Two numbers are randomly selected between $(0,1)$, uniformly and independently distributed. What is the probability that the three resulting line segments, which are obtained by cutting the interval at the two selected numbers, form a triangle?

So, I was thinking I can do something like this:

$X \sim U(0,1), Y \sim U(0,1),$ which are independent. In order to form a triangle, each side cannot be bigger than the sum of the other two sides. So, each segment cannot be greater than $1/2$.

If $X > Y$, the first segment is of length $Y$, the second is of length $X-Y$, and the third is of length $1-X$.

Note: $P(X>Y)=P(X<Y)$.

So, can we obtain the answer by computing $2P(Y,X-Y,1-x < 1/2)$? If so, how would you compute this probability?


1 Answer 1


Consider the case where $x>y$, and sketch the region between the lines $y<1/2$, $x>1/2$, $x-y<1/2$. All possibilities lie in the square $[0,1]\times[0,1]$, with an area of $1$. So, the area of region between the lines and its symmetric around $y=x$ for the case $x<y$ will be the probability you seek for, which is $1/4$.

  • $\begingroup$ Wow, this was incredibly helpful. I was anticipating a crazy use of transformations to solve this problem, but it's just simple geometry and integration. Thank you! $\endgroup$
    – Ron Snow
    Commented Nov 18, 2019 at 0:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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