I would be glad if one of you could help me. I do not even know how to start. The question is:
A symmetrical, six sided, die is thrown independently seven times. What is the probability that the total number of points obtained is 14?
Well if been trying to solve the problem with the generating function method.
I tried to use the random sum formula, using a Random Variable which is almost everywhere 7. Then computed its generating function and inserted the generating function of a thrown dice.
Then I tried to reengineer the term to be an infinte sum, such that I can read out the probiblity.
The probability is calculated by the formula \begin{equation*} \frac{\text{number of favorable outcomes}}{\text{number of possible outcomes}} \end{equation*} The number of possible outcomes is $6^7$ since each of the seven die can give $6$ values. In order to calculate the number of favorable outcomes, denote by $k_1, k_2,\ldots k_7$ the seven values obtained by throwing the die. Then we have \begin{equation*} k_1+k_2+\ldots+k_7=14\Rightarrow \end{equation*} \begin{equation*} (k_1-1)+(k_2-1)+\ldots+(k_7-1)=7 \end{equation*} Denote $\alpha_i=k_i-1$, $i=\overline{1, 7}$. Then, $\alpha_i\in\{0,1,2,3,4,5\}$ for any $i$. A combination of $n$ objects taken $m$ at a time with repetition represents the number of ways we can write the number $n$ as a sum of $m$ nonnegative integers and is denoted by \begin{equation*} \overline{C_{n}^m}=C_{m+n-1}^m={n+m-1\choose m} \end{equation*} For more on combinations with repetition see this. Thus, the number of ways we can choose $7$ nonnegative integers with sum $7$ is \begin{equation*} \overline{C_7^7}=1716 \end{equation*} But this is not equal with the number of ways we can choose $\alpha_1, \alpha_2,\ldots, \alpha_7$ because we included sums which contain integers larger than $5$. These are the sums with one $7$ and six zeroes and the ones with one $6$, one $1$ and fie zeroes. The ones with a seven are in number of $7$ (because we can put the $7$ on every one of the seven positions) and the ones with a $6$ are $42$ because we can place the $6$ on seven positions and the $1$ on six. Thus, we can choose $\alpha_1, \alpha_2, \ldots, \alpha_7$ in $1667$ ways. Since by choosing these numbers we automatically choose $k_1, k_2,\ldots, k_7$, we have that $1667$ is indeed the number of favorable outcomes. So, the desired probability is equal to \begin{equation*} \frac{1667}{279936}\simeq 0.005954 \end{equation*}