<s>[INST] Calculate the probability that the sum of 3 fair dice is at least 13 How would I approach this. I'm more concerned with method than answer. 
 [/INST] Can be done using generating functions which reduces down to polynomial multiplication.
$G(x)=\left(\frac{x}{6}+\frac{x^2}{6}+\frac{x^3}{6}+\frac{x^4}{6}+\frac{x^5}
{6}+\frac{x^6}{6}\right)^3 $
$G(x)$ models throwing 3 fair dice.
When that is expanded.
$\frac{x^3}{216}+\frac{x^4}{72}+\frac{x^5}{36}+\frac{5 \
x^6}{108}+\frac{5 x^7}{72}+\frac{7 x^8}{72}+\frac{25 \
x^9}{216}+\frac{x^{10}}{8}+\frac{x^{11}}{8}+\frac{25 \
x^{12}}{216}+\frac{7 x^{13}}{72}+\frac{5 x^{14}}{72}+\frac{5 \
x^{15}}{108}+\frac{x^{16}}{36}+\frac{x^{17}}{72}+\frac{x^{18}}{216}$
The ones that count are all the x's whose exponent are greater than or equal to 13.
$\frac{7 x^{13}}{72}+\frac{5 x^{14}}{72}+\frac{5 \
x^{15}}{108}+\frac{x^{16}}{36}+\frac{x^{17}}{72}+\frac{x^{18}}{216}$
since the x's are just placeholders and the coefficients are what counts we substitute $x = 1$
$P(\geq 13)=\frac{7}{72}+\frac{5}{72}+\frac{5}{108}+\frac{1}{36}+\frac{1}{72}+\frac{1}{216}=\frac{7}{27}$
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