Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

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

  • This Question is Closed
  1. Akshay_Budhkar
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

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

  2. Ishaan94
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  3. Akshay_Budhkar
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

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

  4. Dalvoron
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    ya i agree

  6. Dalvoron
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  7. Akshay_Budhkar
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    yes agreed

  8. Dalvoron
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

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

  10. Dalvoron
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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. Dalvoron
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    ok

  13. Dalvoron
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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. Dalvoron
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    The answer isn't 0.5 is it?

  15. Akshay_Budhkar
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    BINGO

  16. Akshay_Budhkar
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

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

  17. Dalvoron
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

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

  19. Akshay_Budhkar
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    Dalvoron\?

  20. Dalvoron
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Working on it...

  21. Akshay_Budhkar
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    ok

  22. Dalvoron
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  23. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  24. Akshay_Budhkar
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    plz help satellite!!

  25. Dalvoron
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Maybe an addition series that approximates to 0.5?

  26. Dalvoron
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Or tends towards it, or whatever.

  27. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  28. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  29. Dalvoron
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  30. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    did you first do the triangle problem?

  31. Dalvoron
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    1 Attachment
  34. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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. Dalvoron
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Hah, nice.

  36. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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. Dalvoron
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  38. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh and if you keep reading it gets better

  39. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  40. Akshay_Budhkar
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    yes so what is the non probability proof?

  41. Akshay_Budhkar
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    satelite? you there?

  42. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy