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There is a line the length of which $L$. We throw $2$ random points on it, hence $3$ segments are formed. I've denoted the length of the first segment $x$, the other one $y$, hence the 3rd one $L-x-y$.

I've found that the probability that these $3$ segments would form a triangle is $\frac 14$. Now I need to find what is the probability that the length of the smallest side will be not bigger(<=) than $\frac L3$ if these $3$ segments form a triangle.

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The answer is $1$ but I can't figure out why.

Thanks.

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In any division of the stick into three pieces, if no piece is shorter than $L/3$ then all sticks must be $L/3$ in length, else they would form a stick of length greater than $L$, and this event has measure zero (i.e. it almost never happens). Thus there is probability 1 that the shortest stick is shorter than $L/3$, irrespective of whether the pieces form a triangle or not.

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  • $\begingroup$ If they are all equal to $\frac L3$ then the shortest side is not less than that? $\endgroup$
    – MRobinson
    Oct 26 '18 at 7:42
  • $\begingroup$ Relating to your edit: "almost never happens" is not the same as impossible $\endgroup$
    – MRobinson
    Oct 26 '18 at 7:43
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    $\begingroup$ @MRobinson It may happen, but it is a finite event among an infinity of possibilities. Thus its probability is zero. $\endgroup$ Oct 26 '18 at 7:44
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    $\begingroup$ Yes, sorry you're right. Still too early for me apparently $\endgroup$
    – MRobinson
    Oct 26 '18 at 7:44
  • $\begingroup$ Oh sorry the problem was saying not bigger than , I've editted, which means equal is also considered $\endgroup$
    – user604383
    Oct 26 '18 at 7:55

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