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Math Help - Question about a Question: Inclusion-Exclusion theorm

  1. #1
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    Question about a Question: Inclusion-Exclusion theorm

    QUESTION: Question about a Question: Inclusion-Exclusion theorm-question.jpeg

    Okey.
    So, i'm curious how
    P(A) = 0.5
    Because
    Possible numbers (sum) = 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 = 11 Total
    (because if two dice roll lowest number and we add them, you get two)

    So why, is P(A) = 1/2?
    Surely it should be = 6/11?

    Thanks
    Last edited by AshleyT; January 27th 2011 at 01:31 PM.
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  2. #2
    CSM
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    Did you account for the various ways to throw 7 for example?
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    What do you mean?
    I didn't think the 'way' in which the number's are aquired (aka 6 + 1, 3 + 3) would be required for this question?
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  4. #4
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    Quote Originally Posted by AshleyT View Post
    What do you mean?
    I didn't think the 'way' in which the number's are aquired (aka 6 + 1, 3 + 3) would be required for this question?
    Here are the probabilities.
    \begin{array}{*{20}c}<br />
   {P(2)} &  =  & {\frac{1}<br />
{{36}}}  \\<br />
   {P(4)} &  =  & {\frac{3}<br />
{{36}}}  \\<br />
   {P(6)} &  =  & {\frac{5}<br />
{{36}}}  \\<br />
   {P(8)} &  =  & {\frac{5}<br />
{{36}}}  \\<br />
   {P(10)} &  =  & {\frac{3}<br />
{{36}}}  \\<br />
   {P(12)} &  =  & {\frac{1}<br />
{{36}}}  \\<br /> <br />
 \end{array}

    The ways to get 10 are \{(5,5),~(4,6),~(6,4)\} that is 3 out of 36.
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    Quote Originally Posted by Plato View Post
    Here are the probabilities.
    \begin{array}{*{20}c}<br />
   {P(2)} &  =  & {\frac{1}<br />
{{36}}}  \\<br />
   {P(4)} &  =  & {\frac{3}<br />
{{36}}}  \\<br />
   {P(6)} &  =  & {\frac{5}<br />
{{36}}}  \\<br />
   {P(8)} &  =  & {\frac{5}<br />
{{36}}}  \\<br />
   {P(10)} &  =  & {\frac{3}<br />
{{36}}}  \\<br />
   {P(12)} &  =  & {\frac{1}<br />
{{36}}}  \\<br /> <br />
 \end{array}

    The ways to get 10 are \{(5,5),~(4,6),~(6,4)\} that is 3 out of 36.
    Ah, i see.
    I was thinking: 11 outcomes, therefore P(2) would be 1/11.
    Thankyou.
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  6. #6
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    Hello, AshleyT!

    You've never rolled a pair of dice before, have you?


    You have two dice, each has a choice of six numbers.

    There are 36 possible outcomes:

    . . \begin{array}{cccccc} <br />
(1,1) & (1,2) & (1,3) & (1,4) & (1,5) & (1,6) \\<br />
(2,1) & (2,2) & (2,3) & (2,4) & (2,5) & (2,6) \\<br />
(3,1) & (3,2) & (3,3) & (3,4) & (3,5) & (3,6) \\<br />
(4,1) & (4,2) & (4,3) & (4,4) & (4,5) & (4,6) \\<br />
(5,1) & (5,2) & (5,3) & (5,4) & (5,5) & (5,6) \\<br />
(6,1) & (6,2) & (6,3) & (6,4) & (6,5) & (6,6) \end{array}


    Now count the cases . . .


    Sum is even: . \begin{Bmatrix}(1,1) & (1,3) & (1,5) \\\<br />
(2,2) & (2,4) & (2,6) \\ (3,1) & (3,3) & (3,5) \\ (4,2) & (4,4) & (4,6) \\ (5,1) & (5,3) & (5,5) \\ (6,2) & (6,4) & (6,6) \end{Bmatrix} \quad \text{ 18 cases}

    \displaystyle{\text{Therefore: }\;P(\text{Sum is even}) \;=\;\frac{18}{36} \;=\;\frac{1}{2}

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