amazon.co.ukProbability rules, conditional probability, tree and Venn diagrams, the binomial and normal distributions, and hypothesis testing with critical regions - A-Level Maths statistics (Edexcel 9MA0 Topics 13-15, AQA 7357 M-O) revised with worked examples and mark-scheme logic.
Probability at A-level runs on three laws (Edexcel 9MA0 Topic 13; AQA 7357 M1-M3; OCR A H640 18.01). The addition law: P(A or B) = P(A) + P(B) - P(A and B), simplifying to P(A) + P(B) when the events are mutually exclusive (cannot both happen). The conditional probability: P(A|B) = P(A and B)/P(B) - the probability of A given B occurred, found by restricting the sample space to B. The independence condition: P(A and B) = P(A) x P(B), equivalently P(A|B) = P(A) - knowing B occurred changes nothing about A. Exam questions often ask you to test independence from a table or Venn diagram: compute P(A)P(B) and P(A and B) separately and compare.
Venn diagrams organise overlapping events: fill from the intersection outwards, and read probabilities as region sizes over the total. Tree diagrams organise sequential events: multiply along branches (the multiplication law) and add across branches (the addition law) - and remember the second-level branches are conditional probabilities.
Machine A produces 60% of a factory's items with a 2% defect rate; machine B produces the other 40% with a 5% defect rate. Find P(defective) and P(A | defective).
Line 1: P(A and D) = 0.6 x 0.02 = 0.012 and P(B and D) = 0.4 x 0.05 = 0.020 (M1: multiply along branches).
Line 2: P(D) = 0.012 + 0.020 = 0.032 (A1: add across branches).
Line 3: P(A | D) = P(A and D)/P(D) = 0.012/0.032 = 0.375 (M1 A1: conditional probability).
Of 100 students, 60 study maths, 45 study physics and 20 study both. Find P(maths or physics), P(neither) and P(maths | physics).
Line 1: P(M or P) = 0.60 + 0.45 - 0.20 = 0.85 (M1 A1: addition law).
Line 2: P(neither) = 1 - 0.85 = 0.15 (A1: complement).
Line 3: P(M | P) = P(M and P)/P(P) = 0.20/0.45 = 4/9 (A1: restricted sample space).
The binomial distribution X ~ B(n, p) models a count of successes in n independent trials, each with success probability p and two outcomes (Edexcel 9MA0 Topic 14; AQA 7357 N1-N2; OCR A H640 19.01). The four conditions are examinable and worth a mark: a fixed number of trials n, two outcomes per trial (success/failure), a constant probability p, and independent trials. If any condition fails, the binomial is the wrong model - a question asking why a binomial model may not be appropriate wants the failed condition named in context (e.g. trials not independent when sampling without replacement from a small population).
The probability formula: P(X = r) = nCr p^r (1 - p)^(n-r). The mean and variance: E(X) = np and Var(X) = np(1 - p) - both formulae are in the booklet but are recalled constantly. Probabilities come from the formula, from statistical tables, or from calculator functions (binomial PD for single values, binomial CD for cumulative) - know which your calculator offers and check a known value like P(X = 0) = (1 - p)^n before trusting it.
X ~ B(10, 0.5). Find P(X >= 8).
Line 1: P(X = 8) = 10C8 (0.5)^8 (0.5)^2 = 45/1024 (M1: formula).
Line 2: P(X = 9) = 10C9 (0.5)^9 (0.5) = 10/1024; P(X = 10) = (0.5)^10 = 1/1024 (A1: remaining terms).
Line 3: P(X >= 8) = (45 + 10 + 1)/1024 = 56/1024 = 0.0547 (3 s.f.) (A1).
Watch the inequality direction: P(X >= 8) needs 1 - P(X <= 7) when working from cumulative tables - off-by-one errors at the boundary are the most common binomial slip, so write the boundary explicitly. In context questions, the expected-value result earns interpretation marks: for 50 rolls of a fair die modelled binomially (rolling a six), E(X) = 50 x (1/6) = 8.33 sixes on average, and Var(X) = 50 x (1/6) x (5/6) = 6.94 - quote the units (a count) and interpret in context rather than leaving bare numbers.
The normal distribution X ~ N(mu, sigma^2) is continuous, symmetric and bell-shaped, fully described by its mean mu and variance sigma^2 (Edexcel 9MA0 Topic 14; AQA 7357 N3; OCR A H640 21.01). It is standardised via Z = (X - mu)/sigma, converting any normal to the standard normal Z ~ N(0, 1) read from tables or the calculator's inverse normal function. The 68-95-99.7 rule gives the shape: about 68.3% of values lie within 1 sigma of the mean, 95.4% within 2 sigma and 99.7% within 3 sigma - use it to sanity-check answers.
Forward problems: given X values, standardise and read probabilities. Inverse problems: given a probability, find the X value (or sigma, or mu) by reversing - locate the z-value from the probability, then convert back. Inverse questions are where marks are won: set up the standardisation equation carefully and keep the sign of z consistent with the direction of the inequality.
If X ~ N(100, 15^2), find P(85 < X < 115).
Line 1: Standardise: z = (85 - 100)/15 = -1 and z = (115 - 100)/15 = 1 (M1: both limits).
Line 2: P(-1 < Z < 1) = 0.8413 - 0.1587 = 0.6826 (A1: agreeing with the 68.3% rule).
X ~ N(mu, sigma^2). Given that P(X < 60) = 0.2119 and mu = 70, find sigma.
Line 1: Standardise: P(Z < (60 - 70)/sigma) = 0.2119 (M1).
Line 2: From tables, the z-value with 0.2119 below it is -0.8 (A1: inverse read).
Line 3: (60 - 70)/sigma = -0.8, so -10 = -0.8 sigma, giving sigma = 12.5 (A1).
Since the normal is symmetric, P(X < mu) = 0.5, and probabilities mirror: P(X > mu + a) = P(X < mu - a) - exploiting symmetry halves the table-reading in many questions.
A hypothesis test (Edexcel 9MA0 Topic 15; AQA 7357 O1-O2; OCR A H640 20.01) compares a null hypothesis H0 (no change, the assumed model) against an alternative H1 (the claimed change) using a test statistic. The language is examinable: the significance level is the threshold for rejecting H0 (5% or 1%); a one-tailed test has H1 with a direction (p > 0.5 or p < 0.5) and puts all the significance in one tail; a two-tailed test has H1 without direction (p is not 0.5) and splits the significance between the tails - 5% two-tailed means 2.5% per tail. The critical region is the set of test statistic values for which H0 is rejected, with the boundary the critical value.
The routine: (1) state H0 and H1 in terms of the parameter; (2) state the distribution of the test statistic under H0; (3) compute the p-value or critical region; (4) compare with the significance level; (5) conclude in context - reject or accept H0 with a sentence about what the evidence suggests.
A fair coin is tossed 20 times and 15 heads are observed. Test at the 5% level whether the coin is biased.
Line 1: H0: p = 0.5; H1: p is not 0.5 (two-tailed, so 2.5% per tail) (M1 B1: hypotheses, tail split).
Line 2: Under H0, X ~ B(20, 0.5) (B1: model stated).
Line 3: P(X >= 15) = 1 - P(X <= 14) = 0.0207 (M1 A1: p-value).
Line 4: 0.0207 < 0.025, so the result falls in the critical region: reject H0 (A1: comparison).
Line 5: There is significant evidence at the 5% level that the coin is biased (A1: conclusion in context).
Critical region questions run the test in reverse: find the largest c such that P(X <= c) < 0.025 (or the full region for a two-tailed test), stating both tails - and remember the actual significance may not be exactly 5%, since the binomial is discrete; questions ask for the actual significance level as the sum of both tail probabilities.
Model selection is examinable: a binomial fits a count of successes in fixed independent trials; a normal fits continuous symmetric measurements. When n is large and p is not too close to 0 or 1, the binomial is well approximated by the normal N(np, np(1 - p)) - a standard rule of thumb is that np and n(1 - p) should exceed 5 - so a binomial with n = 50, p = 0.5 can be worked with the normal, while n = 50, p = 0.05 cannot. The central limit theorem extends the idea: for a large random sample from any distribution, the sample mean is approximately N(mu, sigma^2/n) - the idea underpinning tests on sample means in some specifications and in Further Mathematics.
Exam technique across the topic: (1) write conditions for a binomial model in context, not in the abstract - name the trials and the success event; (2) state H0 and H1 with the parameter and the tail direction before computing anything - the hypotheses carry method marks; (3) halve the significance level for two-tailed tests and say so; (4) write probabilities to 4 decimal places from tables or calculator before rounding the final answer to 3 s.f.; (5) conclude in context: reject H0 with a sentence naming what the evidence suggests, not just reject H0; (6) use symmetry in normal questions to halve the work and check answers against the 68-95-99.7 rule. Board coverage: Edexcel 9MA0 Topics 13-15, AQA 7357 M1-O2, OCR A H640 18.01, 19.01, 20.01 and 21.01, WJEC/Eduqas and CCEA AS/A2 statistics - identical techniques, differing wording and table provision.