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Math Help - Loaded Die Probability

  1. #1
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    Loaded Die Probability

    Hi yall...I need some help with this question. It seems to be kicking my trash some. I know how to calculate normal probability, but with the die being loaded I'm not quite sure what to do. Any help would be terrific. Thanks!

    A die is loaded so that each odd number is twice as likely as each even number to appear. Create a probability distribution for such a die. Also find P(G), wehre G is the event that a number greater than 3 occurs on a single roll of the die.

    Again..THANKS!
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by pison View Post
    Hi yall...I need some help with this question. It seems to be kicking my trash some. I know how to calculate normal probability, but with the die being loaded I'm not quite sure what to do. Any help would be terrific. Thanks!

    A die is loaded so that each odd number is twice as likely as each even number to appear. Create a probability distribution for such a die. Also find P(G), wehre G is the event that a number greater than 3 occurs on a single roll of the die.

    Again..THANKS!
    so

    P(odd)=2P(even)

    and

    P(odd)+P(even)=1

    by sub

    3P(even)=1

    or P(even)=1/3 and then P(odd)=2/3

    Then P(2)+P(4)+P(6)=1/3

    ASSUMING that these are equal then

    P(2)=P(4)=P(6)=1/9

    by a simialr argument

    P(1)=P(3)=P(5)=2/9

    so then P(G) = P(4)+P(5)+P(6)=1/9+2/9+1/9=4/9
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  3. #3
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    thank you for responding so quickly...

    could you clarify a little more? i am looking to get a probability distribution out of this. any help?

    thanks!
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  4. #4
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    So if I understand correctly:

    If I say p(x) is the probablity of rolling an x, the problem states p(2)=p(4)=p(6)=\frac{1}{2} p(1)=\frac{1}{2} p(3)=\frac{1}{2} p(5) . Of course  \sum_{i=1}^6 p(i) = 1 . But this implies that:

    <br />
p(1)=p(3)=p(5) = \frac{2}{9}  <br />
,<br />
p(2)=p(4)=p(6) = \frac{1}{9}  <br />

    Then G is the event that a number strictly greater than 3 appears, meaning either a 4 or a 5 or a 6 is rolled. So P(G) = p(4)+p(5)+p(6) = \frac{1}{9} + \frac{2}{9} + \frac{1}{9} = \frac{4}{9}

    EDIT: The distribution is exactly those values p(i) .
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