A passenger arrives uniformly at random within a 60-minute window: what is the probability the arrival falls in the final 10 minutes?

## What the continuous uniform distribution says A continuous uniform distribution on an interval `[a, b]` is the formal way to say "any point in this range is as likely as any other". Its probability density function is flat: ``` f(x) = 1 / (b - a) for a <= x <= b, and 0 outside ``` For a 60-minute window, `a = 0`, `b = 60`, so `f(x) = 1/60` for every minute in the window. The height is chosen so that the total area under the density equals 1, which is the defining requirement of any density: `60 * (1/60) = 1`. ## Why the answer is a length ratio For a continuous variable, probability is area under the density, not the height of the density. Because the height is constant here, area over a sub-interval is just height times width: ``` P(c <= X <= d) = (d - c) / (b - a) ``` The final 10 minutes are the interval `[50, 60]`, so `P = (60 - 50) / (60 - 0) = 10/60 = 1/6 ≈ 0.167`. The same arithmetic gives 1/6 for `[0, 10]` and for `[25, 35]`. Under a uniform law, position is irrelevant and only width matters — that is the whole content of the model. ## The CDF The cumulative distribution function accumulates that area from the left: ``` F(x) = 0 for x < a F(x) = (x - a)/(b - a) for a <= x <= b F(x) = 1 for x > b ``` So `F(x) = x/60` on the window: a straight line from 0 to 1. Interval probabilities are differences of the CDF, `P(50 < X <= 60) = F(60) - F(50) = 1 - 0.8333 = 0.1667`. A linear CDF is the visual signature of a uniform distribution, just as a flat density is. ## Mean, variance and their derivations By symmetry the mean sits at the midpoint: ``` E[X] = (a + b) / 2 = 30 minutes ``` For the variance, the second moment of a uniform is `E[X^2] = (a^2 + ab + b^2) / 3`, and subtracting the squared mean collapses to a clean form: ``` Var(X) = (b - a)^2 / 12 = 3600 / 12 = 300 SD(X) = sqrt(300) ≈ 17.3 minutes ``` The `/12` is worth memorising because it appears constantly: the spread of a uniform grows linearly with the width of its support, and its standard deviation is about 28.9% of that width. ## Two traps **A density is not a probability.** Densities have units of "probability per unit of x", so they can exceed 1. A uniform on `[0, 0.5]` has density 2 everywhere, which is perfectly legal — the area is still `0.5 * 2 = 1`. Only integrals of the density are probabilities, and those never exceed 1. **Single points have probability zero.** `P(X = 30) = 0` for any continuous variable, because a point has zero width and therefore zero area. This is why `P(X < 30)` and `P(X <= 30)` are equal for continuous distributions, unlike the discrete case where the endpoint carries real mass. It does not mean the outcome is impossible — it means probability lives in intervals. ## Where it shows up The uniform is the base case for reasoning about "arrives at random", "rounding error is somewhere in the last digit", or "a scheduled job starts somewhere inside its window". It is also the raw material for generating other distributions: a uniform draw on `[0, 1]` fed through the inverse of any CDF produces a draw from that distribution, which is why a uniform generator is the primitive underneath simulation. In an interview, the expected move is to recognise the flat density immediately, answer with a length ratio rather than an integral, and state mean and variance from memory.