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How do you model the number of successes in a fixed number of independent trials, and find probabilities from that model?

The conditions for a binomial model, the binomial probability formula, calculating individual and cumulative probabilities, the mean of a binomial distribution, and using the model in context.

A focused answer to the AQA A-Level Mathematics binomial distribution content, covering the conditions for a binomial model, the probability formula, individual and cumulative probabilities, the mean, and applying the model in context.

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  1. What this dot point is asking
  2. Conditions for a binomial model
  3. The probability formula
  4. Cumulative probabilities
  5. Mean and variance
  6. Modelling in context

What this dot point is asking

AQA wants you to recognise when a binomial model applies, use the binomial probability formula, calculate individual and cumulative probabilities (using the calculator's binomial functions on Paper 3), find the mean of a binomial distribution, and apply the model to real contexts while stating its assumptions. The binomial is also the model behind one-tailed and two-tailed hypothesis tests for a proportion, so it links directly forward.

Conditions for a binomial model

A classic failure is sampling without replacement from a small population: the probability of success changes after each draw, breaking both the constant-probability and independence conditions. Sampling from a very large population, by contrast, keeps pp effectively constant, so the binomial is a reasonable model.

The probability formula

The three factors each have a meaning: (nr)\binom{n}{r} counts the orderings, prp^r is the probability of the rr successes, and (1p)nr(1-p)^{n-r} is the probability of the nrn - r failures. Understanding this structure helps you avoid dropping the binomial coefficient.

Cumulative probabilities

For ranges such as P(Xr)P(X \le r), P(Xr)P(X \ge r) or P(aXb)P(a \le X \le b), use the cumulative binomial function. Two conversions are essential: P(Xr)=1P(Xr1)P(X \ge r) = 1 - P(X \le r - 1), and P(X>r)=1P(Xr)P(X > r) = 1 - P(X \le r). Read inequalities carefully, since "fewer than 33" means P(X2)P(X \le 2) while "at most 33" means P(X3)P(X \le 3).

Mean and variance

The mean gives a quick sanity check: if you compute a probability for a value of rr far from npnp and it comes out large, you have likely made an error.

Modelling in context

Real questions rarely say "binomial"; you have to recognise it. Look for a fixed number of repeated trials, a clear success or failure for each, a constant success probability, and independence. Phrases such as "88 percent of components are faulty" or "each seed germinates with probability 0.70.7" give you pp, and "a sample of 2020" gives nn. Define your variable explicitly, for example "let XX be the number of faulty components, XB(20,0.08)X \sim B(20, 0.08)", because the examiner rewards a clear statement of the model.

When you are asked to criticise a binomial model, the usual targets are the independence and constant-probability assumptions. Components made on the same machine may fail together if the machine drifts, breaking independence; sampling without replacement from a small batch changes pp on each draw. Stating which specific assumption is doubtful, and why, earns the evaluation marks that a generic "it might not be accurate" does not.

Exam-style practice questions

Practice questions written in the style of AQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AQA 20196 marksPaper 3, Section A. A multiple-choice test has 1515 questions, each with four options. A student guesses every answer at random, so the number correct XX is modelled by a binomial distribution. (a) State the values of nn and pp. (b) Find the probability that the student gets exactly 55 correct. (c) Find the probability that the student gets at least 66 correct. (d) Find the expected number correct.
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For (a), n=15n = 15 and p=0.25p = 0.25, so XB(15,0.25)X \sim B(15, 0.25). For (b), P(X=5)=(155)(0.25)5(0.75)10=3003×0.255×0.75100.165P(X = 5) = \binom{15}{5}(0.25)^5(0.75)^{10} = 3003 \times 0.25^5 \times 0.75^{10} \approx 0.165. For (c), P(X6)=1P(X5)10.852=0.148P(X \ge 6) = 1 - P(X \le 5) \approx 1 - 0.852 = 0.148 using the cumulative binomial function. For (d), E(X)=np=15×0.25=3.75E(X) = np = 15 \times 0.25 = 3.75. Markers reward correct identification of nn and pp, use of the binomial pdf for the exact value, conversion of "at least 66" to 1P(X5)1 - P(X \le 5), and the mean formula.

AQA 20215 marksPaper 3, Section B. A factory finds that 88 percent of components are faulty. A random sample of 2020 components is taken. (a) Calculate the probability that exactly two are faulty. (b) Calculate the probability that fewer than three are faulty. (c) State one assumption needed for the binomial model to be valid.
Show worked answer →

Model XB(20,0.08)X \sim B(20, 0.08). For (a), P(X=2)=(202)(0.08)2(0.92)18=190×0.0064×0.92180.271P(X = 2) = \binom{20}{2}(0.08)^2(0.92)^{18} = 190 \times 0.0064 \times 0.92^{18} \approx 0.271. For (b), "fewer than three" means P(X2)=P(0)+P(1)+P(2)0.787P(X \le 2) = P(0) + P(1) + P(2) \approx 0.787 from the cumulative function. For (c), the faults must occur independently and at a constant rate of 88 percent (a fixed probability per component). Markers reward correct parameters, careful handling of the strict inequality (fewer than three is at most two), and a sensible stated assumption.

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