Mathematics Advanced • Year 12 • Module 5 • Lesson 5

Discrete Probability Distributions

Build fluency in probability-distribution tables, E(X), Var(X), SD(X), and linear transforms aX + b.

Build · Skill Drill

1. Quick recall

Answer each question in the space provided. 1 mark each

Q1.1 Complete the formulas. Let X be a discrete random variable with possible values x and probabilities P(X = x).

E(X) = μ = Σ ____ · ____

Var(X) = E(X²) − ____² where E(X²) = Σ ____ · ____

SD(X) = √____

Q1.2 Valid distribution check. State the two conditions a function P(X = x) must satisfy:

(1) For every x: 0 ____ P(X = x) ____ 1.

(2) Σ P(X = x) = ______.

Q1.3 Linear transforms. If Y = aX + b, fill the formulas: E(Y) = ______ · E(X) + ______; Var(Y) = ______ · Var(X); SD(Y) = ______ · SD(X).

Stuck? Revisit lesson § Copy Into Your Books — all four formulas are listed there.

2. Worked example — biased die

Problem. A biased die has distribution P(1) = 0.1, P(2) = 0.15, P(3) = 0.2, P(4) = 0.2, P(5) = 0.15, P(6) = 0.2. Find E(X), Var(X), SD(X).

Step 1 — Check it is a valid distribution.

ΣP = 0.1 + 0.15 + 0.2 + 0.2 + 0.15 + 0.2 = 1.00 ✓

Step 2 — Compute E(X) = Σ x · P(X = x).

E(X) = 1(0.1) + 2(0.15) + 3(0.2) + 4(0.2) + 5(0.15) + 6(0.2)

= 0.1 + 0.3 + 0.6 + 0.8 + 0.75 + 1.2 = 3.75

Step 3 — Compute E(X²) = Σ x² · P(X = x).

E(X²) = 1(0.1) + 4(0.15) + 9(0.2) + 16(0.2) + 25(0.15) + 36(0.2)

= 0.1 + 0.6 + 1.8 + 3.2 + 3.75 + 7.2 = 16.65

Step 4 — Variance and SD.

Var(X) = E(X²) − [E(X)]² = 16.65 − (3.75)² = 16.65 − 14.0625 = 2.5875

SD(X) = √2.5875 ≈ 1.609

Conclusion. E(X) = 3.75, Var(X) = 2.5875, SD(X) ≈ 1.609. The mean is slightly higher than the fair-die mean (3.5) because the bias favours the larger faces.

3. Faded example — three-value distribution

A random variable Y has distribution P(Y = −2) = 0.3, P(Y = 1) = 0.5, P(Y = 4) = 0.2. Find E(Y) and Var(Y). 4 marks

Step 1 — Validity check. Σ P = ______ + ______ + ______ = ______

Step 2 — Compute E(Y).

E(Y) = (−2)(______) + (1)(______) + (4)(______) = ______ + ______ + ______ = ______

Step 3 — Compute E(Y²).

E(Y²) = (4)(______) + (1)(______) + (16)(______) = ______ + ______ + ______ = ______

Step 4 — Variance.

Var(Y) = ______ − (______)² = ______

Conclusion. E(Y) = ______, Var(Y) = ______, SD(Y) = √______ ≈ ______.

Stuck? Revisit lesson § Worked Example (biased die). The structure is identical.

4. Graduated practice

Foundation — basic E(X) and validity (4 questions)

Q	Question	Working (one line)	Answer
4.1 1	For a fair die, find E(X).
4.2 1	X has P(0) = 0.5, P(1) = 0.3, P(2) = 0.2. Find E(X).
4.3 1	Y has P(Y=1) = k, P(Y=2) = 0.3, P(Y=3) = 0.4. Find k.
4.4 1	E(X) = 5, Var(X) = 4. Find E(3X − 2) and Var(3X − 2).

Standard — typical HSC difficulty (6 questions)

4.5 X has P(X = 0) = 0.2, P(X = 2) = 0.5, P(X = 5) = 0.3. Find E(X), Var(X) and SD(X). 3 marks

4.6 A biased coin has P(H) = 0.6. You win $2 for heads, lose $1 for tails. Let W be your winnings. Find E(W) and Var(W). 3 marks

4.7 A game costs $3 to play; you roll a fair die and win $x dollars where x is the number rolled. Let P = (winnings) − (entry fee). Find E(P) — the expected profit per game. 2 marks

4.8 Two fair dice are rolled; let S be the sum. Find E(S) without enumerating the full distribution. (Hint: E(S) = E(X₁) + E(X₂).) 2 marks

4.9 X has the table below. Find k, then E(X) and Var(X).

x	1	2	3	4	5
P(X=x)	0.1	0.25	k	0.2	0.15

3 marks

4.10 X is the temperature in Celsius with E(X) = 20 and SD(X) = 5. Find E(Y) and SD(Y) where Y = (9/5)X + 32 (Fahrenheit). 2 marks

Extension — combine ideas (2 questions)

4.11 Design a fair dice game. The game costs $c to play; you roll a fair die and win $x. Find the value of c that makes the game fair (zero expected profit for the player). 3 marks

4.12 Prove that Var(aX + b) = a² Var(X) starting from the definition Var(Y) = E(Y²) − [E(Y)]². 3 marks

Stuck on 4.12? Use linearity of expectation: E(Y) = aE(X) + b, then expand (aX + b)² = a²X² + 2abX + b².

5. Self-check the easy 3

Tick the first three once you've verified your method works.

For 4.1 I wrote E(X) = (1+2+3+4+5+6)/6 = 21/6 = 3.5 — recognising this is not a possible roll.

For 4.2 I wrote E(X) = 0(0.5) + 1(0.3) + 2(0.2) = 0.7 — multiplied each x by its own P, not all by the same number.

For 4.3 I noticed Σ P must = 1, so k = 1 − 0.3 − 0.4 = 0.3.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

Q1.1 — Formulas

E(X) = Σ x · P(X = x). Var(X) = E(X²) − [E(X)]² where E(X²) = Σ x² · P(X = x). SD(X) = √Var(X).

Q1.2 — Valid distribution

(1) 0 ≤ P(X = x) ≤ 1. (2) Σ P(X = x) = 1.

Q1.3 — Linear transforms

E(Y) = a · E(X) + b; Var(Y) = a² · Var(X); SD(Y) = |a| · SD(X).

Q3 — Faded example (Y)

Step 1: 0.3 + 0.5 + 0.2 = 1.0 ✓.
Step 2: E(Y) = (−2)(0.3) + (1)(0.5) + (4)(0.2) = −0.6 + 0.5 + 0.8 = 0.7.
Step 3: E(Y²) = (4)(0.3) + (1)(0.5) + (16)(0.2) = 1.2 + 0.5 + 3.2 = 4.9.
Step 4: Var(Y) = 4.9 − (0.7)² = 4.9 − 0.49 = 4.41.
Conclusion: E(Y) = 0.7, Var(Y) = 4.41, SD(Y) = √4.41 = 2.1.

Q4.1 — Fair die

E(X) = (1+2+3+4+5+6)/6 = 21/6 = 3.5.

Q4.2 — Three-value distribution

E(X) = 0(0.5) + 1(0.3) + 2(0.2) = 0.7.

Q4.3 — Solve for k

k + 0.3 + 0.4 = 1, so k = 0.3.

Q4.4 — Linear transform with given E and Var

E(3X − 2) = 3(5) − 2 = 13; Var(3X − 2) = 3² · 4 = 36.

Q4.5 — Three-value (E, Var, SD)

E(X) = 0(0.2) + 2(0.5) + 5(0.3) = 0 + 1 + 1.5 = 2.5.
E(X²) = 0(0.2) + 4(0.5) + 25(0.3) = 0 + 2 + 7.5 = 9.5.
Var(X) = 9.5 − (2.5)² = 9.5 − 6.25 = 3.25; SD(X) = √3.25 ≈ 1.803.

Q4.6 — Biased coin winnings

E(W) = 2(0.6) + (−1)(0.4) = 1.2 − 0.4 = $0.80.
E(W²) = 4(0.6) + 1(0.4) = 2.4 + 0.4 = 2.8.
Var(W) = 2.8 − (0.8)² = 2.8 − 0.64 = 2.16 (in dollars²).

Q4.7 — Die game expected profit

Let X = roll, so E(X) = 3.5. Profit P = X − 3. E(P) = E(X) − 3 = 3.5 − 3 = $0.50. Favours the player by 50 cents per play on average.

Q4.8 — Sum of two dice

E(S) = E(X₁) + E(X₂) = 3.5 + 3.5 = 7 (by linearity of expectation; no need to enumerate the 36 outcomes).

Q4.9 — Solve k, then E(X), Var(X)

Σ P = 1 ⇒ 0.1 + 0.25 + k + 0.2 + 0.15 = 1, so k = 0.3.
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.
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.

Q4.10 — Celsius to Fahrenheit

E(Y) = (9/5)(20) + 32 = 36 + 32 = 68°F. SD(Y) = (9/5)(5) = 9°F (the +32 shift does not change spread).

Q4.11 — Fair die-game cost

Expected winnings = E(X) = 3.5. For zero expected profit, c = 3.5, so the entry fee should be $3.50. (Below $3.50 favours the player; above $3.50 favours the house.)

Q4.12 — Var(aX + b) = a² Var(X)

Let Y = aX + b. Then E(Y) = aE(X) + b (linearity).
Y² = (aX + b)² = a²X² + 2abX + b², so E(Y²) = a²E(X²) + 2abE(X) + b².
Var(Y) = E(Y²) − [E(Y)]² = a²E(X²) + 2abE(X) + b² − [aE(X) + b]²
= a²E(X²) + 2abE(X) + b² − a²[E(X)]² − 2abE(X) − b²
= a²(E(X²) − [E(X)]²) = a² Var(X). ▮ (The shift b cancels — variance is shift-invariant.)