The simplest distribution there is: finitely many values, every one equally likely. It is worth its own lesson for two reasons. It is the model behind every fair die, raffle and random selection you will be asked about, and it is the one students most often assume when it does not hold.
Commit to a prediction about two random experiments.
Two experiments. A: spin a fair spinner with 8 equal sectors numbered 1 to 8. B: roll two fair dice and record the sum. Without looking ahead decide which produces equally likely values, and write down what you think decides the question.
State the definition and derive the probability of each value.
The definition · finitely many values, all equally likely
We just saw two random experiments that feel similar. That raises a question: what exactly has to be true of a distribution before we are entitled to call it uniform? This card answers it → it needs finitely many values and every one of them must carry the same probability, which forces that shared probability to be $\tfrac{1}{n}$.
A discrete random variable is uniformly distributed if it takes finitely many values and all of them have the same probability. Both halves matter: finitely many, and all equal.
Suppose $X$ takes exactly $n$ different values and each has the same probability $c$. The probabilities must sum to 1, so
So the shared probability is not a free choice. Once you know how many values there are, uniformity fixes every probability at once:
A fair six-sided die is the standard example: $n = 6$, so $p(x) = \tfrac{1}{6}$ for $x = 1, 2, 3, 4, 5, 6$. A raffle with 50 tickets is another: $n = 50$, so each ticket wins with probability $\tfrac{1}{50} = 0.02$.
$X$ is discrete uniform when it has finitely many values, all equally likely, and then $p(x) = 1/n$ where $n$ is the number of values.; The value $1/n$ follows from $\sum p(x) = 1$; it is not an extra assumption.
Pause, copy the definition of a discrete uniform distribution and the derivation $nc = 1 \Rightarrow c = 1/n$, into your book.
Quick check: A discrete uniform random variable takes the values $5, 10, 15, 20, 25$. What is $p(15)$?
Separate random from uniform, the distinction that earns the mark.
We just saw that uniformity forces $p(x) = \tfrac{1}{n}$. That raises a question: students routinely assume any "random" process is uniform, so how do you check rather than assume? This card answers it → go back to the equally likely outcomes of the experiment and count how many of them produce each value of $X$.
Return to the hook. Rolling two dice gives 36 equally likely ordered outcomes, and those outcomes are uniform. But $X$, the sum, is not, because different sums are produced by different numbers of outcomes:
| $x$ | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| outcomes | 1 | 2 | 3 | 4 | 5 | 6 | 5 | 4 | 3 | 2 | 1 |
$P(X = 7) = \tfrac{6}{36}$ while $P(X = 12) = \tfrac{1}{36}$. The experiment is random, the outcomes are equally likely, and the random variable is not uniform.
True or false: If an experiment uses fair, unbiased equipment, then any random variable defined on that experiment is uniformly distributed.
Use the uniform model on real selection problems.
We just saw how to test whether a variable is uniform. That raises a question: once you have established uniformity, what does it buy you? This card answers it → every probability question collapses to counting, because each value contributes the same $\tfrac{1}{n}$.
When $X$ is uniform on $n$ values, the probability of any event is just the count of favourable values over $n$:
Example. A random digit $X$ is generated from $0, 1, 2, \dots, 9$, uniformly. Then $n = 10$ and $p(x) = 0.1$ for each digit.
That last line is worth noticing: restricting a uniform variable to a subset leaves it uniform on that subset, which makes conditional probabilities on uniform variables unusually easy.
For a uniform $X$ on $n$ values, $P(X \in A) = |A| / n$, so probability questions become counting questions.; Conditioning a uniform variable on a subset leaves it uniform on that subset.
Pause, copy the counting rule $P(X \in A) = |A| / n$ for a uniform variable, and the note that conditioning on a subset preserves uniformity, into your book.
Fill the blanks: A card is drawn from a well-shuffled standard deck of 52 and $X$ is its suit, so $X$ is uniform on 4 values. Give each answer as a decimal.
$p(\text{hearts}) =$
$P(X \text{ is red}) =$
A 20-ticket raffle: $P(\text{your one ticket wins}) =$
Work three examples, then read the traps.
Worked examples · 3 in a row, reveal as you go
A discrete uniform random variable has $p(x) = 0.04$ for each of its values. How many values does it take?
A position from 1 to 10 is chosen at random in the word STATISTICS and $X$ is the letter at that position. Is $X$ uniformly distributed? Justify your answer.
A class of 24 students is numbered 1 to 24 and one student is selected at random to present. Let $X$ be the selected number. Find $P(X \leq 6)$ and $P(X \text{ is a multiple of } 4)$.
Common errors · traps that cost marks
Odd one out: Three of these random variables are uniformly distributed. Which one is not?
Work the quick-fire set, then revisit your prediction.
Quick-fire practice · recognise, compute, model
An eight-sided die is fair. Write the distribution of the score in tabular form and state $p(x)$.
$X$ is uniform on the integers from 5 to 20 inclusive. Find $n$, then $p(x)$, then $P(X > 16)$.
A letter is chosen by picking a random position in the word BANANA. Is the letter uniformly distributed? Show the probabilities.
Explain in one sentence why there is no discrete uniform distribution on the set of all positive integers.
$X$ is uniform with $p(x) = \dfrac{1}{16}$. A student claims $P(X \text{ takes one of its three smallest values}) = \dfrac{3}{16}$. Is the student right? Explain.
Match each random variable to its probability of taking any one specified value:
Earlier you were asked which of the two experiments produces equally likely values. A, the eight-sector spinner, is uniform: eight values, each from one equal sector, so $p(x) = \tfrac{1}{8}$. B, the sum of two dice, is not: the 36 underlying outcomes are equally likely, but they are shared unevenly among the 11 possible sums, so $P(X = 7) = \tfrac{6}{36}$ while $P(X = 12) = \tfrac{1}{36}$. What decides it is not whether the equipment is fair, but whether each value of the variable is produced by the same number of equally likely outcomes.
The uniform distribution is the cleanest place to meet the two summary numbers that come next: the expected value and the variance. Because every value carries the same weight $\tfrac{1}{n}$, the expected value of a uniform variable is simply the ordinary average of its values, which makes it the ideal first case to compute in the next lesson.
Apply the lesson methods, then compare each response with its comprehensive answer.
Use the visible short-answer practice below during this lesson. After completing the focus area, use Checkpoint 3 for the checkpoint question bank.
Q1. A fair twelve-sided die has faces numbered 1 to 12. Let $X$ be the number rolled.
(a) State the distribution of $X$ and give $p(x)$. (b) Find $P(X \leq 4)$ and $P(X \text{ is a multiple of } 3)$. (c) Find $P(X \text{ is a multiple of } 3 \mid X \leq 7)$, and explain why conditioning changes both the numerator and the denominator. (3 marks)
Q2. Two fair four-sided dice, numbered 1 to 4, are rolled. Let $Y$ be the larger of the two numbers shown (or that number, if both are the same).
(a) Explain why there are 16 equally likely ordered outcomes. (b) Construct the probability distribution of $Y$. (c) State with justification whether $Y$ is uniformly distributed. (3 marks)
Q3. A school runs a prize draw. Every student writes their name on a slip; students in Years 11 and 12 are given two slips each, all other year groups one. The slips are mixed and one is drawn. A student argues: "Every slip is equally likely, so this is a discrete uniform distribution and the draw is fair."
(a) Identify the random variable the student is describing, and one they are not. (b) Explain precisely which of the two is uniform and which is not. (c) Evaluate the claim that the draw is "fair", distinguishing the mathematical property from the everyday word. (3 marks)
Drill 1: $x = 1, 2, \dots, 8$ with $p(x) = \tfrac{1}{8} = 0.125$ for every $x$. The eight probabilities sum to $8 \times \tfrac{1}{8} = 1$ ✓
Drill 2: $n = 20 - 5 + 1 = 16$, so $p(x) = \tfrac{1}{16}$. $P(X > 16)$ counts 17, 18, 19, 20, which is 4 values, giving $\tfrac{4}{16} = \tfrac{1}{4}$.
Drill 3: BANANA has B once, A three times, N twice, over 6 positions. So $p(\text{A}) = \tfrac{3}{6} = \tfrac{1}{2}$, $p(\text{N}) = \tfrac{2}{6} = \tfrac{1}{3}$, $p(\text{B}) = \tfrac{1}{6}$. These are not all equal, so the letter is not uniformly distributed, even though the position was chosen uniformly.
Drill 4: If every positive integer had the same probability $c$, then either $c = 0$ and the total is 0, or $c > 0$ and the total is infinite. Neither can equal 1, so no such distribution exists. This is why the definition insists on finitely many values.
Drill 5: The student is right. Each of the 16 values contributes $\tfrac{1}{16}$, so any three of them together contribute $3 \times \tfrac{1}{16} = \tfrac{3}{16}$. For a uniform variable the probability of an event depends only on how many values it contains, not on which ones.
Q1 (3 marks): (a) $X$ is discrete uniform on $\{1, 2, \dots, 12\}$, so $n = 12$ and $p(x) = \tfrac{1}{12}$ for every $x$ [1]. (b) $P(X \leq 4) = \tfrac{4}{12} = \tfrac{1}{3}$. The multiples of 3 are 3, 6, 9, 12, so $P = \tfrac{4}{12} = \tfrac{1}{3}$ [1]. (c) Conditioning on $X \leq 7$ leaves 7 equally likely values, of which 3 and 6 are multiples of 3, so the answer is $\tfrac{2}{7}$. Conditioning changes both parts of the fraction: the numerator drops from 4 to 2, because 9 and 12 are excluded, and the denominator drops from 12 to 7. Changing only one of them is the standard error — keeping the denominator at 12 gives $\tfrac{2}{12}$, and keeping the numerator at 4 gives $\tfrac{4}{7}$, and both are wrong [1].
Q2 (3 marks): (a) Each die shows one of 4 faces and the two rolls are independent, so there are $4 \times 4 = 16$ equally likely ordered pairs [0.5]. (b) $Y = 1$ only from $(1,1)$: 1 outcome. $Y = 2$ from $(1,2), (2,1), (2,2)$: 3 outcomes. $Y = 3$ from the 5 pairs whose maximum is 3. $Y = 4$ from the 7 pairs whose maximum is 4. So $p(1) = \tfrac{1}{16}$, $p(2) = \tfrac{3}{16}$, $p(3) = \tfrac{5}{16}$, $p(4) = \tfrac{7}{16}$, and these sum to $\tfrac{16}{16} = 1$ ✓ [1.5]. (c) $Y$ is not uniform. Uniformity on 4 values would require every $p(y) = \tfrac{1}{4} = \tfrac{4}{16}$, but the probabilities run from $\tfrac{1}{16}$ to $\tfrac{7}{16}$. The underlying 16 outcomes are equally likely; the maximum is not, because larger values are produced by more of them [1].
Q3 (3 marks): (a) The student is describing the slip drawn. The variable that actually matters to the question of fairness is the student who wins [0.5]. (b) The slip drawn is uniform: every slip is physically identical and equally likely to be selected. The winning student is not uniform: a Year 11 or 12 student holds two of the slips and so is twice as likely to win as a student from any other year [1.5]. (c) Mathematically, a "fair" draw would mean the winning student is uniformly distributed over all students, and it is not. But the design is not an error, it is a deliberate weighting in favour of senior students, and whether that is fair in the everyday sense is a question about the school's intention rather than about the mathematics. The precise statement is: the draw is uniform over slips and non-uniform over students [1].
Retrieve the central ideas, then mark the lesson complete or continue to the module quiz.
Answer from memory first, then return to the matching Learn checkpoint to check and correct your response.
A full module quiz covering every lesson in this module, not just this one. Set aside a decent block of time and treat it like a real assessment.
Start the module quiz →Tick when you've finished the practice and review.
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