Mathematics Advanced • Year 12 • Module 5 • Lesson 11
Introduction to Random Variables
Apply probability functions and the CDF to real contexts: surveys, lotteries, manufacturing, customer queues, and quality control.
Problem 1 — Scratchie lottery (probability function)
A "Mini-Five" scratchie has five prize tiers. The probability function for the prize amount X (in dollars) is:
| x ($) | 0 | 2 | 5 | 20 | 100 |
|---|---|---|---|---|---|
| p(x) | 0.55 | 0.30 | 0.10 | k | 0.005 |
Set up: What are we solving for?
(i) Find k and verify the table defines a valid probability function. 2 marks
(ii) A ticket costs $1. Construct the probability function for net profit Y = X − 1, and verify the new table is also valid. 2 marks
(iii) Compute P(X ≥ 20). State, in one sentence, what this number represents to a customer who has just bought a ticket. 2 marks
Stuck on (ii)? Each ticket buyer pays $1 first; subtract 1 from every prize value, keep the same probabilities.Problem 2 — Customer queue (CDF from data)
A café records how many customers are waiting at 8:00 am every weekday for a month. The relative frequencies (used as probabilities) for the number waiting X are:
| x | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| p(x) | 0.05 | 0.15 | 0.25 | 0.30 | 0.15 | 0.10 |
Set up: What are we solving for?
(i) Build the full CDF table F(0), F(1), …, F(5). Confirm F(5) = 1. 3 marks
(ii) Use F to find P(X ≥ 3), and check the answer agrees with the direct sum p(3) + p(4) + p(5). 2 marks
(iii) The owner wants to know the smallest x such that F(x) ≥ 0.5 ("the queue length most days does not exceed this value"). Find this x. 2 marks
Problem 3 — Defective components (formula for p)
A small batch of n = 4 components is tested. The probability that exactly x of them are defective is modelled by p(x) = c (x + 1) for x = 0, 1, 2, 3, 4, for some constant c.
Set up: What are we solving for?
(i) Find c. 2 marks
(ii) Construct the full p(x) table and verify all probabilities lie in [0, 1] and sum to 1. 2 marks
(iii) A batch is rejected if there are 2 or more defective components. Find the probability that a batch is rejected. 2 marks
Stuck on (i)? Write out c(1) + c(2) + c(3) + c(4) + c(5) = c · (1 + 2 + 3 + 4 + 5) = 1.Problem 4 — Driver reaction time (continuous variable)
The reaction time T of a randomly chosen driver (in seconds) is a continuous random variable. A traffic engineer gives selected cumulative probabilities:
| t (s) | 0.30 | 0.50 | 0.80 | 1.20 | 2.00 |
|---|---|---|---|---|---|
| F(t) = P(T ≤ t) | 0.05 | 0.30 | 0.70 | 0.92 | 1.00 |
Set up: What are we solving for?
(i) State the value of P(T = 0.80) and explain in one sentence why. 1 mark
(ii) Compute P(0.50 ≤ T ≤ 1.20) using the F values. 2 marks
(iii) The engineer flags drivers whose reaction time exceeds 1.20 s as "slow responders". Estimate the probability that a randomly chosen driver is a slow responder, and explain in one sentence why this number matters for road-safety design. 2 marks
Problem 5 — Linear-ramp lottery (HSC-style)
Tickets are numbered 1 to 100 in a charity raffle, and the probability of ticket number x being drawn is modelled by p(x) = x / 5050 for x = 1, 2, …, 100. (Recall 1 + 2 + … + 100 = 5050.)
Set up: What are we solving for?
(i) Verify that p is a valid probability function. 2 marks
(ii) Find F(50) = P(X ≤ 50), giving your answer as an exact fraction and to 3 decimal places. (Use 1 + 2 + … + 50 = 1275.) 2 marks
(iii) A critic argues: "This is unfair because the higher-numbered tickets are far more likely to win." Compute p(100) / p(1), and decide in one sentence whether the criticism is mathematically justified. 2 marks
Stuck on (i)? Use Σ x for x = 1, …, 100 = n(n+1)/2 with n = 100.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Scratchie lottery
Set up. Use Σ p(x) = 1 to find the missing k, then shift to net profit Y = X − 1 and answer the "big win" probability question.
(i) 0.55 + 0.30 + 0.10 + k + 0.005 = 1 ⇒ 0.955 + k = 1 ⇒ k = 0.045. Each value lies in [0, 1] ✓ and the new sum is 1 ✓.
(ii) Y = X − 1, so y ∈ {−1, 1, 4, 19, 99} with the same probabilities 0.55, 0.30, 0.10, 0.045, 0.005. Each p(y) ∈ [0, 1] ✓; Σ p(y) = 1 ✓.
(iii) P(X ≥ 20) = p(20) + p(100) = 0.045 + 0.005 = 0.05. About 1 in 20 tickets returns $20 or more — most tickets do not.
Problem 2 — Customer queue
Set up. Build the CDF by cumulative addition, then use it to compute upper-tail probabilities and the smallest x crossing the 0.5 threshold.
(i) F(0) = 0.05, F(1) = 0.20, F(2) = 0.45, F(3) = 0.75, F(4) = 0.90, F(5) = 1.00 ✓.
(ii) P(X ≥ 3) = 1 − F(2) = 1 − 0.45 = 0.55. Check: p(3) + p(4) + p(5) = 0.30 + 0.15 + 0.10 = 0.55 ✓.
(iii) F(2) = 0.45 < 0.5; F(3) = 0.75 ≥ 0.5. So the smallest x with F(x) ≥ 0.5 is x = 3.
Problem 3 — Defective components
Set up. Solve Σ p(x) = 1 to find c, then use the resulting table to find P(X ≥ 2).
(i) c(1) + c(2) + c(3) + c(4) + c(5) = c × 15 = 1, so c = 1/15.
(ii) p(0) = 1/15, p(1) = 2/15, p(2) = 3/15, p(3) = 4/15, p(4) = 5/15. All in [0, 1] ✓; sum = 15/15 = 1 ✓.
(iii) P(X ≥ 2) = p(2) + p(3) + p(4) = 3/15 + 4/15 + 5/15 = 12/15 = 4/5 = 0.80. (Equivalently 1 − F(1) = 1 − 3/15 = 12/15.)
Problem 4 — Driver reaction time
Set up. Use the CDF table to answer probability questions about T, remembering T is continuous.
(i) P(T = 0.80) = 0. For a continuous random variable, the probability at any single point is zero — all positive probability comes from intervals.
(ii) P(0.50 ≤ T ≤ 1.20) = F(1.20) − F(0.50) = 0.92 − 0.30 = 0.62. (For continuous T, including or excluding the endpoints does not change the value.)
(iii) P(T > 1.20) = 1 − F(1.20) = 1 − 0.92 = 0.08. About 8% of drivers respond slower than 1.2 s — engineers must design braking distances and signal timing to cover this realistic worst case, not just the average driver.
Problem 5 — Linear-ramp lottery
Set up. Verify validity, compute a cumulative probability with a sum-formula shortcut, then quantify the "unfairness" via a single ratio.
(i) All p(x) = x/5050 ≥ 0 (since x ≥ 1) and each p(x) < 1. Σ p(x) = (1 + 2 + … + 100)/5050 = 5050/5050 = 1 ✓. Valid.
(ii) F(50) = (1 + 2 + … + 50)/5050 = 1275/5050 = 51/202 ≈ 0.252.
(iii) p(100)/p(1) = (100/5050) / (1/5050) = 100. Yes — ticket 100 is exactly 100 times as likely to be drawn as ticket 1, so the criticism is mathematically justified: this design is heavily biased toward higher-numbered tickets.