+20 A line segment of length 5 is broken at two random points along its length. What is the probability that the shortest of the three new segments has length longer than 1?
+9465 This is a good question...now I kinda want to know the answer too... 🤔
( Sorry I'm no help..! )
I am in 8th grade, so this is most likely an incorrect conjecture. However, here is what I have started with:
Assuming that the above must be true, then the value of the first point, x, is {1 < x < 5}. Now, we know that the second point cannot be within 1 unit of the edges of the line, or within 1 unit of the point x. So, the second division, y, must agree with the following statements: {1 < y < 5}, and: {{y < (x - 1)} or {y > (x + 1)}}.
Go ahead and calculate the probability for yourselves, I don't know how to calculate it.
YOU'RE WELCOME!!!
+633 Correction on myself: point x parameters: {1 < x < 4}
point y parameters: {1 < y < 4} and {{x+1 < y} or {x-1 > y}}
Probability graph: Click here
Solution: Right under 20%.
Was I right?
(For the graph, the y-value is the chance that the shortest of the line segments is length one or greater.)
+118608 Hi Guest,
I think there is a concensous of oppinion that the answer is 16%
I do not understand what you have graphed ://
+9465 Can you come up with a specific counter-example that disqualifies the other answers?
Is there a specific x and a y that fall outside the range given in the other answers, but that still makes the shortest of the 3 segments longer than 1? Or something like that... 
+118608 Hi Hectictar,
I have actually done that.
It is a probablility question so showing individual combinations that fit or do not fit the condition is not going to help.
Have you tried to work through mine and/or Alan's answer?
Since I stuffed up a little on my wording, would you like me to do it slowly with loads of diagrams to help you understand? (Send me a message if you do)
If you are not super interested I will not bother
but
if you will put a lot of effort into understanding what I do I will be more than happy to present my answer with much more detail. 
+9465 Oh, okay..I see..
And I haven't thoroughly looked at it yet.. 
It's been sitting on my watchlist because I keep meaning to but I just haven't quite had the motivation to yet...I will try to understand it and if I have a question about it I will ask...Thank you!
+118608 Hi Hectictar,
I have a some questions like that too :)
Sometimes I feel bad because I have asked a question and then I do not put as much effort into understanding as I need to. But there is only so much time and there is always a new question to distract my attention.
So I do understand.
Still if you want me to explain more slowly I can do that. :)
This techniqe was new to me but it is not that hard and it is a nice probablility technique to know :)
+633 Sorry about this guys, I just realized today that I made a huge mistake on the graph! I've fixed it, and now the shaded areas show what the possible values for x and y are. I also recalculated some stuff, and the new calculations show that you were right. You can get to the new graph by clicking here.
(Yes, I did the work.)
+118608 I have just leaned a new technique!! Someone very kind taught me :))
A line segment of length 5 is broken at two random points along its length. What is the probability that the shortest of the three new segments has length longer than 1?
Let the lengths be x, 5-y, and y-x
Obviously x and y are between 0 and 5
and y>x
If you add these together you will see that they add to 5.
So the entire region where x and y can be is a triangle, (the big green one in the contour map)
Now
If x is less then 1, the condition is met. x<1
If x is more than 4 the condition is met. x>4
If 5-y>4 or if 5-y<1 the condition is met.
i.e. if y<1 or if y>4 the condition is met
If 0< y-x<1 or if 4
i.e. x< y<1+x or if 4+x
i.e. y>x and y<1+x or if y>4+x and y<5+x the condition is met
Ok lets try mapping this on a boolean contour map.
I've done all that and the only place that the conditions are not met is the little red triangle in middle.
So all possible outcomes are within the green area which is 0.5*5*5=12.5units squared
The area of red triangle represents all triads where the smallest is bigger than 1
Area = 0.5*2*2 = 2 units squared
So the probability that the shortest side is more than 1 is 2/12.5 = 4/25 = 16%
+633 Um... you said that y<1 or y>4
Then, the shortest of the line segments would always have a length less than 1, because y would divide it either in between 0 and 1, meaning the first line segment would not reach that far, or in between 4 and 5, in which case the last line segment would not reach past 4.
So, Alan or Melody, please explain why y<1 or y>4.
+118608 Yes you are right, I have written it back to front but my logic and my answer is still correct.
I shall attempt to fix my wording :)
+633 Did you check the graph that is a link in response #9? The big figure in the middle shows where each point can be.
+33614 Nicely done Melody! Here's a similar, but step by step version. Again, X and Y values are plotted as coordinates within a square box:
.
+633 Alan, you said that one of the points has length 5-x.
If that were true, x could be 3, and y could be 4.5.
The 3 lengths would be 3, 1.5, and 2.
3+1.5+2=6.5, not 5.
+633 Everyone, reread the question. Some of you may have misinterpreted it (Alan, Melody):
A line segment of length 5 is broken at two random points along its length. What is the probability that the shortest of the three new segments has length longer than 1?
Not shorter than 1, but longer.
I think you interpreted it as being shorter than.
Please reread, and see if answer #5 makes more sense.
+118608 Sorry guest, I have graphed the region wherer the shortest piece is less then 1. In that you are correct BUT
All the regions where one of the lengths is less than 1 are been discarded.
The red triangle in the middle of mine is all that is left where none of the peices are less than 1. :)
So our answer is correct but my working is not worded correctly.
I am sorry, I can see it could be very confusing. :/
