- $\dfrac{1}{4}$
- $\dfrac{1}{3}$
- $\dfrac{1}{2}$
- $\dfrac{2}{3}$
- $\dfrac{3}{4}$

Solution :

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.

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.

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.

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