Discrete uniform distributionAQA A-Level Further Maths: Revision notes

Section 1

The discrete uniform distribution

A DRV $X$ has a discrete uniform distribution on $\{1,2,\ldots,n\}$ if each of the $n$ values is equally likely: $P(X=x)=\frac1n,\qquad x=1,2,\ldots,n.$ Example: a fair eight-sided die gives $P(X=x)=\frac18$ for $x=1,\ldots,8$ . Probabilities are found by counting: $P(X>5)=\frac{3}{8}$ , and for a set of $k$ values $P=\frac kn$ .

Key termsdiscrete uniform distribution

Exam tip

Count the favourable integers carefully: $P(X>5)$ for $n=8$ is $6,7,8$ , so three values.

Section 2

When can it be used as a model?

Use the discrete uniform model when:

the outcomes form a finite set of consecutive integers $1$ to $n$ (or can be numbered that way);
every outcome is equally likely, e.g. a fair die, a spinner with equal sectors, a ticket drawn at random from a numbered set. It is not suitable when outcomes are not equally likely (a biased die), when values are sums of random events (the total of two dice is more likely to be $7$ than $2$ ), or when the variable is a count with a different structure (number of heads in several tosses). In an answer, state the assumption in context: for example, 'each ticket is equally likely to be selected'.

Key termsequally likelymodel

Common mistake

Using the uniform model for the total of two dice. The totals are not equally likely.

Section 3

Mean and variance

For $X$ uniform on $\{1,2,\ldots,n\}$ : $E(X)=\frac{n+1}{2},\qquad \mathrm{Var}(X)=\frac{n^2-1}{12}.$ Example: a fair six-sided die has $E(X)=3.5$ and $\mathrm{Var}(X)=\frac{35}{12}=2.92$ . For $n=20$ , $E(X)=10.5$ and $\mathrm{Var}(X)=\frac{399}{12}=33.25$ . The mean is the middle of the range and need not be a possible value. You can solve for $n$ from a given mean or variance: $E(X)=10.5$ gives $n=20$ ; $\mathrm{Var}(X)=2$ gives $n^2=25$ so $n=5$ .

Key termsmean of uniformvariance of uniform

Common mistake

Writing $\frac n2$ for the mean or $\frac{n^2}{12}$ for the variance.

Section 4

Proof of the mean

Start from the definition with $P(X=x)=\frac1n$ : $E(X)=\sum_{x=1}^{n}x\cdot\frac1n=\frac1n\sum_{x=1}^{n}x=\frac1n\cdot\frac{n(n+1)}{2}=\frac{n+1}{2}.$ The key fact is the sum of the first $n$ integers, $\sum x=\frac{n(n+1)}{2}$ .

Key termssum of integers

Exam tip

In a 'prove that' question, every line must follow from the previous one; do not skip the substitution of the standard sum.

Section 5

Proof of the variance

Use $\mathrm{Var}(X)=E(X^2)-[E(X)]^2$ and the standard sum $\sum x^2=\frac{n(n+1)(2n+1)}{6}$ : $E(X^2)=\frac1n\cdot\frac{n(n+1)(2n+1)}{6}=\frac{(n+1)(2n+1)}{6}.$ $\mathrm{Var}(X)=\frac{(n+1)(2n+1)}{6}-\frac{(n+1)^2}{4}=\frac{(n+1)\left[2(2n+1)-3(n+1)\right]}{12}=\frac{(n+1)(n-1)}{12}=\frac{n^2-1}{12}.$ Use a common denominator of $12$ and factor out $(n+1)$ before simplifying.

Key termsvariance proof

Exam tip

Factor $(n+1)$ out of both terms before expanding; it keeps the algebra short.

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Discrete uniform distribution

A fair eight-sided die, numbered $1$ to $8$ , is rolled once. The score $X$ is the number on the face it lands on, and $X$ is modelled by the discrete uniform distribution on $\{1,2,\ldots,8\}$ .
Find the variance of $X$ .2 marks
The discrete random variable $X$ has a discrete uniform distribution on $\{1,2,\ldots,n\}$ , where $n$ is a positive integer. It is given that $E(X)=10.5$ .
Find $P(X>E(X))$ .2 marks
The discrete random variable $X$ has a discrete uniform distribution on $\{1,2,\ldots,n\}$ , so that $P(X=x)=\frac1n$ for $x=1,2,\ldots,n$ .
Prove that $E(X)=\frac{n+1}{2}$ .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).