Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let there be two identical and independent random variables $X$ and $Y$. Both $X$ and $Y$ are uniformly distributed from $0$ to $50$. What is the probability that $|X - Y| < 10$? This is not homework. I am practicing probability.

I think if $|X - Y| < 10$, then either $X$ is greater than $Y$ or $Y$ is greater than $X$, the expected value of $A = |X - Y|$ should be $\int_0^{50} \int_0^y (y - x) \, dx \, dy + \int_0^{50} \int_0^x (x - y) \, dy \, dx = \frac{125000}{3}$, but how to do I use this value to find $P(A < 10)$?

share|cite|improve this question
Think of $(X,Y)$ as being the cooordinates of a point that is uniformly chosen from the square $[0,50] \times [0,50]$. What does it mean that $|X - Y| < 10$? – Hans Engler Nov 4 '12 at 1:17
Thanks! In that case, $Y < 10 + X$ and $Y < X - 10$... ah, so I just set up a relevant integral within the bounds of [0, 50] X [0, 50]? – David Faux Nov 4 '12 at 1:30
up vote 1 down vote accepted

We want to be between the lines $x-y=-10$ and $x-y=10$. The part of the square which is not between these lines consists of two isosceles right triangles with legs of length $40$. Their combined area is therefore $2(40)^2/2=1600$.

Thus the probability that $|X-Y|\ge 10$ is $\dfrac{1600}{2500}$. Subtract this from $1$ to get the asked for probability.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.