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The binomial distributionAQA A-Level Maths: Revision notes

Section 1

When the binomial model applies

A random variable XX has a binomial distribution, written X∼B(n,p)X\sim B(n,p), when it counts the number of successes in nn trials and these conditions hold:

  • there is a fixed number nn of trials;
  • each trial has two outcomes, success or failure;
  • the probability of success pp is the same for every trial;
  • the trials are independent. Example: 10 seeds, each germinating with probability 0.30.3 independently, gives X∼B(10,0.3)X\sim B(10,0.3). State the conditions in the context, such as 'each seed germinates with the same probability, independently of the others'.
Key termsbinomial distributiontrialsuccess
Common mistake

Giving a generic condition such as 'the trials are random'. Link each condition to the context.

Section 2

Calculating probabilities

For X∼B(n,p)X\sim B(n,p), the probability of exactly rr successes is P(X=r)=(nr)pr(1−p)n−r,P(X=r)=\binom{n}{r}p^r(1-p)^{n-r}, where (nr)\binom nr counts the arrangements of rr successes among nn trials. For X∼B(10,0.3)X\sim B(10,0.3): P(X=3)=(103)(0.3)3(0.7)7=120×0.027×0.0823543=0.267P(X=3)=\binom{10}{3}(0.3)^3(0.7)^7=120\times0.027\times0.0823543=0.267. The probability distribution is the set of all P(X=r)P(X=r), which add to 1. The calculation of the mean and variance of XX is not needed at this level; concentrate on probabilities.

Key termsbinomial coefficientprobability distribution
Common mistake

Leaving out (nr)\binom nr. (0.3)3(0.7)7(0.3)^3(0.7)^7 is the probability of one particular order only.

Section 3

Using the calculator

Use the binomial probability function for P(X=r)P(X=r) and the cumulative binomial function for P(X≤r)P(X\le r). Enter nn, pp and rr. Cumulative functions only give 'at most', so rewrite other probabilities in that form:

  • P(X<r)=P(X≤r−1)P(X<r)=P(X\le r-1)
  • P(X≥r)=1−P(X≤r−1)P(X\ge r)=1-P(X\le r-1)
  • P(X>r)=1−P(X≤r)P(X>r)=1-P(X\le r)
  • P(a≤X≤b)=P(X≤b)−P(X≤a−1)P(a\le X\le b)=P(X\le b)-P(X\le a-1) Example: X∼B(20,0.25)X\sim B(20,0.25). P(X>7)=1−P(X≤7)=1−0.898=0.102P(X>7)=1-P(X\le7)=1-0.898=0.102. For X∼B(12,0.15)X\sim B(12,0.15): P(2≤X≤4)=P(X≤4)−P(X≤1)=0.9761−0.4435=0.533P(2\le X\le4)=P(X\le4)-P(X\le1)=0.9761-0.4435=0.533.
Key termscumulative probability
Exam tip

Write each probability as 'at most' first, then use the calculator. Check the inequality is strict (>>) or not (≥\ge).

Section 4

Using the model: repeated and combined situations

The probability from a binomial calculation can itself be used in another probability. If a box is rejected with probability p=0.135p=0.135 and 5 boxes are independent, P(at least one rejected)=1−(1−p)5=0.516P(\text{at least one rejected})=1-(1-p)^5=0.516. If 12 patients are booked and there are 10 slots, more patients attend than slots when at most 1 does not attend: P(X≤1)=0.443P(X\le1)=0.443. Watch the wording: decide what counts as a success before choosing pp. 'Does not attend' might be success with p=0.15p=0.15, so 'more than 10 attend' becomes X≤1X\le1.

Key termsat least one
Exam tip

Define XX and its distribution first, then translate the question into a statement about XX.

Section 5

The binomial distribution as a model

The binomial distribution is a model: real situations only approximately meet its conditions. Evaluate the model by questioning the conditions. In a clinic, patients from one family may attend or miss together, so independence may not hold. Quality-control examples may fail the constant probability condition if a machine drifts over time. If the conditions are not met, the calculated probabilities may be inaccurate, so state the limitation and, where possible, the likely effect. A model using a sample can also estimate pp from real data, but that value then carries its own uncertainty.

Key termsmodelindependence
Exam tip

In 'state an assumption' questions, say how the assumption could fail in this particular situation.

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Exam questions on The binomial distribution

  1. A gardener plants 10 seeds. Each seed germinates with probability 0.30.3, independently of the others. The number of seeds that germinate is XX, where X∼B(10,0.3)X\sim B(10,0.3).
    State two conditions that must hold for XX to be modelled by a binomial distribution in this context.2 marks
  2. A multiple-choice test has 20 questions, each with four options of which exactly one is correct. A student guesses every answer. The number of correct answers is XX, where X∼B(20,0.25)X\sim B(20,0.25).
    Find the probability that the student gets more than 7 questions correct.2 marks
  3. A factory makes light bulbs and 8% of them are defective, independently of one another. A box contains 25 bulbs. The number of defective bulbs in a box is YY, where Y∼B(25,0.08)Y\sim B(25,0.08).
    Find the probability that a box contains exactly 2 defective bulbs.3 marks
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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).