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

- Akshay_Budhkar

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- schrodinger

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- Akshay_Budhkar

truthfolly i just have the answer.. i dont have the solution too

- anonymous

what do you mean by quad please explain a little : )

- Akshay_Budhkar

lol! ishaan a quadrilateral of course!!!! not a square

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## More answers

- anonymous

For a quadrilateral, each length has to be less than the sum of the other 3 lengths, i.e. \(a

- Akshay_Budhkar

ya i agree

- anonymous

So we rule out the cases where 3 cuts appear in one half of the rod.

- Akshay_Budhkar

yes agreed

- 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

wait
how is it 0.5*0.5*0.5???????????

- 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

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

ok

- 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

The answer isn't 0.5 is it?

- Akshay_Budhkar

BINGO

- Akshay_Budhkar

how did you do it???????????????????????????????????????????????????????????????????????????????????

- 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

nice guess sir.. i still want the solution though...

- Akshay_Budhkar

Dalvoron\?

- anonymous

Working on it...

- Akshay_Budhkar

ok

- anonymous

Nah, I'm stumped. Integration was never my strong point.

- anonymous

actually you do not need integration for this problem, but as i recall it is a bear

- Akshay_Budhkar

plz help satellite!!

- anonymous

Maybe an addition series that approximates to 0.5?

- anonymous

Or tends towards it, or whatever.

- anonymous

i have seen this for a triangle, but there it is just in two dimensions. here you are in three.

- anonymous

so it is going to be a volume problem rather than an area one. hold on

- anonymous

I assumed it was in one dimension. Things just got harder.

- anonymous

did you first do the triangle problem?

- 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

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

##### 1 Attachment

- 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

Hah, nice.

- 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

Well got to go to work now anyway. I'll come back to this later.

- anonymous

oh and if you keep reading it gets better

- anonymous

much much better. so perhaps we go to the end an mimic the "simple proof' for the case n = 4

- Akshay_Budhkar

yes so what is the non probability proof?

- Akshay_Budhkar

satelite? you there?

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