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Two points are randomly selected on a line of length $1$. What is the probability that one of the segments is greater than $\frac{1}{2}$? Points can be placed anywhere between [0, 1], for example. Thanks!

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  • $\begingroup$ Notice that a line segment will only be greater than 1/2 if both chosen points are either both to the left or to the right of the middle point. So knowing this I guess you could work out the details $\endgroup$ – Tom Ultramelonman May 30 '16 at 21:07
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    $\begingroup$ @TomUltramelonman What if the first point is at $.1$ and the second at $.9$? Then the segment between them has length $.8$. $\endgroup$ – lulu May 30 '16 at 21:09
  • $\begingroup$ Oh damn, how could I overlook this ^^ Yea sorry $\endgroup$ – Tom Ultramelonman May 30 '16 at 21:10
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    $\begingroup$ @TomUltramelonman That's not correct. Two points divide a segment into three parts and any of them may be longer than $\frac 12$... $\endgroup$ – CiaPan May 30 '16 at 21:10
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Ok suppose you take some point $x\in[0,1/2]$. Now taking a second point $y\in[0,1]$, there are two situations where you obtain a segment of length at least $1/2$. Firstly if $y\le 1/2$. Secondly if $y\ge x+1/2$. So for taking a first point $x$, the chance that you have a segment of desired length is $1/2+(1-1/2-x)=1 - x$. Now integrating this over $[0,1/2]$ you get $3/8$. Yu can do the same for $x\in[1/2,1]$. So this gives you a chance of $6/8$.

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  • $\begingroup$ Thanks to lulu and CiaPan for the comments. $\endgroup$ – Tom Ultramelonman May 30 '16 at 21:28
  • $\begingroup$ This is solid (+1). My variant, below, does the same thing geometrically...without referring to an integral. But the underlying principle is the same. $\endgroup$ – lulu May 30 '16 at 21:31
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Probability is $3/4$ if points $x$ and $y$ are chosen with uniform probability. That corresponds to the area in color in the picture below.

enter image description here

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With probability $\frac 12$ both points are on the same side of the midpoint, so we are guaranteed success.

If the points are on opposite sides of the midpoint(a probability $\frac 12$ event, with $P<\frac 12< Q$ say, then again with probability $\frac 12$ we have $Q$ is nearer $1$ than $P$ is near $\frac 12$,so the segment between them has length greater than $\frac 12$.

Thus the total probability Is $$\frac 12+\frac 12\times \frac 12=\frac 34$$

Note: this is equivalent to asking how probable it is that the three segments formed by the two points can form a triangle (the above shows that the answer is $\frac 14$). Many proofs for that can be found e.g. here

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