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Math Help - Dice

  1. #1
    Member GAdams's Avatar
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    Dice

    I worked out that (a) the chances of the same score is 1/6 for the ordinary die and 4/6 for the other die.

    Independent events, so multiplied together gives 1/9.



    I am having problems with (b).
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  2. #2
    Eater of Worlds
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    It appears there are only two ways to get a sum of 9. The ordinary die has a 6, then the other must have a 3. The ordinary die has a 5, then the other must have a 4.

    (\frac{1}{6})(\frac{3}{6})+(\frac{1}{6})(\frac{1}{  6})
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  3. #3
    Member GAdams's Avatar
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    Quote Originally Posted by galactus View Post
    It appears there are only two ways to get a sum of 9. The ordinary die has a 6, then the other must have a 3. The ordinary die has a 5, then the other must have a 4.

    (\frac{1}{6})(\frac{3}{6})+(\frac{1}{6})(\frac{1}{  6})
    Thanks, the first part is clear 1/6 * 3/6 but I don't understand why you are adding the 1/6 * 1/6
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by GAdams View Post
    Thanks, the first part is clear 1/6 * 3/6 but I don't understand why you are adding the 1/6 * 1/6
    The (1/6)*(3/6) is the probability of getting a 9 with a 6 on the first die and
    a 3 on the second.

    The (1/6)*(1/6) is the probability of getting a 0 with a 5 on the first die and
    a 4 on the second.

    These are mutualy exclusive outcomes so you add the probabilities.

    RonL
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