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Show that a sample space is valid by verifying properties

  1. Mar 30, 2016 #1
    1. The problem statement, all variables and given/known data
    http://puu.sh/nYQqE/2b0eaf2720.png [Broken]

    2. Relevant equations
    http://puu.sh/nYSjQ/e48cad3a8b.png [Broken]

    3. The attempt at a solution
    http://puu.sh/nYYjW/174ad8267c.png [Broken]

    My main issue is with part b) and part d). I think that part b) is mostly right, but part d) is definitely wrong and incomplete, and I have no idea how to do it.
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Mar 30, 2016 #2

    andrewkirk

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    (d) is not too bad. The suggestions I would make are:
    - in iii you need to state that A and B are disjoint, otherwise the equalities will not hold. Also the first ##Pr## should be ##Pr'##. You need to be careful with your symbols in a problem like this, as misplaced or wrong symbols lead to confusion.
    - in ii, the statement ##\sum_{i\in S}\frac{n_i}{36}## should be ##\sum_{i\in S'}\frac{n_i}{36}## (care with symbols again). You'll probably find it easier if you split it into two sums ##\sum_{i=2}^6## and ##\sum_{i=7}^{12}##. Then use the above definitions of ##n_i## in terms of ##i##.
    - the proof that ##P(\emptyset)=0## is not correct because dice have nothing to do with it. What is ##\sum_{i\in\emptyset}\frac{n_i}{36}## (how many terms are there in the sum?)?
     
    Last edited: Mar 30, 2016
  4. Mar 30, 2016 #3
    I have used your suggestions and have fixed my answers to this:
    http://puu.sh/nZZZl/d243480f50.png [Broken]

    but I did not know what you mean about splitting the sums for d)ii, and I'm still unsure how to do d)i
     
    Last edited by a moderator: May 7, 2017
  5. Mar 30, 2016 #4

    andrewkirk

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    What are you trying to prove in (d)(i)? I think all the requirements of a probability space are contained in (d)ii and iii.

    For splitting sums, replace ##\sum_{j\in S'}## by ##\sum_{j=2}^6+\sum_{j=7}^{12}## (I prefer to use ##j## rather than ##i## as an index variable because MathJax insists on inappropriately autocorrecting ##i## to ##I##).
     
  6. Mar 30, 2016 #5
    I'm trying to prove property 1 for d)i.

    I don't really understand how you get j = 2 and j = 7 though for the sums or where you get the two sums from to begin with.
     
  7. Mar 30, 2016 #6

    andrewkirk

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    You started with ##\sum_{j\in S'}##
    What is ##S'##?
    Assuming you prove ##Pr(S')=1## in ii, you just have to prove that ##A\subseteq S'\Rightarrow Pr(A)\leq Pr(S')##. Try splitting ##S'## into ##A## and ##S'-A##.
     
  8. Mar 30, 2016 #7
    S' is 36, but I still don't understand how you got 2 and 7 specifically..

    If I split S' into A and S'-A, then I say something like \sum{i \in A} + \sum{i\in S'-A} = \sum{i \in S}?
    I'm not sure how splitting could help me for part d)i.
     
  9. Mar 31, 2016 #8

    andrewkirk

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    No it isn't. Look at the OP where ##S'## is defined.
     
  10. Mar 31, 2016 #9
    Oh, i see what you are saying. So then I evaluate for n_j/36 for the two sums?

    I did 2(1+2+3+4+5+6+7)/36 but this does not equal 1

    I have modified my answers to this
    http://puu.sh/o111N/1376d907c3.png [Broken]
     
    Last edited by a moderator: May 7, 2017
  11. Mar 31, 2016 #10

    andrewkirk

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    That sum has fourteen terms (2 x 7). How many terms are there supposed to be in the sum?
     
  12. Mar 31, 2016 #11
    10...

    so it's 2(1 + 2 + 3 + 4 + 5)/36? That's 30/36 though
     
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