Mathematics Advanced • Year 12 • Module 5 • Lesson 11
Introduction to Random Variables
Build fluency in classifying random variables, verifying probability functions, and computing values of the cumulative distribution function.
1. Quick recall
Answer each question in the space provided. 1 mark each
Q1.1 Complete the two properties that every probability function p(x) of a discrete random variable must satisfy:
(i) For every x in the range: ____________ ≤ p(x) ≤ ____________
(ii) Sum over all x in the range: Σ p(x) = ____________
Q1.2 Define the cumulative distribution function F(x) in terms of the probability function p(t):
F(x) = P( ____________ ) = Σ ____________ for all t ____________ x
Q1.3 For a continuous random variable, state the value of P(X = a) for any specific value a:
P(X = a) = ____________
2. Worked example — verifying a probability function and computing F
Follow each step. The reasoning is shown on the right of each line.
Problem. A discrete random variable X has p(1) = 0.10, p(2) = 0.25, p(3) = 0.30, p(4) = 0.20, p(5) = 0.15.
Step 1 — Check property 1: every probability lies in [0, 1].
0.10, 0.25, 0.30, 0.20, 0.15 — all between 0 and 1 ✓
Step 2 — Check property 2: probabilities sum to 1.
Σ p(x) = 0.10 + 0.25 + 0.30 + 0.20 + 0.15 = 1.00 ✓
Reason: both properties hold, so this is a valid probability function.
Step 3 — Build the CDF by cumulative addition from the left.
F(1) = p(1) = 0.10
F(2) = F(1) + p(2) = 0.10 + 0.25 = 0.35
F(3) = F(2) + p(3) = 0.35 + 0.30 = 0.65
F(4) = F(3) + p(4) = 0.65 + 0.20 = 0.85
F(5) = F(4) + p(5) = 0.85 + 0.15 = 1.00
Reason: F(x) accumulates probability for all values ≤ x. F at the largest value must equal 1.
Step 4 — Use the CDF to find P(2 ≤ X ≤ 4).
P(2 ≤ X ≤ 4) = F(4) − F(1) = 0.85 − 0.10 = 0.75
Reason: F(4) includes X = 1; subtract F(1) to remove it. (X is integer-valued.)
Conclusion. p is a valid probability function, F(5) = 1, and P(2 ≤ X ≤ 4) = 0.75.
3. Faded example — fill in the missing steps
A discrete random variable Y has p(0) = 0.15, p(1) = 0.30, p(2) = k, p(3) = 0.20, p(4) = 0.10. Fill in each blank. 4 marks
Step 1 — Use Σ p(x) = 1 to find k:
0.15 + 0.30 + k + 0.20 + 0.10 = 1
⇒ 0.75 + k = 1 ⇒ k = ____________
Step 2 — Build the CDF:
F(0) = ____________
F(1) = F(0) + p(1) = ____________ + 0.30 = ____________
F(2) = F(1) + p(2) = ____________ + ____________ = ____________
F(3) = F(2) + p(3) = ____________ + 0.20 = ____________
F(4) = ____________
Step 3 — Use the CDF to find P(Y > 2):
P(Y > 2) = 1 − P(Y ≤ 2) = 1 − F(2) = 1 − ____________ = ____________
Conclusion. k = ____________, F(4) = ____________, P(Y > 2) = ____________.
4. Graduated practice
Show the line of working for each part — examiners reward setting up Σ p(x) = 1 even before you reach the final answer.
Foundation — classify and compute single values (4 questions)
| Q | Task | Answer |
|---|---|---|
| 4.1 1 | Classify as discrete or continuous: number of text messages a student sends in a day. | |
| 4.2 1 | Classify as discrete or continuous: weight of a randomly chosen apple at a market. | |
| 4.3 1 | For X with p(1) = 0.2, p(2) = 0.5, p(3) = 0.3, state F(2). | |
| 4.4 1 | For the same X, state P(X > 2). |
Standard — typical HSC difficulty (6 questions)
Show working in the space below each question.
4.5 A discrete X has p(0) = 0.1, p(1) = 0.3, p(2) = 0.4, p(3) = c. Find c and verify Σ p(x) = 1. 2 marks
4.6 A discrete X has p(x) = kx for x = 1, 2, 3, 4. Find k and write out p(1), p(2), p(3), p(4). 2 marks
4.7 X has p(x) = k x² for x = 1, 2, 3. Find k. 2 marks
4.8 A random variable X has CDF values F(1) = 0.2, F(2) = 0.5, F(3) = 0.8, F(4) = 1.0. Compute p(2), p(3), p(4). 2 marks
4.9 For the biased die p(x) = x / 21 (x = 1, …, 6), compute P(X ≥ 4). 2 marks
4.10 X has p(1) = 0.10, p(2) = 0.20, p(3) = 0.30, p(4) = 0.25, p(5) = 0.15. Find P(2 ≤ X ≤ 4) using F. 2 marks
Extension — combine concepts (2 questions)
4.11 A game charges $2 to play and pays $x dollars where x is the result of rolling a fair six-sided die. Let Y = net profit. Write the probability function of Y in a table, and verify it is valid. 3 marks
4.12 Explain in two sentences why, for a continuous random variable, P(3 < X < 5) = P(3 ≤ X ≤ 5). Reference the value of P(X = a) for a single point a in your explanation. 3 marks
5. Self-check the easy 3
Tick the first three once you've checked your method works.
How did this worksheet feel?
What I'll revisit before next class:
Q1.1 — Properties of a probability function
(i) 0 ≤ p(x) ≤ 1. (ii) Σ p(x) = 1.
Q1.2 — Definition of the CDF
F(x) = P(X ≤ x) = Σ p(t) for all t ≤ x.
Q1.3 — Continuous P(X = a)
P(X = a) = 0 for any specific value a. (For continuous variables only intervals carry positive probability.)
Q3 — Faded example, Y
Step 1: k = 0.25.
Step 2: F(0) = 0.15; F(1) = 0.15 + 0.30 = 0.45; F(2) = 0.45 + 0.25 = 0.70; F(3) = 0.70 + 0.20 = 0.90; F(4) = 1.00.
Step 3: P(Y > 2) = 1 − F(2) = 1 − 0.70 = 0.30.
Q4.1 — Text messages per day
Discrete — a count taking values 0, 1, 2, …
Q4.2 — Weight of an apple
Continuous — weight is a measurement that can take any value in an interval (e.g. 142.7 g).
Q4.3 — F(2)
F(2) = p(1) + p(2) = 0.2 + 0.5 = 0.7.
Q4.4 — P(X > 2)
P(X > 2) = 1 − F(2) = 1 − 0.7 = 0.3. (Equivalently, P(X = 3) = 0.3.)
Q4.5 — c such that 0.1 + 0.3 + 0.4 + c = 1
c = 1 − 0.8 = 0.2. Verify: 0.1 + 0.3 + 0.4 + 0.2 = 1.0 ✓, and all values are in [0, 1].
Q4.6 — p(x) = kx for x = 1, 2, 3, 4
k(1 + 2 + 3 + 4) = 1 ⇒ 10k = 1 ⇒ k = 1/10. So p(1) = 0.1, p(2) = 0.2, p(3) = 0.3, p(4) = 0.4.
Q4.7 — p(x) = kx² for x = 1, 2, 3
k(1 + 4 + 9) = 1 ⇒ 14k = 1 ⇒ k = 1/14.
Q4.8 — Recover p from F
p(2) = F(2) − F(1) = 0.5 − 0.2 = 0.3. p(3) = F(3) − F(2) = 0.8 − 0.5 = 0.3. p(4) = F(4) − F(3) = 1.0 − 0.8 = 0.2.
Q4.9 — Biased die p(x) = x/21
P(X ≥ 4) = p(4) + p(5) + p(6) = 4/21 + 5/21 + 6/21 = 15/21 = 5/7 ≈ 0.714.
Q4.10 — P(2 ≤ X ≤ 4) via F
Build F first: F(1) = 0.10, F(2) = 0.30, F(3) = 0.60, F(4) = 0.85, F(5) = 1.00. Then P(2 ≤ X ≤ 4) = F(4) − F(1) = 0.85 − 0.10 = 0.75. (Check by summing: 0.20 + 0.30 + 0.25 = 0.75 ✓.)
Q4.11 — Net profit Y from $2 die game
Y = x − 2, so Y ∈ {−1, 0, 1, 2, 3, 4} each with probability 1/6:
y | −1 0 1 2 3 4
p(y) | 1/6 1/6 1/6 1/6 1/6 1/6
Verify: each p(y) = 1/6 ∈ [0, 1] ✓; Σ p(y) = 6 × 1/6 = 1 ✓.
Q4.12 — Why endpoints don't matter for continuous X
For a continuous random variable, P(X = a) = 0 for every specific value a. Therefore including or excluding the endpoints 3 and 5 changes the probability by 0, so P(3 < X < 5) = P(3 ≤ X ≤ 5).