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All the stars heros sensi champions and other great ppl.. try this a tough one a rod is divided into 4 parts.. what is the the probability that the parts constitute a quadrilateral? especially for people like dalvoron,satellite,ishaan,joe,myininimya,and the other great ppl.. lol

Mathematics
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truthfolly i just have the answer.. i dont have the solution too
what do you mean by quad please explain a little : )
lol! ishaan a quadrilateral of course!!!! not a square

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Other answers:

For a quadrilateral, each length has to be less than the sum of the other 3 lengths, i.e. \(a
ya i agree
So we rule out the cases where 3 cuts appear in one half of the rod.
yes agreed
The probability that 3 cuts appear in the left half of the rod is \(0.5\times0.5\times0.5=0.125\), likewise for the right half of the rod.
wait how is it 0.5*0.5*0.5???????????
So at most, the odds of the 4 parts constituting a quad is 0.75. A middle piece may be greater than half the length of the rod also though...
Cut 1 has a 0.5 chance of being in the left part of the rod. Cut 2 has a 0.5 chance, Cut 3 has a 0.5 chance. The chance that all 3 are in the same half therefore is the product of those 3 probabilities.
ok
I think I'm going about it the wrong way though, because a piece of 1/2 rod may exist anywhere along the length. Need to find the probability that the distance between any adjacent cuts is greater than 1/2 the rod length.
The answer isn't 0.5 is it?
BINGO
how did you do it???????????????????????????????????????????????????????????????????????????????????
Hah, boom. Not exactly sure why that's the answer. I think you'd need to do an integral to prove it. I just figure there's a 1/2 chance that at least half the rod will be intact.
nice guess sir.. i still want the solution though...
Dalvoron\?
Working on it...
ok
Nah, I'm stumped. Integration was never my strong point.
actually you do not need integration for this problem, but as i recall it is a bear
plz help satellite!!
Maybe an addition series that approximates to 0.5?
Or tends towards it, or whatever.
i have seen this for a triangle, but there it is just in two dimensions. here you are in three.
so it is going to be a volume problem rather than an area one. hold on
I assumed it was in one dimension. Things just got harder.
did you first do the triangle problem?
Never did a triangle problem. I did a similar stick based problem by Richard Wiseman before, but it didn't help much http://richardwiseman.wordpress.com/2011/01/07/its-the-friday-puzzle-93/#comments
answer their is easier to come by. i think we can cheat if we want, because this is a pretty standard problem. i am not sure i am up to it but the idea will have some sort of region in it
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god but do i love the internet. if they had had this when i was in school i would have had 5 degrees by now
Hah, nice.
answer is worked out there, but i would recommend starting at problem 2 first so the idea is more clear. your answer is #4
Well got to go to work now anyway. I'll come back to this later.
oh and if you keep reading it gets better
much much better. so perhaps we go to the end an mimic the "simple proof' for the case n = 4
yes so what is the non probability proof?
satelite? you there?

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