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If a line segment is cut into 3 parts, what is the probability that those 3 parts can form a triangle ?

Probability
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50% ??
For example, if I have a length of rope 10. I can cut it into 1-1-8 (no triangle) Or 2-3-5 (no triangle) Or 4-4-2 (triangle) Or 3-3-4 (triangle).
just apply triangle inequality.

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Other answers:

your point is , sum of two sides of triangle must be > 3rd ?
yaaa
used that, solved it, got 50% want to verify whether its correct....
verify- 3,3,3(forms triangle) sum of two sides of triangle must be > 3rd(6>3)
i would like to restate : sum of two sides of triangle must be > or = 3rd
right
6>3(3+3>3) PROVE IT!
thus, it proves it forms a triangle!!! @hartnn can you understand??
yes, that forms equilateral triangle. but i want to find the probability.
i think probability is outcome/parts=10/3
probability can never be >1 -_-
You want that 1 of the segments is not >0.5 of the stick.
I think the probability here may be dependent on the decision procedure that you choose the 3 parts in. Can I assume you throw 2 darts simultaneously at the stick, and these points are the points that you cut it?
even i can't say what can be assumed, this was the question...and i found it mathematically that this probability is 50% , just want to see new approaches and verify my answer. so , go ahead and try it...
Well, using my method it's less than 50% that form a triangle. Call where the first dart lands X|dw:1354373456702:dw|, in the half of the stick called A Disregard the 'A' half of the stick. Now, the chance that the second dart lands in A, leaving there a segment of >0.5 is 50%. However...
It is possible that the second dart lands in B AND the second dart lands in A AND STILL there would be 1 segment >0.5 length of the stick.|dw:1354373603207:dw| So the chance of a non-triangle, because of this, is LESS than 50%. But it absolutely matters which procedure you use to cut.
For example, you could choose the longer of the 2 parts that the first dart cuts in half, and throw the second dart against this part of the stick.
i didn't get u entirely, but got some idea on what u saying.... moreover, i think there's a numerical answer to this, rather than inequality.(like >0.5)....
Yes, there is (not sure how to yet). Do you understand my reasoning that 1 side of the triangle cannot be >0.5 of the original stick (that is, cannot be > than the sum of the other 2 sides)?
ofcourse, i got that when i solved, and that gave me 50%.
By what logic?
ok, i'll show what i had done in test.
I'll type up my try to work out a numerical answer at the same time
|dw:1354374054772:dw|
@UnkleRhaukus can u have a look and share any ideas ?? same for Callisto.
Assume dart 1 has an equal probability of landing anywhere on the stick. By definition, it lands on the half of the stick called A. Equal probability of landing anywhere on there. Now, your question is equivalent to saying: 'what is the probability that the 2nd dart will not leave a part-stick of >0.5 length?' There are 2 ways of making a part-stick of >0.5 length: -dart 2 lands in half A also -this happens|dw:1354374314148:dw|
The second way works as there is a hole of length >=0.5 in the middle. Working it out: Prob(dart 2 lands in B)=0.5 Prob(x <= y)=(y/0.5) where y is a length measured from the right, and 0.5 is a length.
isn't this getting complicated ? for me, yes.... what wrong in my answer ?
Have I misunderstood the question? It looks like for you a is the random variable, and x, y and z are fixed. How are you making your triangle?
yeah, a is fixed. i will make a triangle with sides x,y,z.
so on of the triangles lengths must be greater than the sum of the other two , you need both darts to cut the string on different sides of the half way point , assuming the first dart land one side, the probability that second dart land on the opposite half of the string is 50%
one*
so, what @henpen said here 'Prob(dart 2 lands in B)=0.5' its straightaway 50% ?
My problem is that there are ways of the too large side being formed even if they land in the opposite sides, for example:|dw:1354375141124:dw|
|dw:1354375288008:dw| We want \[p+q<0.5\]
By definition, \[p<0.5\] and \[q<0.5\] So it seems in the second case it is 50% (not 50% overall)|dw:1354375464988:dw|
So overall I'd say 75% (as in the first case it's 100% not going to form a triangle (100% of 50%=50% overall) and in the second case it's 50% not going to form a triangle (50% of 50%=25%)).
hmm,
huh ?
25% sure ?
fairly sure

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