Choose two numbers at random from $[0, 1]$. What is the chance they are within $\frac{1}{4}$ of each other? Such questions have a beautiful answer: the pair is a random point in a square, and the probability is an area. This lesson extends densities to two dimensions in the simplest case, a uniform point.
Key idea
If a point is uniform in a region $R$, then $$P(\text{point lands in } E) = \frac{\text{area}(E \cap R)}{\text{area}(R)}.$$ If $X$ and $Y$ are independent and uniform on $[0, 1]$, the pair $(X, Y)$ is a uniform point in the unit square, so the probability of an event is simply its area.
Why it works
For independent uniforms, $$P(a \le X \le b,\ c \le Y \le d) = (b - a)(d - c)$$ the probability of every rectangle is its area. Areas of other shapes are built from rectangles (that is what an integral does), so every event's probability is its area. This is also the joint density $f(x, y) = 1$ on the square, a preview of joint densities.
Worked examples
Worked example: Within a quarter
$X$ and $Y$ are independent and uniform on $[0, 1]$. Find $P(\lvert X - Y \rvert < \frac{1}{4})$.
- The event is the band around the diagonal. Its complement is two corner triangles, each with legs $1 - \frac{1}{4} = \frac{3}{4}$.
- Their total area: $2 \cdot \frac{1}{2}\left(\frac{3}{4}\right)^{2} = \frac{9}{16}$.
- $P = 1 - \frac{9}{16} = \frac{7}{16}$.
Answer: $\frac{7}{16}$
Worked example: Meeting for coffee
Two friends arrive at random, independently, between noon and 1 p.m. Each waits 10 minutes and then leaves. Find the probability that they meet.
- In hours, they meet when $\lvert X - Y \rvert \le \frac{1}{6}$.
- The two missing corners have legs $\frac{5}{6}$: area $\left(\frac{5}{6}\right)^{2} = \frac{25}{36}$.
- $P(\text{meet}) = 1 - \frac{25}{36} = \frac{11}{36} \approx 0.31$.
Answer: $\frac{11}{36}$
Worked example: Under a curve
Find $P(X + Y < \frac{1}{2})$ and $P(XY < \frac{1}{2})$.
- $x + y < \frac{1}{2}$ is a corner triangle with legs $\frac{1}{2}$: area $\frac{1}{8}$.
- $xy < \frac{1}{2}$: for $x \le \frac{1}{2}$ every $y$ works; for $x > \frac{1}{2}$, $y < \frac{1}{2x}$. Area $$= \frac{1}{2} + \int_{1/2}^{1} \frac{dx}{2x} = \frac{1}{2} + \frac{\ln 2}{2} \approx 0.847$$
Answer: $\frac{1}{8}$ and $\frac{1}{2} + \frac{\ln 2}{2}$
Common mistake
Working with lengths on one axis, for example saying "$\lvert X - Y \rvert < \frac{1}{4}$ is like an interval of length $\frac{1}{2}$, so the chance is $\frac{1}{2}$". The event depends on both numbers, so it is a region in the square. Check: draw the square of all pairs, shade the event, and find its area.