S5 Mathematics Paper: Probability, Random Variables And Distributions (100 Marks)

S5 Mathematics Paper: Probability, Random Variables and Distributions (100 marks)

2026-11-11

S5 Mathematics – probability and statistics

S5 Maths paper: 6 questions, 24 parts, 100 marks, 75 minutes. Conditional probability, expectation and variance, normal and binomial distributions, dispersion.

Download　Question paper PDF　|　Answer key PDF

Paper details　S5 Mathematics practice paper on conditional probability, expectation and variance, the normal and binomial distributions and comparing dispersion, with full answers.

Question 1: Conditional probability and a two-way table

Of 200 S5 students in a school, 120 attend tutorial classes and the other 80 do not. Among those attending tutorials 90 pass, and among those not attending 45 pass.

#	Question	Answer and marking notes
1	Find the probability that a randomly chosen student passes.	Answer: the number passing is 90 + 45 = 135, so the probability is 135 / 200 = 0.675.
2	Given that the student passes, find the probability that the student attends tutorials.	Answer: P(tutorial \| pass) = 90 / 135 = 2/3, about 0.667.
3	Find the pass probability separately for students with and without tutorials, and compare them.	Answer: with tutorials 90 / 120 = 0.75; without tutorials 45 / 80 = 0.5625. The former is higher by 0.1875, that is 18.75 percentage points.
4	Explain whether these data prove that tutorials cause a higher pass rate.	Answer: no. These are merely observational data with no random assignment; prior ability, motivation and home support may affect both tutorial attendance and results, so the data show an association only, not causation.

Question 2: Independent events and the addition rule

In a school the probability that an S5 student takes Extended Module 2 (M2) is 0.6 and the probability of taking Physics is 0.5. The two choices are independent.

#	Question	Answer and marking notes
5	Find the probability that a student takes both M2 and Physics.	Answer: P(M2 and Physics) = 0.6 x 0.5 = 0.30.
6	Find the probability that a student takes at least one of the two subjects.	Answer: P(at least one) = 0.6 + 0.5 – 0.30 = 0.80.
7	Find the probability that a student takes neither subject.	Answer: P(neither) = (1 – 0.6)(1 – 0.5) = 0.4 x 0.5 = 0.20, consistent with 1 – 0.80.
8	Find the probability that a student takes exactly one of the two subjects.	Answer: M2 only is 0.6 x 0.5 = 0.30 and Physics only is 0.4 x 0.5 = 0.20, giving 0.50.

Question 3: Expectation and variance of a discrete random variable

The number X of notebooks a shop sells each day has the distribution P(X = 0) = 0.1, P(X = 1) = 0.3, P(X = 2) = 0.4 and P(X = 3) = 0.2. Each notebook sells for 40 dollars and the daily fixed cost is 20 dollars.

#	Question	Answer and marking notes
9	Find E(X).	Answer: E(X) = 0 x 0.1 + 1 x 0.3 + 2 x 0.4 + 3 x 0.2 = 0 + 0.3 + 0.8 + 0.6 = 1.7 notebooks.
10	Find Var(X).	Answer: E(X squared) = 0 + 0.3 + 4 x 0.4 + 9 x 0.2 = 0.3 + 1.6 + 1.8 = 3.7; Var(X) = 3.7 – 1.7 squared = 3.7 – 2.89 = 0.81.
11	Find the standard deviation of X.	Answer: the standard deviation is the square root of 0.81, that is 0.9 notebooks.
12	Find the expected daily profit.	Answer: profit = 40X – 20, so E(profit) = 40 x 1.7 – 20 = 68 – 20 = 48 dollars.

Question 4: The normal distribution and standardisation

The marks of a subject follow a normal distribution with mean 62 and standard deviation 8. Given P(Z below 1) = 0.8413, P(Z below 2) = 0.9772 and P(Z below -1) = 0.1587.

#	Question	Answer and marking notes
13	Find the probability that a mark is below 70.	Answer: z = (70 – 62) / 8 = 1, so the probability is 0.8413, that is 84.13%.
14	Find the probability that a mark is above 78.	Answer: z = (78 – 62) / 8 = 2, so the probability is 1 – 0.9772 = 0.0228, that is 2.28%.
15	Find the probability that a mark lies between 54 and 70.	Answer: the z values are -1 and 1, so the probability is 0.8413 – 0.1587 = 0.6826, that is 68.26%.
16	If there are 500 candidates, estimate how many score above 78.	Answer: 500 x 0.0228 = 11.4, that is about 11 candidates.

Question 5: The binomial distribution

In a multiple-choice test each question has four options with exactly one correct. A student guesses every answer on 8 questions. Let X be the number of correct answers.

#	Question	Answer and marking notes
17	State the distribution of X and find E(X).	Answer: X follows the binomial distribution B(8, 0.25) and E(X) = 8 x 0.25 = 2 questions.
18	Find the variance and standard deviation of X.	Answer: Var(X) = 8 x 0.25 x 0.75 = 1.5; the standard deviation is the square root of 1.5, about 1.22 questions.
19	Find the probability of getting no answer correct.	Answer: P(X = 0) = 0.75 raised to the power 8, about 0.1001, that is 10.01%.
20	Find the probability of getting at least one answer correct.	Answer: P(X at least 1) = 1 – 0.1001 = 0.8999, that is 89.99%.

Question 6: Standard deviation, coefficient of variation and data transformation

Class A has 30 students with a mean mathematics mark of 62 and a standard deviation of 9. Class B has 30 students with a mean of 68 and a standard deviation of 5.

#	Question	Answer and marking notes
21	Calculate the coefficient of variation for both classes and judge which class is more dispersed.	Answer: class A 9 / 62 x 100% is about 14.52% and class B 5 / 68 x 100% is about 7.35%. Class A is larger, so class A is relatively more dispersed.
22	Explain why both the standard deviation and the mean are needed when comparing dispersion.	Answer: the standard deviation carries units and depends on the size of the mean, so comparing it directly misleads when the means differ by 6 marks. The coefficient of variation expresses the standard deviation as a percentage of the mean, removing the scale effect and allowing a fair comparison.
23	If 5 marks are added to every student in class A, find the new mean and standard deviation.	Answer: the mean becomes 62 + 5 = 67 and the standard deviation stays 9, since a shift of all values does not change the gaps between marks.
24	If every mark in class A is instead multiplied by 1.1, find the new mean and standard deviation.	Answer: the mean is 62 x 1.1 = 68.2 and the standard deviation is 9 x 1.1 = 9.9.

Find a tutor — free Tutor Registration →

A TutorZone practice paper; original questions and passages, not publisher material.