1. ## Help with probability question please

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.

2. Originally Posted by clockingly

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:

$
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:

$
G_{S_5}(z)=[G(z)]^5
$

and so the required prob is:

$
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:

$
P(S_5=15) =651/6^5 \approx 0.08372
$

RonL

3. 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.

4. Thank you! But question - how exactly did you get that the coefficient is 651?

5. 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:

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

$(1-x^{6})^{5}=1-\begin{pmatrix}5\\1\end{pmatrix}x^{6}+\begin{pmatr ix}5\\2\end{pmatrix}x^{12}-\begin{pmatrix}5\\3\end{pmatrix}x^{18}+\begin{pmat rix}5\\4\end{pmatrix}x^{24}-\begin{pmatrix}5\\5\end{pmatrix}x^{30}$

6. 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?

7. 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