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Probability Question_cutting one into two .

  1. Sep 26, 2014 #1
    Hi! I am a new member, and I'm posting b/c I have a question from the practice list I have received as a practice for an upcoming quiz.
    I've done many questions where mass x (or y) is dependent of x. However, I've come across one where the length of a piece x (of two) is completely independent of x:

    Suppose you have a stick of length L that you would like to cut into two pieces. Assume that the cutting is completely random, that the (1) probability density p(x) for having either one of the pieces be of mass x is completely independent of x.
    Find the function p(x).

    So, I did integral (from x =0 to x=L) p(x)dx = integral (x=0 to x = L) C dx = 1 for normalization. (I put C as a constant).
    Then, I equivelated 1 = C integral (x=0 to x = L) dx = C x/(x=0 to x = L)
    which becomes:

    1 = C (L - 0) = CL
    C = 1/L
    ...and, from this, how am I to get the function p(x)? Do I just put p(x) = 1/L?

    Also, in case I want to find the (2) average length of one piece (preferably a larger one? w/out L/2 value), how do I solve it? For this, I did the similar thing as above, except:

    integral (from y = L/2 to L) p(y) dy = integral (y = L/2 to L) C_new dy = 1 for normalization (I put y = larger piece)
    THen I equivalted 1 = C_new integral (y = L/2 to L) dy

    1 = C_new (L - L/2)
    1 = C_new (L/2)
    C_new = 2/L....

    So, I wonder if average is 2/L, from 2/L integral (y=L/2 to L) y*dy? Or, is it L/2?

    Any suggestions or replies would be welcome!
  2. jcsd
  3. Sep 26, 2014 #2


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    Ask yourself a quick sanity check of the two answers. When the length of the original rod gets longer, should the average length of the cut sections get shorter ( 2/L ) or should they get longer ( L/2 )?

    The average length is the "expected value". Have you studied that? If so, you should check the definition. I think you are missing a factor in your integral.

    P.S. The decision to calculate the average length of the longer section is an interesting one. You should think carefully about what average you want to calculate and how that changes the answer.
  4. Sep 26, 2014 #3
    Thank you!!! Now I realize I forgot "a factor" in my integral. Now, the result I have makes sense!
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