Math GRE - #21

Let $x$ and $y$ be uniformly distributed, independent random variables on [0, 1]. The probability that the distance between $x$ and $y$ is less than $\frac{1}{2}$ is:

• $\dfrac{1}{4}$


• $\dfrac{1}{3}$


• $\dfrac{1}{2}$


• $\dfrac{2}{3}$


• $\dfrac{3}{4}$

Solution :

The answer is $\frac{3}{4}$.

If $x$ and $y$ are uniformly distributed and independent random variables on [0, 1], this means that they are in the unit square.

We require $|x-y| < \frac{1}{2}$, i.e. we need \[-\frac{1}{2} < x - y < \frac{1}{2}\].
On the unit square, we can draw two regions which represent the inequality above: \[y < x+\frac{1}{2}\] \[y > x - \frac{1}{2}.\]
If we can find the area of this region (shown below), then we are done the question.

The region we are looking at.

I'll leave it as an exercise to show that the filled out region above is $\frac{3}{4}$ of the unit square as we wanted.