Statistics · Statistical Distributions · Year 1

Discrete Probability Distributions

A random variable turns the outcome of chance into a number, and a probability distribution lists every value it can take alongside how likely each one is.

9 min read 4 worked examples

What a random variable is

A random variable (usually a capital letter, like $X$) assigns a number to every outcome of a random experiment. A discrete random variable takes only a finite, or countable, list of possible values — the number rolled on a die, or the number of heads in three coin flips, rather than something measured on a continuous scale.

The probability distribution

A discrete random variable's probability distribution lists every value it can take, alongside the probability of each. It's normally shown as a table. Every valid probability distribution obeys two rules:

Every probability satisfies $0 \le P(X=x) \le 1$.

The probabilities sum to exactly $1$: $\displaystyle\sum P(X=x) = 1$.

Why the probabilities must sum to 1

The listed values of $X$ are mutually exclusive (the variable can only take one value at a time) and exhaustive (every possible outcome corresponds to one of the listed values) — exactly the conditions from the previous lesson that force probabilities to add to $1$. A probability distribution is really just an exhaustive set of mutually exclusive events, relabelled with numbers.

The discrete uniform distribution

The simplest discrete probability distribution is the discrete uniform distribution: every possible value of $X$ is equally likely. If $X$ can take any one of $n$ equally likely values,

$P(X=x) = \dfrac{1}{n} \quad \text{for each of the } n \text{ possible values.}$

The score on a fair die is the standard example: $X$ takes the values $1,2,3,4,5,6$, each with probability $\tfrac16$. You should be able to recognise this distribution from a description or a table — every entry in the $P(X=x)$ row is identical.

Image pending

discrete-random-variable-diagram.png

A probability bar chart for the score on a fair die: six bars at $x=1,\dots,6$, each of identical height $\tfrac16$ — the flat profile that makes a distribution "discrete uniform".

Recognising the shape

A distribution table with every probability equal is discrete uniform; a table that rises to a peak and falls away (like the binomial distribution in the next lesson) is not. Spotting this from the table alone, before doing any calculation, is often exactly what a question is testing.

Choosing and critiquing a model

In practice, a discrete probability distribution is used to model a real situation — and part of working with distributions is judging whether the model is actually appropriate, not just calculating with it. A discrete uniform distribution assumes every outcome is equally likely; a binomial distribution (the next lesson) assumes a fixed number of independent trials with a constant probability of success. Choosing the wrong model, or using one where its assumptions clearly don't hold, is a common way a "solution" can go wrong even when the arithmetic inside it is correct.

Questions to ask before trusting a model

Are all the listed outcomes genuinely equally likely, or does the context suggest some are more likely than others? Is there a fixed, known number of possible values? Could real-world factors — bias, dependence between trials, changing conditions — break the assumptions the distribution relies on? Being able to raise these questions, in context, is itself an examinable skill.

Worked example 1

Finding an unknown probability

The random variable $X$ has the distribution shown. Find the value of $k$.
$x$: 1, 2, 3, 4 · $P(X=x)$: 0.15, 0.3, $k$, 0.25

MethodUse ΣP(X=x) = 1

$0.15+0.3+k+0.25 = 1$.

All probabilities must sum to 1.

$0.7+k=1 \Rightarrow k=0.3$.

Solve for k.

▪ solved

Examiner's eye

This is the standard first step whenever a distribution table has a missing entry — always check the table sums to $1$ before doing anything else with it, since an error here invalidates every later calculation.

Worked example 2

Identifying a discrete uniform distribution

A spinner is divided into 5 equal sectors, numbered $1$ to $5$. Let $X$ be the number the spinner lands on. Write down the probability distribution of $X$, and state its name.

MethodEqual sectors mean equal probabilities

The 5 sectors are equal in size, so each of the 5 outcomes is equally likely.

Equal areas give equal probabilities.

$P(X=x) = \dfrac15$ for $x=1,2,3,4,5$.

n = 5 equally likely outcomes.

Every entry in the distribution table is identical, so $X$ follows a discrete uniform distribution.

Matches the definition exactly.

▪ solved

Examiner's eye

Check $\sum P(X=x) = 5\times\tfrac15 = 1$ as a quick sanity check — it always should, but it's especially easy to verify for a uniform distribution since every term is the same.

Worked example 3

A distribution defined by a formula

A discrete random variable $X$ has $P(X=x) = \dfrac{x}{10}$ for $x=1,2,3,4$. Show this is a valid probability distribution, and find $P(X\ge 3)$.

MethodCheck the sum, then add the relevant probabilities

$P(X=x)$ for $x=1,2,3,4$: $\dfrac{1}{10}, \dfrac{2}{10}, \dfrac{3}{10}, \dfrac{4}{10}$.

Substitute each value of x.

Sum: $\dfrac{1+2+3+4}{10} = \dfrac{10}{10}=1$, so it is a valid distribution.

Confirms ΣP(X=x)=1.

$P(X\ge 3) = P(X=3)+P(X=4) = \dfrac{3}{10}+\dfrac{4}{10} = \dfrac{7}{10}$.

Add the probabilities of every qualifying value.

▪ solved

Common mistake

When a distribution is given by a formula rather than a table, still write out the individual probabilities before summing — trying to check $\sum P(X=x)=1$ or add probabilities "in your head" from the formula alone is where arithmetic slips creep in.

Worked example 4

Critiquing a model · Challenge

A teacher models the number of days per week (out of 5 school days) that a particular student arrives late, $X$, using a discrete uniform distribution on $\{0,1,2,3,4,5\}$. Explain why this is probably not an appropriate model, and suggest a more suitable distribution.

MethodTest the model's assumptions against the real context

A discrete uniform distribution assumes every outcome from $0$ to $5$ is equally likely — including being late all $5$ days, which is just as likely as being late $0$ days.

States the assumption being tested.

In reality, most students are late on relatively few days, and being late every single day is far less likely than being on time most days — the outcomes are not equally likely.

Checks the assumption against real behaviour.

A more suitable model is the binomial distribution: 5 independent school days, each with some fixed probability $p$ of being late, giving $X\sim B(5,p)$.

Binomial fits a fixed number of trials with constant success probability better than uniform.

▪ critiqued

Look deeper — this is what "select an appropriate distribution" means

Exam questions often describe a real situation and ask you to name a distribution and justify it, or to spot why a suggested distribution is wrong. The justification always comes back to the model's assumptions — equally likely outcomes for discrete uniform; a fixed number of independent trials with constant probability for binomial — matched against what the context actually describes.

Common mistake

Don't assume a distribution is uniform just because you aren't told the exact probabilities — "equally likely" has to be justified by the context (equal areas, a fair die, a well-shuffled deck), not assumed by default when information is missing.

Check yourself

A fair coin is tossed twice, and $X$ is the number of heads obtained. By listing all four equally likely outcomes, show that $X$ does not follow a discrete uniform distribution, and write out the full probability distribution of $X$.
(Hint: list HH, HT, TH, TT first, then count heads in each.)

Method

The four equally likely outcomes are HH, HT, TH, TT, each with probability $\tfrac14$.

Number of heads: HH → 2, HT → 1, TH → 1, TT → 0.

Count heads in each outcome.

$P(X=0)=\tfrac14$, $P(X=1)=\tfrac24=\tfrac12$ (from HT and TH), $P(X=2)=\tfrac14$.

Combine outcomes giving the same X value.

The probabilities are $\tfrac14, \tfrac12, \tfrac14$ — not all equal, so $X$ is not uniform, even though the four underlying outcomes were.

Equally likely outcomes don't guarantee an equally likely X.

▪ solved

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Tree Diagrams

The Binomial Distribution