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Math Help - Dice Probability

  1. #1
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    Dice Probability


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  2. #2
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    The propability is not my strength but, I remember that we have 1/36 to have 2 and 1/36 to have 12 as overall outcome.

    so for not 2 could be 1 - \frac{1}{36}

    for 4 \frac{3}{36}

    and for 6 \frac{5}{36}

    => think to 2 you have 1/6 for the first dice and 1/6 for the second so 1/36

    Hope that it is useful
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  3. #3
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    Hello, ndcruz!

    It's just matter of counting, isn't it?



    There are 36 possible outcomes.

    There are: . 1 + 3 + 5 \:=\:9 ways to get a sum of 2, 4 or 6.

    So there are: . 36 - 9 \:=\:27 ways for some other sum.

    Therefore: . P(\text{sum not 2, 4, or 6}) \;=\;\frac{27}{36} \;=\;\frac{3}{4}

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