## Akshay_Budhkar 4 years ago 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

1. Akshay_Budhkar

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

2. anonymous

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

3. Akshay_Budhkar

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

4. anonymous

For a quadrilateral, each length has to be less than the sum of the other 3 lengths, i.e. $$a<b+c+d$$. No piece of the rod may therefore be greater than half the full length of the rod.

5. Akshay_Budhkar

ya i agree

6. anonymous

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

7. Akshay_Budhkar

yes agreed

8. 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.

9. Akshay_Budhkar

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

10. 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...

11. 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.

12. Akshay_Budhkar

ok

13. 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.

14. anonymous

The answer isn't 0.5 is it?

15. Akshay_Budhkar

BINGO

16. Akshay_Budhkar

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

17. 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.

18. Akshay_Budhkar

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

19. Akshay_Budhkar

Dalvoron\?

20. anonymous

Working on it...

21. Akshay_Budhkar

ok

22. anonymous

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

23. anonymous

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

24. Akshay_Budhkar

plz help satellite!!

25. anonymous

Maybe an addition series that approximates to 0.5?

26. anonymous

Or tends towards it, or whatever.

27. anonymous

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

28. anonymous

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

29. anonymous

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

30. anonymous

did you first do the triangle problem?

31. 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

32. 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

33. anonymous

34. 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

35. anonymous

Hah, nice.

36. 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

37. anonymous

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

38. anonymous

oh and if you keep reading it gets better

39. anonymous

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

40. Akshay_Budhkar

yes so what is the non probability proof?

41. Akshay_Budhkar

satelite? you there?