Uniform

Constant density over an interval, the natural model for equally likely continuous outcomes.

1. The idea

A Uniform $(a, b)$ random variable has constant density $f (x) = 1/ (b - a)$ on $[a, b]$ , so probabilities are proportional to length: $P (X \in [c, d]) = (d - c) / (b - a)$ .

This makes it the natural tool for geometric probability problems, where you compare areas or lengths directly instead of integrating.

Order statistics of iid Uniform $(0, 1)$ samples have clean closed forms — the $k$ -th smallest of $n$ samples follows a Beta $(k, n - k + 1)$ distribution, with mean $k / (n + 1)$ .

2. Why it matters

Many "random point on a segment/disk" and "random arrival time in an interval" questions reduce to comparing lengths or areas once you set up a Uniform model, and order statistics recur in "largest/smallest of $n$ random values" questions.

3. Core mathematics

Parameters

a < b

Support

x \in [a, b]

Expectation

E [X] = \frac{a + b}{2}

Variance

Var (X) = \frac{( b - a ) ^{2}}{12}

Key property

P (X \in [c, d]) = \frac{d - c}{b - a}

probability equals proportion of length

Order statistics

E [X_{(k)}] = \frac{k}{n + 1} (b - a) + a

for $n$ iid Uniform $(a, b)$ samples

4. How to recognize it

Think of this when…

"A point is chosen uniformly at random on $[0, 1]$ " or on a geometric region
Questions phrased in terms of lengths, areas, or "equally likely anywhere in the interval"
"Expected value of the maximum/minimum/ $k$ -th largest of $n$ random numbers"
Meeting-time or scheduling problems with independent uniform arrival times

5. Worked example

Worked example

Two people arrive independently at a meeting point at uniformly random times in $[0, 60]$ minutes. Find the probability they arrive within $10$ minutes of each other.

1
Let $X, Y \sim Uniform (0, 60)$ independent; the sample space is the square $[0, 60]^{2}$ with uniform density.
2
We want $P (∣ X - Y ∣ \leq 10)$ , a geometric probability: the region is a band around the diagonal.
3
The complement (two triangles) has total area $2 \times \frac{1}{2} (50)^{2} = 2500$ out of total area $6 0^{2} = 3600$ .
4
So the band's probability is one minus that ratio.
$P (∣ X - Y ∣ \leq 10) = 1 - \frac{2500}{3600} = \frac{1100}{3600} = \frac{11}{36}$
Answer: $P \approx 0.306$ , i.e. $11/36$ .

6. Common mistakes

Forgetting that Uniform probabilities are ratios of length/area, not counts
Using the wrong order-statistic formula direction (mixing up $k$ -th smallest vs. $k$ -th largest)
Setting up a 2D geometric probability problem but computing area with the wrong region boundaries
Assuming uniformity over an interval implies uniformity over a transformed variable (e.g. $X^{2}$ is not Uniform even if $X$ is)

7. Quick check

If $X \sim Uniform (0, 10)$ , what is $Var (X)$ ?

For $5$ iid Uniform $(0, 1)$ samples, what is the expected value of the maximum?

8. Practice

35 QuantSolve questions use this concept.

Practice this concept

9. Related methods

Geometric Probability Order Statistics

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