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hartnn

If a line segment is cut into 3 parts, what is the probability that those 3 parts can form a triangle ?

  • one year ago
  • one year ago

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  1. hartnn
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    50% ??

    • one year ago
  2. mayankdevnani
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    For example, if I have a length of rope 10. I can cut it into 1-1-8 (no triangle) Or 2-3-5 (no triangle) Or 4-4-2 (triangle) Or 3-3-4 (triangle).

    • one year ago
  3. mayankdevnani
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    just apply triangle inequality.

    • one year ago
  4. hartnn
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    your point is , sum of two sides of triangle must be > 3rd ?

    • one year ago
  5. mayankdevnani
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    yaaa

    • one year ago
  6. hartnn
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    used that, solved it, got 50% want to verify whether its correct....

    • one year ago
  7. mayankdevnani
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    verify- 3,3,3(forms triangle) sum of two sides of triangle must be > 3rd(6>3)

    • one year ago
  8. hartnn
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    i would like to restate : sum of two sides of triangle must be > or = 3rd

    • one year ago
  9. mayankdevnani
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    right

    • one year ago
  10. mayankdevnani
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    6>3(3+3>3) PROVE IT!

    • one year ago
  11. mayankdevnani
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    thus, it proves it forms a triangle!!! @hartnn can you understand??

    • one year ago
  12. hartnn
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    yes, that forms equilateral triangle. but i want to find the probability.

    • one year ago
  13. mayankdevnani
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    i think probability is outcome/parts=10/3

    • one year ago
  14. hartnn
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    probability can never be >1 -_-

    • one year ago
  15. henpen
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    You want that 1 of the segments is not >0.5 of the stick.

    • one year ago
  16. henpen
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    I think the probability here may be dependent on the decision procedure that you choose the 3 parts in. Can I assume you throw 2 darts simultaneously at the stick, and these points are the points that you cut it?

    • one year ago
  17. hartnn
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    even i can't say what can be assumed, this was the question...and i found it mathematically that this probability is 50% , just want to see new approaches and verify my answer. so , go ahead and try it...

    • one year ago
  18. henpen
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    Well, using my method it's less than 50% that form a triangle. Call where the first dart lands X|dw:1354373456702:dw|, in the half of the stick called A Disregard the 'A' half of the stick. Now, the chance that the second dart lands in A, leaving there a segment of >0.5 is 50%. However...

    • one year ago
  19. henpen
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    It is possible that the second dart lands in B AND the second dart lands in A AND STILL there would be 1 segment >0.5 length of the stick.|dw:1354373603207:dw| So the chance of a non-triangle, because of this, is LESS than 50%. But it absolutely matters which procedure you use to cut.

    • one year ago
  20. henpen
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    For example, you could choose the longer of the 2 parts that the first dart cuts in half, and throw the second dart against this part of the stick.

    • one year ago
  21. hartnn
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    i didn't get u entirely, but got some idea on what u saying.... moreover, i think there's a numerical answer to this, rather than inequality.(like >0.5)....

    • one year ago
  22. henpen
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    Yes, there is (not sure how to yet). Do you understand my reasoning that 1 side of the triangle cannot be >0.5 of the original stick (that is, cannot be > than the sum of the other 2 sides)?

    • one year ago
  23. hartnn
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    ofcourse, i got that when i solved, and that gave me 50%.

    • one year ago
  24. henpen
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    By what logic?

    • one year ago
  25. hartnn
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    ok, i'll show what i had done in test.

    • one year ago
  26. henpen
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    I'll type up my try to work out a numerical answer at the same time

    • one year ago
  27. hartnn
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    |dw:1354374054772:dw|

    • one year ago
  28. hartnn
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    @UnkleRhaukus can u have a look and share any ideas ?? same for Callisto.

    • one year ago
  29. henpen
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    Assume dart 1 has an equal probability of landing anywhere on the stick. By definition, it lands on the half of the stick called A. Equal probability of landing anywhere on there. Now, your question is equivalent to saying: 'what is the probability that the 2nd dart will not leave a part-stick of >0.5 length?' There are 2 ways of making a part-stick of >0.5 length: -dart 2 lands in half A also -this happens|dw:1354374314148:dw|

    • one year ago
  30. henpen
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    The second way works as there is a hole of length >=0.5 in the middle. Working it out: Prob(dart 2 lands in B)=0.5 Prob(x <= y)=(y/0.5) where y is a length measured from the right, and 0.5 is a length.

    • one year ago
  31. hartnn
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    isn't this getting complicated ? for me, yes.... what wrong in my answer ?

    • one year ago
  32. henpen
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    Have I misunderstood the question? It looks like for you a is the random variable, and x, y and z are fixed. How are you making your triangle?

    • one year ago
  33. hartnn
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    yeah, a is fixed. i will make a triangle with sides x,y,z.

    • one year ago
  34. UnkleRhaukus
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    so on of the triangles lengths must be greater than the sum of the other two , you need both darts to cut the string on different sides of the half way point , assuming the first dart land one side, the probability that second dart land on the opposite half of the string is 50%

    • one year ago
  35. UnkleRhaukus
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    one*

    • one year ago
  36. hartnn
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    so, what @henpen said here 'Prob(dart 2 lands in B)=0.5' its straightaway 50% ?

    • one year ago
  37. henpen
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    My problem is that there are ways of the too large side being formed even if they land in the opposite sides, for example:|dw:1354375141124:dw|

    • one year ago
  38. henpen
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    |dw:1354375288008:dw| We want \[p+q<0.5\]

    • one year ago
  39. henpen
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    By definition, \[p<0.5\] and \[q<0.5\] So it seems in the second case it is 50% (not 50% overall)|dw:1354375464988:dw|

    • one year ago
  40. henpen
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    So overall I'd say 75% (as in the first case it's 100% not going to form a triangle (100% of 50%=50% overall) and in the second case it's 50% not going to form a triangle (50% of 50%=25%)).

    • one year ago
  41. UnkleRhaukus
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    hmm,

    • one year ago
  42. hartnn
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    huh ?

    • one year ago
  43. hartnn
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    25% sure ?

    • one year ago
  44. henpen
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    fairly sure

    • one year ago
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