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.
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:
$\displaystyle
G(z) = z/6 + z^2/6 + z^3/6 + z^4/6 + z^5/6 + z^6/6
$
So the generating function of the sum of 5 such RV's (independent) is:
$\displaystyle
G_{S_5}(z)=[G(z)]^5
$
and so the required prob is:
$\displaystyle
P(S_5=15) = \left. \frac{1}{15!}~\frac{d^{15}}{dz^{15}}G_{S_5}(z)\rig ht|_{z=0}
$
Now you won't get me doing that by hand!
Doing it by machine gives:
$\displaystyle
P(S_5=15) =651/6^5 \approx 0.08372
$
RonL
You can use:
$\displaystyle \left(\sum_{k=1}^{6}x^{k}\right)^{5}$
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 $\displaystyle 6^{5}=7776$
Therefore, the probability of getting a sum of 15 is
$\displaystyle \frac{651}{7776}$
Oops, CB beat me. Oh well, same conclusion. That's good.
I would just use tech to expand it, but you can do it with poly expansions.
$\displaystyle \left(x+x^{2}+x^{3}+x^{4}+x^{5}+x^{6}\right)^{5}$
=$\displaystyle x^{5}\left(1+x+x^{2}+x^{3}+x^{4}+x^{5}\right)^{5}$
=$\displaystyle x^{5}\left(\frac{1-x^{6}}{1-x}\right)^{5}$
You can then use the expansion identities:
$\displaystyle \frac{1}{(1-x)^{5}}=1+\begin{pmatrix}5\\4\end{pmatrix}x$$\displaystyle +\begin{pmatrix}6\\4\end{pmatrix}x^{2}$$\displaystyle +\begin{pmatrix}7\\4\end{pmatrix}x^{3}$$\displaystyle +\begin{pmatrix}8\\4\end{pmatrix}x^{4}$$\displaystyle +\begin{pmatrix}9\\4\end{pmatrix}x^{5}+......+\beg in{pmatrix}14\\4\end{pmatrix}x^{10}+............$
$\displaystyle (1-x^{6})^{5}=1-\begin{pmatrix}5\\1\end{pmatrix}x^{6}+\begin{pmatr ix}5\\2\end{pmatrix}x^{12}-\begin{pmatrix}5\\3\end{pmatrix}x^{18}+\begin{pmat rix}5\\4\end{pmatrix}x^{24}-\begin{pmatrix}5\\5\end{pmatrix}x^{30}$
The thing to do is use a calculator that'll expand. But, if you must do it the long way.
Since, $\displaystyle \frac{1}{(1-x)^{5}}=\sum_{k=0}^{\infty}\begin{pmatrix}5-1+k\\n-1\end{pmatrix}x^{k}$
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:
$\displaystyle \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}+$$\displaystyle ......+1001x^{10}+....)$
Looking at this we can see the terms:
$\displaystyle (-5x^{6})(70x^{4})+(1)(1001)x^{10}$ are the terms which have the x^10 term when we multiply.
(-5)(70)+1001=651