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The binomial distributionEdexcel International A Level Further Maths: Revision notes

Section 1

When a binomial model applies

A binomial distribution B(n,p)\mathrm{B}(n,p) models the number of successes XX in nn trials provided:

  • there is a fixed number of trials, nn;
  • each trial has just two outcomes, success or failure;
  • the probability of success pp is the same for every trial;
  • the trials are independent. Write X∼B(n,p)X\sim\mathrm{B}(n,p) and q=1−pq=1-p. Example: the number of sixes in 1010 rolls of a fair die is B(10,16)\mathrm{B}\left(10,\frac16\right).
Key termsbinomial distributionindependent trials
Exam tip

In a context question, state the model first: 'Let XX be the number of ... then X∼B(n,p)X\sim\mathrm{B}(n,p)'.

Section 2

Calculating probabilities

P(X=r)=(nr)pr(1−p)n−r,r=0,1,…,n.\mathrm{P}(X=r)=\binom nr p^r(1-p)^{n-r},\quad r=0,1,\dots,n. The coefficient (nr)=n!r!(n−r)!\binom nr=\frac{n!}{r!(n-r)!} counts the arrangements of the rr successes among the nn trials. Example: X∼B(8,0.4)X\sim\mathrm{B}(8,0.4). P(X=3)=(83)(0.4)3(0.6)5=56×0.064×0.07776=0.279\mathrm{P}(X=3)=\binom83(0.4)^3(0.6)^5=56\times0.064\times0.07776=0.279.

Key termsprobability function
Common mistake

Leaving out (nr)\binom nr, or swapping pp and qq. Check that the powers add up to nn.

Section 3

Cumulative probabilities and tables

Tables give P(X≤r)\mathrm{P}(X\le r). Convert every probability into this form:

  • P(X≥r)=1−P(X≤r−1)\mathrm{P}(X\ge r)=1-\mathrm{P}(X\le r-1)
  • P(X>r)=1−P(X≤r)\mathrm{P}(X>r)=1-\mathrm{P}(X\le r)
  • P(a≤X≤b)=P(X≤b)−P(X≤a−1)\mathrm{P}(a\le X\le b)=\mathrm{P}(X\le b)-\mathrm{P}(X\le a-1) Example: X∼B(8,0.4)X\sim\mathrm{B}(8,0.4). P(X≥5)=1−P(X≤4)=1−0.8263=0.1737\mathrm{P}(X\ge5)=1-\mathrm{P}(X\le4)=1-0.8263=0.1737 and P(3≤X≤5)=0.9502−0.3154=0.6348\mathrm{P}(3\le X\le5)=0.9502-0.3154=0.6348. Tables stop at p=0.5p=0.5. For p>0.5p>0.5, let Y=n−X∼B(n,1−p)Y=n-X\sim\mathrm{B}(n,1-p). Example: X∼B(10,0.7)X\sim\mathrm{B}(10,0.7), P(X≤4)=P(Y≥6)=1−0.9527=0.0473\mathrm{P}(X\le4)=\mathrm{P}(Y\ge6)=1-0.9527=0.0473. A calculator can give the same values.
Key termscumulative probability
Common mistake

Reading P(X≤r)\mathrm{P}(X\le r) for P(X≥r)\mathrm{P}(X\ge r). For 'at least rr' use 1−P(X≤r−1)1-\mathrm{P}(X\le r-1).

Section 4

Mean and variance

For X∼B(n,p)X\sim\mathrm{B}(n,p), no derivation needed: E(X)=np,Var(X)=np(1−p).\mathrm{E}(X)=np,\qquad\mathrm{Var}(X)=np(1-p). The standard deviation is np(1−p)\sqrt{np(1-p)}. The variance is always less than the mean. Example: B(30,0.4)\mathrm{B}(30,0.4) has mean 1212 and variance 7.27.2. To find nn and pp from the mean and variance, divide: np(1−p)np=1−p\frac{np(1-p)}{np}=1-p. If the mean is 1212 and the variance is 7.27.2, then 1−p=0.61-p=0.6, p=0.4p=0.4 and n=30n=30.

Key termsmeanvariance

Section 5

Commenting on the model

Say whether each condition is reasonable in the context.

  • Independence may fail when trials affect each other, for example sampling without replacement from a small population, or members of a group influenced by one another.
  • Constant pp may fail when the probability changes with time, person or batch.
  • Two outcomes: a trial with several outcomes needs to be simplified to success or failure.
  • Fixed nn: the number of trials must not depend on the outcomes. Compare data with the model: for binomial counts the variance is smaller than the mean. If observed variance is much larger, trials are probably not independent or pp varies.
Key termsappropriate model
Exam tip

Give a reason set in the context, not a general statement. 'Seeds in the same tray share watering' is better than 'they are not independent'.

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Exam questions on The binomial distribution

  1. The discrete random variable XX has the binomial distribution B(12, 0.25)\mathrm{B}(12,\,0.25).
    Find P(X≥2)\mathrm{P}(X\ge2).2 marks
  2. A machine makes components. Each component is independently defective with probability 0.040.04. A random sample of 2020 components is taken, and the number of defective components in the sample is XX.
    Find the probability that more than 22 components in the sample are defective.2 marks
  3. A multiple-choice test has 1515 questions. Each question has four options, of which exactly one is correct. A student chooses an option at random for every question, independently. The number of correct answers is XX.
    Find the probability that the student gets exactly 55 questions correct.3 marks
See the full worksheet

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).