Cambridge International AS & A Level Mathematics 9709: Probability & Statistics 1 – Revision Notes

These notes condense topic 5, Probability & Statistics 1, of the Cambridge International AS & A Level Mathematics 9709 syllabus for exams in 2026 and 2027 (Version 4): sections 5.1 to 5.5. All of it is examined on Paper 5 (1 hour 15 minutes, 50 marks, 40% of the AS Level and 20% of the A Level). For full explanations and worked examples, use the Probability & Statistics 1 study guide.

Other links: Probability & Statistics 1 practice questions, the A Level Mathematics hub, the printable 9709 checklist and the free AS Level diagnostic.

Formulas

Result	Formula	In MF19?
Mean (raw)	x̄ = Σx / n	Yes
Standard deviation (raw)	√(Σx²/n − x̄²)	Yes
Mean (grouped)	Σxf / Σf, x = mid-point	Yes
Standard deviation (grouped)	√(Σx²f/Σf − x̄²)	Yes
Combinations	ⁿCᵣ = n! / (r!(n − r)!)	Yes
Conditional probability	P(A | B) = P(A ∩ B) / P(B)	No
Independence test	P(A ∩ B) = P(A) × P(B)	No
Discrete expectation	E(X) = Σxp	Yes
Discrete variance	Var(X) = Σx²p − {E(X)}²	Yes
Binomial B(n, p)	P(X = r) = ⁿCᵣ pʳ(1 − p)ⁿ⁻ʳ; mean np; variance np(1 − p)	Yes
Geometric Geo(p)	P(X = r) = p(1 − p)ʳ⁻¹; mean 1/p	Yes
Geometric tail	P(X > r) = (1 − p)ʳ	No (derive it)
Standardising	Z = (X − μ)/σ	No
Negative z	Φ(−z) = 1 − Φ(z)	Yes (with the table)

5.1 Representation of data

Diagrams. Stem-and-leaf keeps raw data (needs a key; back-to-back compares two sets). Box-and-whisker shows five values and makes skew visible. Histogram: bar area ∝ frequency, height = frequency density = frequency ÷ class width. Cumulative frequency: plot at upper class boundaries.

Reading a cumulative frequency graph (n values).

1. Median at n/2, Q₁ at n/4, Q₃ at 3n/4 on the vertical axis.
2. A percentile, e.g. the 90th, at 0.9n.
3. “How many above 40?” Read the cumulative frequency at 40 and subtract from n.

Coded data. If y = x − a: x̄ = a + ȳ and the standard deviation of x equals the standard deviation of y.

Two data sets. Add n, Σx and Σx² across the sets, then use the formulas once.

Comparing data. One comment on centre (median or mean), one on spread (IQR or standard deviation), both in context.

5.2 Permutations and combinations

Method in steps.

1. Order matters? Permutation. Order does not matter? Combination.
2. Repeats: divide by k! for each letter repeated k times.
3. “Together”: glue into one block, arrange blocks, multiply by internal arrangements.
4. “Not together”: total − together, or arrange the rest and choose gaps.
5. “At least / at most”: list exact cases and add.
6. Two rows: arrange each row and multiply.

Small reminder: 7 people in a line with A and B not next to each other: 7! − 2 × 6! = 5040 − 1440 = 3600.

5.3 Probability

Must-know distinctions.

Mutually exclusive	Independent
Cannot happen together	One does not affect the other
P(A ∩ B) = 0	P(A ∩ B) = P(A) × P(B)
P(A or B) = P(A) + P(B)	P(A and B) = P(A) × P(B)

Two events with non-zero probabilities cannot be both exclusive and independent.

Conditional probability from a tree.

1. Draw the tree; later branches may depend on earlier ones (no replacement).
2. Numerator: the branch (or branches) where both A and B happen.
3. Denominator: every branch where B happens.
4. Divide. Leave as a fraction or give 3 s.f.

The general formula P(A ∪ B) = P(A) + P(B) − P(A ∩ B) is not required explicitly by the syllabus; a Venn diagram or a count of outcomes is enough.

5.4 Discrete random variables

Probability distribution table. List all values of X; check Σp = 1; E(X) = Σxp; Var(X) = Σx²p − {E(X)}². An unknown probability is found from Σp = 1, and a second unknown from a given E(X).

Worked reminder (two unknowns). X takes the values 1, 2, 3, 4 with probabilities 0.1, a, b, 0.3, and E(X) = 2.7.

Σp = 1:     a + b = 0.6
E(X) = 2.7: 0.1 + 2a + 3b + 1.2 = 2.7  →  2a + 3b = 1.4
solve:      b = 0.2, a = 0.4
E(X²) = 0.1 + 1.6 + 1.8 + 4.8 = 8.3
Var(X) = 8.3 − 2.7² = 1.01

Binomial or geometric?

B(n, p)	Geo(p)
Fixed number of trials n	Trials continue until the first success
X = number of successes, 0 to n	X = trial on which the first success occurs, 1, 2, 3, …
Mean np, variance np(1 − p)	Mean 1/p (variance not required)

Both need independent trials and a constant probability p.

Useful rewrites. P(X ≥ 1) = 1 − P(X = 0). P(X > r) for Geo(p) = (1 − p)ʳ. “Before the kth trial” for Geo(p) means X ≤ k − 1.

5.5 The normal distribution

Method in steps (probability from a value).

1. Sketch the curve; mark μ and shade the region.
2. z = (x − μ)/σ, at least 3 d.p. when possible.
3. Φ(z) from the table; for a negative z, use 1 − Φ(|z|).
4. Right-hand tail: 1 − Φ(z). Between two values: Φ(z₂) − Φ(z₁).

Method in steps (value from a probability).

1. Turn the probability into an area to the left.
2. Use the critical values table (e.g. 0.95 → 1.645) or read the main table backwards.
3. Give z the correct sign: negative if the area to the left is below 0.5.
4. Solve (x − μ)/σ = z; with two unknowns, solve two equations simultaneously.

Worked reminder (μ and σ both unknown). P(X < 40) = 0.05 and P(X > 70) = 0.1.

(40 − μ)/σ = −1.645      (70 − μ)/σ = 1.282
subtract:  30 = 2.927σ  →  σ = 10.2 (3 s.f.)
μ = 40 + 1.645 × 10.249 = 56.9 (3 s.f.)

Keep σ unrounded when you substitute back for μ.

Normal approximation to B(n, p). Conditions: np > 5 and nq > 5. Use N(np, npq). Continuity corrections:

Binomial	Normal
P(X ≤ 15)	P(Y < 15.5)
P(X < 15)	P(Y < 14.5)
P(X ≥ 15)	P(Y > 14.5)
P(X = 15)	P(14.5 < Y < 15.5)

Quick self-test

1. How many different arrangements are there of the letters of CASSETTE?
2. In how many ways can 4 books be chosen from 9 different books?
3. For 10 values, Σx = 85 and Σx² = 790. Find the mean and standard deviation.
4. P(A) = 0.4, P(B) = 0.5 and P(A ∩ B) = 0.2. Are A and B independent? Find P(A | B).
5. X ~ B(8, 0.4). Find P(X = 3).
6. X ~ Geo(0.25). Find P(X = 3) and E(X).
7. X ~ Geo(0.25). Find P(X > 4).
8. Find P(Z < −0.84) where Z ~ N(0, 1).
9. X ~ N(30, 4²). Find P(X < 25).
10. Can the normal approximation be used for B(40, 0.1)? Give a reason.
11. A class 15 ≤ x < 25 has frequency 34. Find its frequency density.
12. Two fair dice are thrown. Find P(total is at least 10).

Answers

1. 8! ÷ (2! × 2! × 2!) = 5040 (S, T and E each appear twice).
2. ⁹C₄ = 126.
3. Mean = 8.5; standard deviation = √(79 − 8.5²) = √6.75 = 2.60 (3 s.f.).
4. 0.4 × 0.5 = 0.2 = P(A ∩ B), so independent. P(A | B) = 0.2 ÷ 0.5 = 0.4.
5. ⁸C₃ × 0.4³ × 0.6⁵ = 56 × 0.064 × 0.07776 = 0.279 (3 s.f.).
6. 0.75² × 0.25 = 0.141 (3 s.f.); E(X) = 1 ÷ 0.25 = 4.
7. 0.75⁴ = 0.316 (3 s.f.).
8. 1 − Φ(0.84) = 1 − 0.7995 = 0.2005.
9. z = −1.25; 1 − Φ(1.25) = 1 − 0.8944 = 0.1056.
10. No: np = 4, which is not greater than 5.
11. 34 ÷ 10 = 3.4.
12. Totals 10, 11, 12 come from 3 + 2 + 1 = 6 outcomes; 6/36 = 1/6.

Where marks are usually lost

• Frequency plotted as bar height on a histogram with unequal class widths.
• Combined mean found by averaging two means from sets of different sizes.
• Standard deviation given when the question asked for variance, or the reverse.
• An “at least” selection done as one product that double-counts.
• Letters that repeat not divided out, or divided out when they are different.
• Independence claimed without comparing P(A ∩ B) with P(A) × P(B) numerically.
• Conditional probability left as P(A ∩ B), without dividing by P(B).
• Geometric P(X ≤ r) worked out term by term with an arithmetic slip, instead of 1 − (1 − p)ʳ.
• Continuity correction missed, or applied in the wrong direction.
• The z-value for “top 10%” given as −1.282 instead of +1.282.

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.