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Math Help - Help with probability question please

  1. #1
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    Help with probability question please

    I'm not really sure how to go about this, so any help would be greatly appreciated!

    Suppose you throw a six-sided die five times. Find the probability that the sum of the outcomes of the throws is 15 using generating functions.
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  2. #2
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    Quote Originally Posted by clockingly View Post
    I'm not really sure how to go about this, so any help would be greatly appreciated!

    Suppose you throw a six-sided die five times. Find the probability that the sum of the outcomes of the throws is 15 using generating functions.
    Well the generating function for a single throw is:

    <br />
G(z) = z/6 + z^2/6 + z^3/6 + z^4/6 + z^5/6 + z^6/6<br />

    So the generating function of the sum of 5 such RV's (independent) is:

    <br />
G_{S_5}(z)=[G(z)]^5<br />

    and so the required prob is:

    <br />
P(S_5=15) = \left. \frac{1}{15!}~\frac{d^{15}}{dz^{15}}G_{S_5}(z)\rig  ht|_{z=0}<br />

    Now you won't get me doing that by hand!

    Doing it by machine gives:

    <br />
P(S_5=15) =651/6^5 \approx 0.08372<br />

    RonL
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  3. #3
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    You can use:

    \left(\sum_{k=1}^{6}x^{k}\right)^{5}

    When you expand it out, the coefficient of x^15 is your answer.

    Doing so, we see the coefficient of x^15 is 651.

    The total number of outcomes is 6^{5}=7776

    Therefore, the probability of getting a sum of 15 is

    \frac{651}{7776}


    Oops, CB beat me. Oh well, same conclusion. That's good.
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  4. #4
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    Thank you! But question - how exactly did you get that the coefficient is 651?
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  5. #5
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    I would just use tech to expand it, but you can do it with poly expansions.

    \left(x+x^{2}+x^{3}+x^{4}+x^{5}+x^{6}\right)^{5}

    = x^{5}\left(1+x+x^{2}+x^{3}+x^{4}+x^{5}\right)^{5}

    = x^{5}\left(\frac{1-x^{6}}{1-x}\right)^{5}

    You can then use the expansion identities:

    \frac{1}{(1-x)^{5}}=1+\begin{pmatrix}5\\4\end{pmatrix}x +\begin{pmatrix}6\\4\end{pmatrix}x^{2} +\begin{pmatrix}7\\4\end{pmatrix}x^{3} +\begin{pmatrix}8\\4\end{pmatrix}x^{4} +\begin{pmatrix}9\\4\end{pmatrix}x^{5}+......+\beg  in{pmatrix}14\\4\end{pmatrix}x^{10}+............

    (1-x^{6})^{5}=1-\begin{pmatrix}5\\1\end{pmatrix}x^{6}+\begin{pmatr  ix}5\\2\end{pmatrix}x^{12}-\begin{pmatrix}5\\3\end{pmatrix}x^{18}+\begin{pmat  rix}5\\4\end{pmatrix}x^{24}-\begin{pmatrix}5\\5\end{pmatrix}x^{30}
    Last edited by galactus; October 20th 2007 at 10:40 AM.
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  6. #6
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    Thanks! But I'm having trouble seeing how the coefficient of x^15 can be determined from the expansions. Could you possibly show how that works?
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  7. #7
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    The thing to do is use a calculator that'll expand. But, if you must do it the long way.

    Since, \frac{1}{(1-x)^{5}}=\sum_{k=0}^{\infty}\begin{pmatrix}5-1+k\\n-1\end{pmatrix}x^{k}

    We take note that we factored out a x^5. Therefore, we need only look for x^10 terms, instead of x^15.

    When we multiply we get:

    \frac{(1-x^{6})^{5}}{(1-x)^{5}}=(1-5x^{6}+10x^{12}-10x^{18}+5x^{24}-x^{30})(1+5x+15x^{2}+35x^{3}+70x^{4}+ ......+1001x^{10}+....)

    Looking at this we can see the terms:

    (-5x^{6})(70x^{4})+(1)(1001)x^{10} are the terms which have the x^10 term when we multiply.

    (-5)(70)+1001=651
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