Mathematics Advanced • Year 12 • Module 5 • Lesson 5
Discrete Probability Distributions
Past-paper style: probability distribution tables, E(X), Var(X), and an extended response on designing and justifying a game.
1. Short-answer questions
1.1 The random variable X has the following distribution:
| x | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| P(X = x) | 0.1 | 0.25 | k | 0.2 | 0.15 |
(a) Find k. (b) Find E(X). (c) Find Var(X). 3 marks Band 3
1.2 The random variable X represents the number of tails when 3 fair coins are flipped. Construct the probability distribution table for X and find E(X). 3 marks Band 3-4
1.3 If E(X) = 4 and Var(X) = 9, find E(2X − 3) and SD(2X − 3). Justify each step with the correct rule. 3 marks Band 4
Stuck on 1.3? Var(aX + b) = a²Var(X) and SD(aX + b) = |a|·SD(X) — the shift b cancels in both.2. Extended response
2.1 You are designing a charity-night dice game. Customers pay an entry fee of $c per play. They roll two fair dice. Let S = sum of the two dice. The game pays:
- $25 if S = 12
- $10 if S ∈ {2, 11}
- $3 if S ∈ {3, 4, 10}
- $0 otherwise
(a) Construct the probability distribution for the payout W, including all distinct outcomes.
(b) Find E(W).
(c) Find Var(W) and SD(W).
(d) Determine the entry fee c that makes the game fair (E(profit) = 0 for the player). Then explain in 1-2 sentences why a charity might choose c = $3.50 (favourable to the charity), what expected charity profit results from 500 plays, and what role variance plays in the player's experience even when the game is unfavourable on average. 8 marks Band 5-6
Explicit marking criteria
Part (a) — 2 marks
• 1 mark — uses the 36-outcome sample space of two dice; counts each sum correctly: P(S=2)=1/36, P(S=11)=2/36, P(S=12)=1/36, P(S=3)=2/36, P(S=4)=3/36, P(S=10)=3/36.
• 1 mark — combines into payout table: W = 25 with P = 1/36; W = 10 with P = (1+2)/36 = 3/36; W = 3 with P = (2+3+3)/36 = 8/36; W = 0 with P = 24/36. Σ P = 36/36 ✓.
Part (b) — 1 mark
• E(W) = 25(1/36) + 10(3/36) + 3(8/36) + 0(24/36) = (25 + 30 + 24)/36 = 79/36 ≈ $2.19.
Part (c) — 2 marks
• 1 mark — E(W²) = 625(1/36) + 100(3/36) + 9(8/36) + 0 = (625 + 300 + 72)/36 = 997/36 ≈ 27.69.
• 1 mark — Var(W) = 27.69 − (2.19)² ≈ 27.69 − 4.81 ≈ 22.88; SD(W) ≈ $4.78.
Part (d) — 3 marks
• 1 mark — fair fee c = E(W) ≈ $2.19.
• 1 mark — at c = $3.50, E(profit per play for charity) = 3.50 − 2.19 = $1.31; across 500 plays, expected charity income ≈ $655.
• 1 mark — variance comment: even with negative player E(profit), high SD ($4.78) means individual players can still win $20+ on a single play, which creates excitement, encourages repeat plays, and is exactly the design pattern casinos and charity games rely on.
Your response:
When computing E(W²), square the payouts, not the original dice sums.How did this worksheet feel?
What I'll revisit before next class:
1.1 — Find k, E(X), Var(X) (3 marks)
Sample response. (a) Σ P = 1 ⇒ 0.1 + 0.25 + k + 0.2 + 0.15 = 1 ⇒ k = 0.3.
(b) E(X) = 1(0.1) + 2(0.25) + 3(0.3) + 4(0.2) + 5(0.15) = 0.1 + 0.5 + 0.9 + 0.8 + 0.75 = 3.05.
(c) E(X²) = 1(0.1) + 4(0.25) + 9(0.3) + 16(0.2) + 25(0.15) = 0.1 + 1.0 + 2.7 + 3.2 + 3.75 = 10.75. Var(X) = 10.75 − (3.05)² = 10.75 − 9.3025 = 1.4475.
Marking notes. 1 mark — k = 0.3 with the Σ P = 1 condition stated. 1 mark — E(X) correct using Σ x · P. 1 mark — Var(X) correct using E(X²) − [E(X)]². Common error: subtracting E(X) instead of [E(X)]² in the variance.
1.2 — Three-coin flips: tails distribution (3 marks)
Sample response. Sample space has 2³ = 8 equally likely outcomes (HHH, HHT, HTH, …, TTT). Counting tails per outcome:
| x | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| P(X = x) | 1/8 | 3/8 | 3/8 | 1/8 |
E(X) = 0(1/8) + 1(3/8) + 2(3/8) + 3(1/8) = 0 + 3/8 + 6/8 + 3/8 = 12/8 = 1.5.
Marking notes. 1 mark — correct distribution table with all four probabilities summing to 1. 1 mark — correct E(X) calculation. 1 mark — final answer 1.5 with comment that this matches the binomial mean np = 3 × 0.5. Common error: forgetting that x = 0 contributes 0 to the sum but still belongs in the table.
1.3 — Linear transform of E and SD (3 marks)
Sample response. By linearity of expectation, E(2X − 3) = 2 · E(X) − 3 = 2(4) − 3 = 5. By the variance scaling rule, Var(2X − 3) = 2² · Var(X) = 4 · 9 = 36, so SD(2X − 3) = √36 = 6.
Marking notes. 1 mark — E(2X − 3) = 5 using E(aX + b) = aE(X) + b. 1 mark — Var(2X − 3) = 36 using Var(aX + b) = a²Var(X). 1 mark — SD = 6 with the note that the −3 shift does not change spread. Common error: writing SD(2X − 3) = 2 · SD(X) − 3 = 0 — must use SD(aX + b) = |a| · SD(X), shift cancels.
2.1 — Charity dice game (8 marks): sample Band-6 response with annotations
Sample Band-6 response.
(a) Distribution of W. Sample space is 36 ordered pairs from two dice; sum counts are well-known: P(2) = P(12) = 1/36; P(3) = P(11) = 2/36; P(4) = P(10) = 3/36; P(5) = P(9) = 4/36; P(6) = P(8) = 5/36; P(7) = 6/36. [1 mark — uses correct sum-counts on 36 outcomes.]
Mapping to payouts:
| W | 25 | 10 | 3 | 0 |
|---|---|---|---|---|
| P(W) | 1/36 | 3/36 | 8/36 | 24/36 |
Σ P = (1 + 3 + 8 + 24)/36 = 36/36 = 1 ✓. [1 mark — payout table assembled and validated.]
(b) E(W).
E(W) = 25(1/36) + 10(3/36) + 3(8/36) + 0(24/36)
= (25 + 30 + 24)/36 = 79/36 ≈ $2.19. [1 mark]
(c) Var(W) and SD(W).
E(W²) = 625(1/36) + 100(3/36) + 9(8/36) + 0 = (625 + 300 + 72)/36 = 997/36 ≈ 27.69. [1 mark]
Var(W) = 27.69 − (2.19)² ≈ 27.69 − 4.81 = 22.88; SD(W) = √22.88 ≈ $4.78. [1 mark]
(d) Entry fee, charity profit, variance role.
Fair fee: The game is fair when E(player profit) = 0, i.e. c = E(W) ≈ $2.19. [1 mark]
Charity choice c = $3.50: E(profit per play for charity) = 3.50 − 2.19 = $1.31. Over 500 plays, expected charity income ≈ 500 × 1.31 ≈ $655. [1 mark]
Role of variance: Even though the game is unfavourable on average for the player, SD(W) ≈ $4.78 is large relative to the entry fee. This means individual players can still walk away with $25 or $10 wins, which creates excitement, generates word-of-mouth, and encourages repeat plays. Charity (and casino) games exploit this gap between low expected value and high variance to extract revenue while maintaining a sense of "skill / luck / fun" for players. [1 mark — variance comment ties to player experience and design intent.] ▮
Total: 8/8.
Band descriptors for marker.
Band 3: Distribution table constructed but counts incorrect (often misses that S = 3 and S = 11 each occur twice); E(W) computed using flawed table. ≈ 3-4 marks.
Band 4: Table correct, E(W) ≈ $2.19; Var(W) correct numerically; (d) gives fair fee but no commentary. ≈ 5-6 marks.
Band 5: All numerical parts complete; (d) mentions fair fee and charity profit; variance comment present but generic. ≈ 6-7 marks.
Band 6: All numerical parts complete with units ($); variance comment links explicitly to player psychology / repeat plays / casino-style design rationale. 8/8.