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Thread: I'm missing something

  1. #1
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    I'm missing something

    so I've got this question:
    A die with faces numbered 1 to 6 is biased so that $\displaystyle P(score=r)=kr$, (r=1,...,6). Find the value of k.

    Now my problem is i don't know which numbers it is biased to. supposing that (r=1,...,6) means the numbers from 1 to 6.
    Thanks
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  2. #2
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    Quote Originally Posted by arze View Post
    so I've got this question:
    A die with faces numbered 1 to 6 is biased so that $\displaystyle P(score=r)=kr$, (r=1,...,6). Find the value of k.

    Now my problem is i don't know which numbers it is biased to. supposing that (r=1,...,6) means the numbers from 1 to 6.
    Thanks
    You don't need to, you only need to know that:

    $\displaystyle P(score=1)+ P(score=2)+ ...+ P(score=6)=1$

    CB
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  3. #3
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    I'm really sorry i don't get it.
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by arze View Post
    I'm really sorry i don't get it.
    When you roll the die it will come up with some face, so the probability of a result is 1, which is the sum of the probabilities of all the possible outcomes.

    CB
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  5. #5
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    ok. so i know that r can be any number, and P(any number)=1. and what's next?
    P(score is r)=kr=1?
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  6. #6
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    The sum of the probabilities of all posssible outcomes must be $\displaystyle 1$.

    So:

    $\displaystyle
    \sum_{r=1}^6 P(score=r)=\sum_{r=1}^6 k r = 21k=1
    $

    Hence $\displaystyle k=1/21$.

    CB
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  7. #7
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    Hello, arze!

    You have to understand what was given.


    A die with faces numbered 1 to 6 is biased so that $\displaystyle P(\text{score}=r)\:=\:kr,\;\;(r\,=\,1,\,2,\,\hdots ,\,6)$
    Find the value of $\displaystyle k.$

    We are told that the probability of rolling a number $\displaystyle r$ is a constant $\displaystyle k$ times $\displaystyle r$

    So we have these probabilities:

    . . . . $\displaystyle \begin{array}{c|c}
    r & P(r)\\ \hline
    1 & k \\
    2 & 2k \\
    3 & 3k \\
    4 & 4k \\
    5 & 5k \\
    6 & 6k \end{array}$


    The probability that one of the numbers turn up is 100%.

    So we have: .$\displaystyle k + 2k + 3k + 4k + 5k + 6k \:=\:1 \quad\Rightarrow\quad 21k \:=\:1$

    Therefore: .$\displaystyle k \:=\:\frac{1}{21}$

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  8. #8
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    Thank you! i understand now!
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