These are original Marlbridge practice questions, not past exam questions. All data, simulations and settings are fictional. The rubrics are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. Assumptions for every question: X is a discrete random variable; give probabilities to 4 decimal places unless they are exact. Show the rule you use, the values you substitute and the answer.
Question 1 (multiple choice · foundation)
Which table shows a valid probability distribution?
- (A) x: 0, 1, 2, 3 with P(X = x): 0.3, 0.4, 0.2, 0.1
- (B) x: 0, 1, 2, 3 with P(X = x): 0.4, 0.3, 0.2, 0.2
- (C) x: 0, 1, 2, 3 with P(X = x): 0.5, 0.4, 0.2, −0.1
- (D) x: 0, 1, 2, 3 with P(X = x): 0.25, 0.25, 0.25, 0.15
Answer and explanation
Answer: (A). Every probability is between 0 and 1, and 0.3 + 0.4 + 0.2 + 0.1 = 1.
- (B) adds to 1.1. The probabilities of all possible values must add to exactly 1.
- (C) adds to 1, but it contains −0.1. A probability can never be negative.
- (D) adds to 0.9, so 0.1 of the probability is missing.
Question 2 (multiple choice · core)
The number of goals, X, scored by a fictional youth football team in a match has this distribution:
| x | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| P(X = x) | 0.15 | 0.25 | 0.30 | 0.20 | 0.10 |
What is P(1 < X ≤ 3)?
- (A) 0.30
- (B) 0.50
- (C) 0.75
- (D) 0.90
Answer and explanation
Answer: (B). “1 < X ≤ 3” means X = 2 or X = 3, because X = 1 is excluded by the strict inequality. 0.30 + 0.20 = 0.50.
- (A) is P(X = 2) only. It leaves out X = 3, which is allowed by “≤ 3”.
- (C) includes X = 1: 0.25 + 0.30 + 0.20 = 0.75. That is P(1 ≤ X ≤ 3).
- (D) is P(X ≤ 3), which also includes X = 0 and X = 1.
Question 3 (multiple choice · core)
A random variable X takes the values 1, 2, 3 and 4. Its cumulative distribution is:
| x | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| P(X ≤ x) | 0.20 | 0.45 | 0.80 | 1.00 |
What is P(X = 3)?
- (A) 0.20
- (B) 0.35
- (C) 0.55
- (D) 0.80
Answer and explanation
Answer: (B). P(X = 3) = P(X ≤ 3) − P(X ≤ 2) = 0.80 − 0.45 = 0.35.
- (A) is 1 − 0.80 = P(X > 3), which is P(X = 4).
- (C) is 1 − 0.45 = P(X > 2), which includes both X = 3 and X = 4.
- (D) reads P(X ≤ 3) from the table as if it were P(X = 3). Cumulative values are running totals.
Question 4 (constructed response · core)
A fictional café has two coffee machines. On a given morning, machine A works with probability 0.95 and machine B works with probability 0.85. Assume the machines work or fail independently. Let X = the number of machines that work on a given morning.
(a) Construct the probability distribution of X as a table. Show your working. (b) Show that your distribution is valid. (c) The café can serve customers if at least one machine works. Find this probability.
Worked solution and suggested Marlbridge rubric
(a) X can be 0, 1 or 2.
- P(X = 0) = P(A fails) · P(B fails) = 0.05 × 0.15 = 0.0075.
- P(X = 1) = P(A works, B fails) + P(A fails, B works) = 0.95 × 0.15 + 0.05 × 0.85 = 0.1425 + 0.0425 = 0.1850.
- P(X = 2) = 0.95 × 0.85 = 0.8075.
| x | 0 | 1 | 2 |
|---|---|---|---|
| P(X = x) | 0.0075 | 0.1850 | 0.8075 |
(b) Each probability is between 0 and 1, and 0.0075 + 0.1850 + 0.8075 = 1.
(c) P(X ≥ 1) = 1 − P(X = 0) = 1 − 0.0075 = 0.9925. On about 99.25% of mornings, at least one machine works.
| Point | What earns it |
|---|---|
| 1 | P(X = 0) and P(X = 2) correct, using the multiplication rule for independent events |
| 1 | P(X = 1) correct, with both ways of getting exactly one working machine added |
| 1 | Validity checked: all probabilities in [0, 1] and sum = 1 |
| 1 | P(X ≥ 1) = 0.9925 with the complement (or addition) shown, interpreted in context |
The most common error is P(X = 1) = 0.95 × 0.15 = 0.1425 only; this does not earn point 2, and the sum check in (b) would then fail.
Question 5 (constructed response · core)
In a fictional puzzle app, a player has at most 4 attempts to clear a level. For players who clear it, X = the number of attempts used, with
P(X = x) = k(5 − x) for x = 1, 2, 3, 4.
(a) Find k. (b) Write the probability distribution and the cumulative distribution of X as tables. (c) Find the probability that a player who clears the level needs at least 2 attempts.
Worked solution and suggested Marlbridge rubric
(a) The probabilities must add to 1: k(4) + k(3) + k(2) + k(1) = 10k = 1, so k = 0.1.
(b)
| x | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| P(X = x) | 0.4 | 0.3 | 0.2 | 0.1 |
| P(X ≤ x) | 0.4 | 0.7 | 0.9 | 1.0 |
(c) P(X ≥ 2) = 1 − P(X ≤ 1) = 1 − 0.4 = 0.6. Or 0.3 + 0.2 + 0.1 = 0.6.
| Point | What earns it |
|---|---|
| 1 | Sets the sum equal to 1 and finds k = 0.1 |
| 1 | Correct distribution table |
| 1 | Correct cumulative table (running totals ending at 1) |
| 1 | P(X ≥ 2) = 0.6 with working shown |
Question 6 (constructed response · stretch)
A student guesses the answers to 3 true-or-false questions on a fictional quiz. Each guess is correct with probability 0.5, independently. Let X = the number of correct guesses. A class ran a simulation: for each trial, a random digit generator produced three digits, with 0–4 meaning “correct” and 5–9 meaning “wrong”. The results of 400 trials were:
| x | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| Number of trials | 60 | 140 | 147 | 53 |
(a) Explain why the digit rule is a valid way to simulate one guess. (b) Use the simulation to estimate the probability distribution of X and P(X ≥ 2). (c) Find the exact distribution of X using probability rules, and compare it with your estimate. (d) How could the class make the estimates closer to the exact values?
Worked solution and suggested Marlbridge rubric
(a) Each digit 0 to 9 is equally likely, so 5 of the 10 digits (0–4) give “correct”: probability 5/10 = 0.5. The generator’s digits are independent, just like the guesses.
(b) Divide each count by 400:
| x | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| Estimated P(X = x) | 0.15 | 0.35 | 0.3675 | 0.1325 |
Estimated P(X ≥ 2) = 0.3675 + 0.1325 = 0.50.
(c) There are 2 × 2 × 2 = 8 equally likely outcomes (each has probability 0.5³ = 1/8). One outcome has 0 correct, three have 1, three have 2 and one has 3.
| x | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| Exact P(X = x) | 1/8 = 0.125 | 3/8 = 0.375 | 3/8 = 0.375 | 1/8 = 0.125 |
The estimates are close to the exact values (each within 0.03), and both give P(X ≥ 2) = 0.5. They are not identical because a simulation result varies by chance.
(d) Run many more trials. With more trials, relative frequencies tend to be closer to the true probabilities.
| Point | What earns it |
|---|---|
| 1 | (a) Links 5 of 10 equally likely digits to probability 0.5 |
| 1 | (b) All four estimates as relative frequencies, and estimated P(X ≥ 2) |
| 1 | (c) Exact distribution from the 8 equally likely outcomes |
| 1 | (c) and (d) Compares estimate and exact, explains the difference as chance variation and suggests more trials |
Question 7 (explanation · stretch)
In a fictional board game, two fair spinners are spun. Each has four equal sectors numbered 1 to 4. Let X = the larger of the two numbers shown (if they are equal, X is that number). A student writes: “X can be 1, 2, 3 or 4, so P(X = x) = 0.25 for each value.”
(a) Explain why the student is wrong. (b) Construct the correct probability distribution of X. (c) Construct the cumulative distribution and show that P(X ≤ x) = x² ÷ 16 for x = 1, 2, 3, 4. (d) Find P(X ≤ 3).
Model answer and suggested Marlbridge rubric
(a) Having 4 possible values does not make them equally likely. The 16 equally likely outcomes are the ordered pairs (first spinner, second spinner). Only one of them, (1, 1), gives X = 1, but many give X = 4.
(b) Count the pairs whose larger value is x:
- X = 1: (1, 1). 1 pair.
- X = 2: (1, 2), (2, 1), (2, 2). 3 pairs.
- X = 3: (1, 3), (2, 3), (3, 1), (3, 2), (3, 3). 5 pairs.
- X = 4: the remaining 16 − 9 = 7 pairs.
| x | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| P(X = x) | 1/16 = 0.0625 | 3/16 = 0.1875 | 5/16 = 0.3125 | 7/16 = 0.4375 |
Check: (1 + 3 + 5 + 7)/16 = 16/16 = 1.
(c)
| x | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| P(X ≤ x) | 1/16 | 4/16 | 9/16 | 16/16 |
These are 1²/16, 2²/16, 3²/16 and 4²/16. Reason: X ≤ x means both spinners show x or less. Each spinner does this with probability x/4, independently, so P(X ≤ x) = (x/4)² = x²/16.
(d) P(X ≤ 3) = 9/16 = 0.5625.
| Point | What earns it |
|---|---|
| 1 | (a) Explains that the values are not equally likely, referring to the 16 equally likely outcomes |
| 1 | (b) Correct distribution with counts shown and a sum check |
| 1 | (c) Correct cumulative table and a link to x²/16 (by matching values or by the independence argument) |
| 1 | (d) 0.5625 |
How did you do?
- Q1 wrong: re-read “The probability distribution of a discrete random variable” in the study guide. Check both rules.
- Q2 wrong: revisit the word-to-symbol table in “The cumulative distribution”.
- Q3 or Q5(b) wrong: revisit Worked example 2. Subtract cumulative values to get a single probability.
- Q4 or Q7 wrong: work through Worked example 1 again. Add every outcome that gives the same value of X.
- Q6 wrong: revisit Worked example 3 on simulation.
Then tick off the topic checklist.