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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?

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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 $1/4$.

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  • $\begingroup$ 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 Nov 18 '19 at 0:53

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