Akshay_Budhkar
  • Akshay_Budhkar
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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Akshay_Budhkar
  • Akshay_Budhkar
truthfolly i just have the answer.. i dont have the solution too
anonymous
  • anonymous
what do you mean by quad please explain a little : )
Akshay_Budhkar
  • Akshay_Budhkar
lol! ishaan a quadrilateral of course!!!! not a square

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anonymous
  • anonymous
For a quadrilateral, each length has to be less than the sum of the other 3 lengths, i.e. \(a
Akshay_Budhkar
  • Akshay_Budhkar
ya i agree
anonymous
  • anonymous
So we rule out the cases where 3 cuts appear in one half of the rod.
Akshay_Budhkar
  • Akshay_Budhkar
yes agreed
anonymous
  • anonymous
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.
Akshay_Budhkar
  • Akshay_Budhkar
wait how is it 0.5*0.5*0.5???????????
anonymous
  • anonymous
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...
anonymous
  • anonymous
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.
Akshay_Budhkar
  • Akshay_Budhkar
ok
anonymous
  • anonymous
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.
anonymous
  • anonymous
The answer isn't 0.5 is it?
Akshay_Budhkar
  • Akshay_Budhkar
BINGO
Akshay_Budhkar
  • Akshay_Budhkar
how did you do it???????????????????????????????????????????????????????????????????????????????????
anonymous
  • anonymous
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.
Akshay_Budhkar
  • Akshay_Budhkar
nice guess sir.. i still want the solution though...
Akshay_Budhkar
  • Akshay_Budhkar
Dalvoron\?
anonymous
  • anonymous
Working on it...
Akshay_Budhkar
  • Akshay_Budhkar
ok
anonymous
  • anonymous
Nah, I'm stumped. Integration was never my strong point.
anonymous
  • anonymous
actually you do not need integration for this problem, but as i recall it is a bear
Akshay_Budhkar
  • Akshay_Budhkar
plz help satellite!!
anonymous
  • anonymous
Maybe an addition series that approximates to 0.5?
anonymous
  • anonymous
Or tends towards it, or whatever.
anonymous
  • anonymous
i have seen this for a triangle, but there it is just in two dimensions. here you are in three.
anonymous
  • anonymous
so it is going to be a volume problem rather than an area one. hold on
anonymous
  • anonymous
I assumed it was in one dimension. Things just got harder.
anonymous
  • anonymous
did you first do the triangle problem?
anonymous
  • anonymous
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
anonymous
  • anonymous
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
anonymous
  • anonymous
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anonymous
  • anonymous
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
anonymous
  • anonymous
Hah, nice.
anonymous
  • anonymous
answer is worked out there, but i would recommend starting at problem 2 first so the idea is more clear. your answer is #4
anonymous
  • anonymous
Well got to go to work now anyway. I'll come back to this later.
anonymous
  • anonymous
oh and if you keep reading it gets better
anonymous
  • anonymous
much much better. so perhaps we go to the end an mimic the "simple proof' for the case n = 4
Akshay_Budhkar
  • Akshay_Budhkar
yes so what is the non probability proof?
Akshay_Budhkar
  • Akshay_Budhkar
satelite? you there?

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