Cambridge International AS & A Level Mathematics 9709: Probability & Statistics 1 – Study Guide

This guide teaches topic 5, Probability & Statistics 1, of the Cambridge International AS & A Level Mathematics 9709 syllabus for exams in 2026 and 2027 (Version 4). It covers syllabus sections 5.1 to 5.5. This content is examined on Paper 5: 1 hour 15 minutes, 50 marks, 6 to 8 structured questions. Paper 5 is worth 40% of the AS Level and 20% of the A Level, and it is the foundation for Paper 6 (Probability & Statistics 2). A scientific calculator is allowed, and you get the formula list and statistical tables (MF19).

Related pages: the Probability & Statistics 1 revision notes, the existing Probability & Statistics 1 practice questions, the A Level Mathematics hub, the printable 9709 checklist and the free AS Level diagnostic.

What this unit covers

Section	What you must be able to do	Paper
5.1 Representation of data	Choose and interpret stem-and-leaf (including back-to-back), box-and-whisker, histograms, cumulative frequency; mean, median, mode, range, IQR, standard deviation; use Σx, Σx² and coded totals, for up to two data sets	5
5.2 Permutations and combinations	Selections; arrangements in a line with repetition and restriction (not circles)	5
5.3 Probability	Enumeration, addition and multiplication rules, exclusive and independent events, conditional probability	5
5.4 Discrete random variables	Probability distribution tables, E(X), Var(X); B(n, p) and Geo(p)	5
5.5 The normal distribution	N(μ, σ²) using tables, finding μ or σ, normal approximation to the binomial	5

The syllabus says Paper 5 questions are mainly numerical and use no algebra beyond Pure Mathematics 1.

5.1 Representation of data

Choosing a diagram

• Stem-and-leaf: keeps every raw value; good for small data sets. A back-to-back diagram compares two sets on one shared stem. Always include a key, for example “3 | 7 means 37”.
• Box-and-whisker plot: shows minimum, lower quartile, median, upper quartile and maximum. Good for comparing spread and skew; the raw values are lost.
• Histogram: for grouped continuous data. The area of each bar is proportional to frequency, so the height is the frequency density = frequency ÷ class width.
• Cumulative frequency graph: plot cumulative frequency against the upper class boundary, then read off medians, quartiles, percentiles, or the number above or below a value.

Example: a class 10 ≤ t < 20 with frequency 30 has frequency density 30 ÷ 10 = 3. A class 20 ≤ t < 25 with frequency 20 has frequency density 20 ÷ 5 = 4. The second bar is taller even though it holds fewer values.

Measures of centre and spread

Mean, median and mode measure centre. Range, interquartile range (IQR = Q₃ − Q₁) and standard deviation measure spread. The median and IQR are not affected by extreme values; the mean and standard deviation are. When you compare two data sets, make one comment on centre and one on spread, each in context.

Mean and standard deviation from totals

These formulas are in MF19:

mean x̄ = Σx / n        standard deviation = √( Σx²/n − x̄² )
grouped: x̄ = Σxf / Σf   standard deviation = √( Σx²f/Σf − x̄² )

For grouped data, use the class mid-points as x.

Coded totals. If you are given Σ(x − a) and Σ(x − a)², work with y = x − a. The mean shifts by a; the standard deviation does not change.

Worked example. For 12 values, Σ(x − 20) = 30 and Σ(x − 20)² = 210.

mean of (x − 20) = 30/12 = 2.5          so x̄ = 20 + 2.5 = 22.5
variance = 210/12 − 2.5² = 17.5 − 6.25 = 11.25
standard deviation = √11.25 = 3.35 (3 s.f.)

Two data sets. Set A: n = 8, Σx = 96, Σx² = 1240. Set B: n = 12, Σx = 168, Σx² = 2460. Add the totals, not the means:

n = 20,  Σx = 264,  Σx² = 3700
combined mean = 264/20 = 13.2
combined variance = 3700/20 − 13.2² = 185 − 174.24 = 10.76
combined standard deviation = √10.76 = 3.28 (3 s.f.)

5.2 Permutations and combinations

A permutation is an arrangement, so order matters. A combination is a selection, so order does not matter. The formula ⁿCᵣ = n! / (r!(n − r)!) is in MF19.

• n different objects in a line: n! ways.
• Repeated objects: divide by the factorial of each repeat count.
• “Must be together”: glue the group into one unit, arrange the units, then arrange inside the group.
• “Must not be together”: total minus together, or arrange the others and place the separated items in the gaps.
• Two rows: the syllabus allows questions on people seated in two or more rows. Treat each row as its own line and multiply.

Circular arrangements are not examined.

Worked example. The letters of PARALLEL are arranged in a line. There are 8 letters with A twice and L three times.

all arrangements = 8! / (2! × 3!) = 40320 / 12 = 3360
As together: treat AA as one unit → 7 units, L three times
            = 7! / 3! = 840
As not together = 3360 − 840 = 2520

Worked example (selection). A committee of 5 is chosen from 6 men and 4 women. At least 2 women must be included.

2 women, 3 men: ⁴C₂ × ⁶C₃ = 6 × 20 = 120
3 women, 2 men: ⁴C₃ × ⁶C₂ = 4 × 15 = 60
4 women, 1 man: ⁴C₄ × ⁶C₁ = 1 × 6  = 6
total = 186

Split “at least” into exact cases and add. Multiplying ⁴C₂ by ⁸C₃ counts some committees more than once.

5.3 Probability

Equally likely outcomes

For two fair dice there are 36 equally likely outcomes. P(total 8) = 5/36, because (2,6), (3,5), (4,4), (5,3), (6,2) give 8. For balls drawn from a bag, you can count with combinations: P(event) = (ways to get the event) ÷ (total ways).

Addition and multiplication

• Mutually exclusive events cannot both happen: P(A ∩ B) = 0, so P(A or B) = P(A) + P(B).
• Independent events: one does not change the probability of the other. Test by checking whether P(A ∩ B) = P(A) × P(B).

Worked example. Two fair dice are thrown. A is “the first die shows an even number” and B is “the total is 7”. P(A) = 1/2 and P(B) = 6/36 = 1/6. The outcomes in A ∩ B are (2,5), (4,3), (6,1), so P(A ∩ B) = 3/36 = 1/12. Since 1/2 × 1/6 = 1/12, A and B are independent. They are not exclusive, because P(A ∩ B) ≠ 0.

Conditional probability

P(A | B) = P(A ∩ B) / P(B)

This formula is not in MF19; learn it. Tree diagrams handle most questions.

Worked example. On 30% of days it rains. On a rainy day, P(Ali is late) = 0.25. On a dry day, P(Ali is late) = 0.1.

P(late) = 0.3 × 0.25 + 0.7 × 0.1 = 0.075 + 0.07 = 0.145
P(rain | late) = 0.075 / 0.145 = 15/29 = 0.517 (3 s.f.)

The numerator is one branch; the denominator is every branch that ends in “late”.

5.4 Discrete random variables

Probability distribution tables

List every value of X with its probability. The probabilities must add to 1. From MF19:

E(X) = Σxp        Var(X) = Σx²p − {E(X)}²

Worked example. Two counters are taken without replacement from a bag of 3 red and 2 blue. X is the number of red counters.

P(X = 0) = 2/5 × 1/4 = 1/10
P(X = 2) = 3/5 × 2/4 = 3/10
P(X = 1) = 1 − 1/10 − 3/10 = 3/5

E(X)  = 0 × 1/10 + 1 × 3/5 + 2 × 3/10 = 6/5
E(X²) = 1 × 3/5 + 4 × 3/10 = 9/5
Var(X) = 9/5 − (6/5)² = 9/25 = 0.36

Binomial distribution B(n, p)

Use it when there is a fixed number n of independent trials, each with two outcomes and a constant probability p of success. From MF19:

P(X = r) = ⁿCᵣ pʳ (1 − p)ⁿ⁻ʳ     E(X) = np     Var(X) = np(1 − p)

Worked example. X ~ B(10, 0.3). Find P(X ≥ 2).

P(X ≥ 2) = 1 − P(X = 0) − P(X = 1)
         = 1 − 0.7¹⁰ − 10 × 0.3 × 0.7⁹
         = 1 − 0.028248 − 0.121061 = 0.851 (3 s.f.)
E(X) = 3,  Var(X) = 2.1

Geometric distribution Geo(p)

Use it when independent trials with constant p are repeated until the first success, and X is the trial number of that success (X = 1, 2, 3, …). From MF19:

P(X = r) = p(1 − p)ʳ⁻¹       E(X) = 1/p

Two results follow directly: P(X > r) = (1 − p)ʳ (the first r trials all fail) and P(X ≤ r) = 1 − (1 − p)ʳ. The syllabus asks only for the expectation of the geometric distribution, not its variance.

Worked example. A fair die is thrown until a six appears. P(first six on the 4th throw) = (5/6)³ × 1/6 = 125/1296 = 0.0965. The expected number of throws is 1 ÷ (1/6) = 6.

5.5 The normal distribution

If X ~ N(μ, σ²), standardise with Z = (X − μ)/σ and use the table of Φ(z) = P(Z ≤ z). For negative z, Φ(−z) = 1 − Φ(z). The syllabus requires full working for standardisation, and sketches may be required. A quick sketch with the mean marked and the region shaded stops most sign errors.

Worked example. X ~ N(120, 15²).

P(X > 140):  z = (140 − 120)/15 = 1.333
             P = 1 − Φ(1.333) = 1 − 0.9088 = 0.0912 (3 s.f.)

P(100 < X < 130):  z₁ = −1.3333, z₂ = 0.6667
             P = Φ(0.6667) − Φ(−1.3333) = 0.7475 − 0.0912 = 0.656 (3 s.f.)

value exceeded by 5%:  P(Z ≤ z) = 0.95 → z = 1.645 (critical values table)
             x = 120 + 1.645 × 15 = 145 (3 s.f.)

Finding σ. X ~ N(80, σ²) and P(X > 90) = 0.2. Then Φ(z) = 0.8, so z = 0.842, and (90 − 80)/σ = 0.842, giving σ = 11.9 (3 s.f.). With both μ and σ unknown you get two equations of this form and solve them simultaneously.

Normal approximation to the binomial

If X ~ B(n, p) and n is large enough that np > 5 and nq > 5 (q = 1 − p), use X ≈ N(np, npq) with a continuity correction.

Worked example. X ~ B(80, 0.25). np = 20 and nq = 60, both above 5, so use N(20, 15).

P(X ≤ 15) ≈ P(Y < 15.5)
z = (15.5 − 20)/√15 = −1.162
P = 1 − Φ(1.162) = 1 − 0.8774 = 0.123 (3 s.f.)

Write the discrete inequality first (“X ≤ 15”), then decide which side of 15 the half goes.

Common errors

• Histogram heights drawn as frequency instead of frequency density.
• Cumulative frequency plotted at class mid-points instead of upper boundaries.
• Combined standard deviation found by averaging the two standard deviations.
• Forgetting to divide by the factorial of repeated letters.
• Treating “exclusive” and “independent” as the same thing.
• Using a geometric model for a fixed number of trials (that is binomial).
• Using n − 1 instead of n in the standard deviation formula for Paper 5 data.
• z-values from the table read to 2 d.p. when 3 d.p. are needed for 3 s.f. accuracy.

Next steps

Condense this unit with the revision notes and quick self-test, then try the practice questions. Paper 6 builds on this unit: see the Probability & Statistics 2 study guide.

Official syllabus

Cambridge International AS & A Level Mathematics 9709 syllabus, for exams in 2026 and 2027 (Version 4), Cambridge University Press & Assessment. Topic 5, Probability & Statistics 1 (for Paper 5): sections 5.1 to 5.5.