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Math Help - statistics

  1. #1
    Junior Member
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    Oct 2007
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    statistics

    hey

    i'm slightly confused with how i go about doing this question

    A cubical dice is biased so that the probability of any particular score between 1 and 6 (inclusive) being obtained is proportional to that score. Find the probability of scoring a 1.

    There is another question exactly the same except the score being obtained is inversely proportional.

    Thanks. any help would be great
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  2. #2
    Member
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    Quote Originally Posted by lra11 View Post
    hey

    i'm slightly confused with how i go about doing this question

    A cubical dice is biased so that the probability of any particular score between 1 and 6 (inclusive) being obtained is proportional to that score. Find the probability of scoring a 1.

    There is another question exactly the same except the score being obtained is inversely proportional.

    Thanks. any help would be great
    For some constant k, you have the following probabilities:

    p(1)=k;~~~<br />
p(2)=2k;~~~<br />
p(3)=3k;~~~<br />
p(4)=4k;~~~<br />
p(5)=5k;~~~<br />
p(6)=6k

    The sum of the probabilities is 1, so k+2k+3k+4k+5k+6k=1 \implies 21k=1 \implies k=\frac{1}{21}. Therefore p(1)=\frac{1}{21}.

    As for the second, you get:

    p(1)=\frac{k}{1};~~~<br />
p(2)=\frac{k}{2};~~~<br />
p(3)=\frac{k}{3};~~~<br />
p(4)=\frac{k}{4};~~~<br />
p(5)=\frac{k}{5};~~~<br />
p(6)=\frac{k}{6}

    and then \frac{k}{1}+\frac{k}{2}+\frac{k}{3}+\frac{k}{4}+\f  rac{k}{5}+\frac{k}{6}=1 \implies \frac{147k}{60}=21 \implies k=\frac{20}{49}. Therefore, p(1)=\frac{20}{49}
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  3. #3
    Junior Member
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    Oct 2007
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    thanks

    thanks, i really understand now
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