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Math Help - probabilty of mutually exclusive events and trouble with understanding notation

  1. #1
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    probabilty of mutually exclusive events and trouble with understanding notation

    Question. Consider three mutually exclusive and exhaustive events A_0, A_1 and A_2 where
    P(A_0\cup{A_1})=p_0
    P(A_1\cup{A_2})=p_1
    P(A_2\cup{A_0})=p_2.

    What condition on p_0, p_1, p_2 must hold?

    Now generalise to n mutually exclusive and exhaustive events A_0,... A_{n-1} where

    P(\bigcup^{r+k-1}_{i=r}A_{i (mod n)})=p_r,

    for r=0,..., n-1 and 0<k<n. What condition on p_0,..., p_n-1 must hold?

    Note i(mod n) is the remainder when i is divided by n.

    My problems:

    - why the events are called not only 'mutually exclusive' but also 'exhaustive'? What does 'exhaustive' add to 'mutually exclusive'? Does it mean the three events 'take up' the whole sample space?

    - what condition must hold on p_0, p_1, p_2? I know that P(A_0\cup{A_1}\cup{A_2})=1, and the sum of p_0, p_1, p_2 includes at least this amount (1) and more (because A_1 and A_2 are included twice. So far I can only say that sum of p_0, p_1, p_2 is bigger than 1.

    - I have trouble understanding the 'big cup' notation and how does it relate to the example with three As in the beginning of the question. If I cannot understand the notation how can I start thinking of answering the question... It looks like probability of a union of events A_i, with i running from 0 to n-1 (i=r), but where does k come from and what is its role?
    Last edited by Volga; January 23rd 2011 at 06:45 PM.
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  2. #2
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    Hello, Volga!

    Here is part of the solution . . .


    Consider three mutually exclusive and exhaustive events A_0,\,A_1 and A_2 where:
    . . P(A_0\cup{A_1})=p_0
    . . P(A_1\cup{A_2})=p_1
    . . P(A_2\cup{A_3})=p_2.

    What condition on p_0, p_1, p_2 must hold?

    You are correct.

    "Mutually exclusive" and "mutually exhaustive" means the sets have
    . . no overlap, and together they "take up" the entire sample space.

    The Venn diagram looks like this:

    Code:
          * - - - - - - - *
          |               |
          *      A0       *
          | *           * |
          |   *       *   |
          |     *   *     |
          |  A1   *   A2  |
          |       |       |
          * - - - * - - - *

    Let a_0,a_1,a_2 represent the areas of their respective regions in the unit square.

    \text{Then: }\;\begin{Bmatrix}P(A_0) \:=\:a_0 \\ P(A_1) \:=\:a_1 \\ P(A_2) \:=\:a_2 \end{Bmatrix}\quad\Rightarrow\quad \begin{Bmatrix}P(A_0\cup A_1)\:=\:a_0+a_1 \:=\:p_0 & [1] \\ P(A_1\cup A_2) \:=\:a_1+a_2 \:=\:p_2 & [2] \\ P(A_2\cup A_0) \:=\:a_2+a_0 \:=\:p_2 & [3]\end{Bmatrix}


    \text{Add [1], [2] and [3]: }\;2a_0 + 2a_1 + 2a_2 \:=\:p_0 + p_1+p_2

    . . . . . \text{and we have: }\;2\underbrace{(a_0 + a_1 + a_2)}_{\text{This is 1}} \:=\:p_0 + p_1 + p_2

    Therefore: . p_0+p_1+p_2 \;=\;2

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  3. #3
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    Thanks to you, I realised I misread the question ))) there is no A_3 so I should have been able to answer the 'what condition on p' part. I corrected the first post.

    Do you have any pointers on how to intepret the second, 'general' part of the question?...
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