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hartnn
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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 ?
 one year ago
 one year ago
hartnn Group Title
If a line segment is cut into 3 parts, what is the probability that those 3 parts can form a triangle ?
 one year ago
 one year ago

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mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
For example, if I have a length of rope 10. I can cut it into 118 (no triangle) Or 235 (no triangle) Or 442 (triangle) Or 334 (triangle).
 one year ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
just apply triangle inequality.
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
your point is , sum of two sides of triangle must be > 3rd ?
 one year ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
yaaa
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
used that, solved it, got 50% want to verify whether its correct....
 one year ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
verify 3,3,3(forms triangle) sum of two sides of triangle must be > 3rd(6>3)
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
i would like to restate : sum of two sides of triangle must be > or = 3rd
 one year ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
right
 one year ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
6>3(3+3>3) PROVE IT!
 one year ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
thus, it proves it forms a triangle!!! @hartnn can you understand??
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
yes, that forms equilateral triangle. but i want to find the probability.
 one year ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
i think probability is outcome/parts=10/3
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
probability can never be >1 _
 one year ago

henpen Group TitleBest ResponseYou've already chosen the best response.2
You want that 1 of the segments is not >0.5 of the stick.
 one year ago

henpen Group TitleBest ResponseYou've already chosen the best response.2
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?
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
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...
 one year ago

henpen Group TitleBest ResponseYou've already chosen the best response.2
Well, using my method it's less than 50% that form a triangle. Call where the first dart lands Xdw: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...
 one year ago

henpen Group TitleBest ResponseYou've already chosen the best response.2
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 nontriangle, because of this, is LESS than 50%. But it absolutely matters which procedure you use to cut.
 one year ago

henpen Group TitleBest ResponseYou've already chosen the best response.2
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.
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
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)....
 one year ago

henpen Group TitleBest ResponseYou've already chosen the best response.2
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)?
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
ofcourse, i got that when i solved, and that gave me 50%.
 one year ago

henpen Group TitleBest ResponseYou've already chosen the best response.2
By what logic?
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
ok, i'll show what i had done in test.
 one year ago

henpen Group TitleBest ResponseYou've already chosen the best response.2
I'll type up my try to work out a numerical answer at the same time
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
dw:1354374054772:dw
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
@UnkleRhaukus can u have a look and share any ideas ?? same for Callisto.
 one year ago

henpen Group TitleBest ResponseYou've already chosen the best response.2
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 partstick of >0.5 length?' There are 2 ways of making a partstick of >0.5 length: dart 2 lands in half A also this happensdw:1354374314148:dw
 one year ago

henpen Group TitleBest ResponseYou've already chosen the best response.2
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.
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
isn't this getting complicated ? for me, yes.... what wrong in my answer ?
 one year ago

henpen Group TitleBest ResponseYou've already chosen the best response.2
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?
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
yeah, a is fixed. i will make a triangle with sides x,y,z.
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
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 year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
so, what @henpen said here 'Prob(dart 2 lands in B)=0.5' its straightaway 50% ?
 one year ago

henpen Group TitleBest ResponseYou've already chosen the best response.2
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
 one year ago

henpen Group TitleBest ResponseYou've already chosen the best response.2
dw:1354375288008:dw We want \[p+q<0.5\]
 one year ago

henpen Group TitleBest ResponseYou've already chosen the best response.2
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
 one year ago

henpen Group TitleBest ResponseYou've already chosen the best response.2
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%)).
 one year ago

henpen Group TitleBest ResponseYou've already chosen the best response.2
fairly sure
 one year ago
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