks
Stuck on (iii)? "At least one" is a complement signal — count hands with no ace and subtract from 1.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Subject survey
Set up. We need P(E ∪ G) and P(exactly one of E or G) from a 2-circle Venn diagram with sample-space size 200.
(i) E only = 110 − 40 = 70; G only = 90 − 40 = 50; both = 40; neither = 200 − (70 + 50 + 40) = 40. Total = 200 ✓.
(ii) P(E ∪ G) = 110/200 + 90/200 − 40/200 = 160/200 = 4/5 = 0.80.
(iii) P(exactly one) = P(E only) + P(G only) = 70/200 + 50/200 = 120/200 = 3/5 = 0.60. Algebraically, P(E ∪ G) − P(E ∩ G) = 160/200 − 40/200 = 120/200 ✓. Venn check: 70 + 50 = 120 ✓.
Problem 2 — Lottery
Set up. Total possible tickets is n(S) = 62 891 499, equally likely; we are finding P(win) and P(at least one win from many tickets).
(i) P(win) = 1 / 62 891 499 ≈ 1.59 × 10⁻⁸.
(ii) P(at least one winning ticket among 1 000) = 1 − P(none win) = 1 − (1 − 1/62 891 499)¹⁰⁰⁰.
Approximating with (1 − p)ⁿ ≈ 1 − np when np is small: P ≈ 1 000 / 62 891 499 ≈ 1.59 × 10⁻⁵, so P(at least one win) ≈ 0.0000159 (about 1 in 62 900). Even with 1 000 tickets the chance is roughly 1.6 in a hundred thousand.
(iii) Expected winnings ≈ 1.59 × 10⁻⁸ × $25 000 000 ≈ $0.40 per ticket, but tickets cost $1.30 — the syndicate loses roughly $0.90 per ticket on average, so it's a poor investment.
Problem 3 — Weekend weather
Set up. Two events R and S with given individual and joint probabilities; we are computing union and complement.
(i) P(R ∪ S) = 0.30 + 0.40 − 0.15 = 0.55 (55% chance of rain on at least one day).
(ii) P((R ∪ S)′) = 1 − 0.55 = 0.45. Relationship: P(R ∪ S) + P((R ∪ S)′) = 1 (an event and its complement partition the sample space).
(iii) The error: the app used overlap = 0.10 instead of the given 0.15, giving 30 + 40 − 10 = 60%. Using the correct overlap 0.15 gives 55%. (Aside: the larger the true overlap, the smaller the union — so under-estimating overlap inflates the predicted chance of rain.)
Problem 4 — Phone quality control
Set up. A 2 × 2 contingency table; we need single-event, joint, and union probabilities, then a yes/no judgement against a tolerance.
(i) P(F) = 40/500 = 0.08; P(B) = 30/500 = 0.06; P(F ∩ B) = 12/500 = 0.024.
(ii) Addition rule: P(F ∪ B) = 0.08 + 0.06 − 0.024 = 0.116. Complement: P(no defect) = 442/500 = 0.884, so P(at least one defect) = 1 − 0.884 = 0.116 ✓.
(iii) P(any defect) = 11.6%, which is well above 5% — the process is not meeting the target.
Problem 5 — Cards
Set up. Single-card probabilities use 52 outcomes; multi-card probabilities use combinations.
(i) P(H) = 13/52 = 1/4; P(K) = 4/52 = 1/13; P(H ∩ K) = 1/52 (king of hearts).
(ii) P(H ∪ K) = 13/52 + 4/52 − 1/52 = 16/52 = 4/13.
(iii) P(no ace) = C(48,5) / C(52,5) = 1 712 304 / 2 598 960 ≈ 0.659. So P(at least one ace) = 1 − 0.659 ≈ 0.341.