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