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

- hartnn

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- hartnn

50% ??

- mayankdevnani

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).

- mayankdevnani

just apply triangle inequality.

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## More answers

- hartnn

your point is ,
sum of two sides of triangle must be > 3rd ?

- mayankdevnani

yaaa

- hartnn

used that, solved it, got 50%
want to verify whether its correct....

- mayankdevnani

verify-
3,3,3(forms triangle)
sum of two sides of triangle must be > 3rd(6>3)

- hartnn

i would like to restate :
sum of two sides of triangle must be > or = 3rd

- mayankdevnani

right

- mayankdevnani

6>3(3+3>3) PROVE IT!

- mayankdevnani

thus, it proves it forms a triangle!!!
@hartnn can you understand??

- hartnn

yes, that forms equilateral triangle. but i want to find the probability.

- mayankdevnani

i think probability is outcome/parts=10/3

- hartnn

probability can never be >1 -_-

- anonymous

You want that 1 of the segments is not >0.5 of the stick.

- anonymous

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?

- hartnn

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...

- anonymous

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...

- anonymous

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.

- anonymous

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.

- hartnn

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)....

- anonymous

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)?

- hartnn

ofcourse, i got that when i solved, and that gave me 50%.

- anonymous

By what logic?

- hartnn

ok, i'll show what i had done in test.

- anonymous

I'll type up my try to work out a numerical answer at the same time

- hartnn

|dw:1354374054772:dw|

- hartnn

@UnkleRhaukus can u have a look and share any ideas ??
same for Callisto.

- anonymous

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|

- anonymous

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.

- hartnn

isn't this getting complicated ?
for me, yes....
what wrong in my answer ?

- anonymous

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?

- hartnn

yeah, a is fixed.
i will make a triangle with sides x,y,z.

- UnkleRhaukus

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%

- UnkleRhaukus

one*

- hartnn

so, what @henpen said here
'Prob(dart 2 lands in B)=0.5'
its straightaway 50% ?

- anonymous

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|

- anonymous

|dw:1354375288008:dw| We want \[p+q<0.5\]

- anonymous

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|

- anonymous

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%)).

- UnkleRhaukus

hmm,

- hartnn

huh ?

- hartnn

25% sure ?

- anonymous

fairly sure

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