<s>[INST] Dice roll probability, at least 9 total? If I have two dice with $6$ sides each, what is the probability of me rolling atleast $9$ total? I think I'm correct when thinking that the probability of rolling a $9$ is $\frac{4}{36}$, that is $11.1...\%$, but how do I go from here to calculate the "at least" part?
 [/INST] $\begin{array}{|c|c|c|c|}
\hline
 &\overrightarrow{ D2} & \color{blue}{1} & \color{blue}{2} & \color{blue}{3} & \color{blue}{4} & \color{blue}{5} & \color{blue}{6} \\ \hline D1\downarrow\\\hline
\color{blue}{1} &&2 &3 &4 & 5 & 6 & 7\\ \hline
 \color{blue}{2}&& 3 & 4&5 & 6 & 7 & 8\\ \hline
\color{blue}{3} &&4  &5 &6 & 7 & 8 & \color{red}{9}\\ \hline
\color{blue}{4} &&5 & 6 & 7 & 8 &\color{red}{9} &\color{red}{10}\\ \hline
\color{blue}{5} &&6 &7 &8 &\color{red}{9} &\color{red}{10} &\color{red}{11} \\ \hline
\color{blue}{6} && 7 & 8 & \color{red}{9} &\color{red}{10} &\color{red}{11}&\color{red}{12}\\\hline
\end{array}$
Every cell containing a number in red, satisfies: $(D1+D2) \ge 9$
Assuming that each die is a fair die, the probability of obtaining any number from 1-6 on each of the two dice is $\frac{1}{6}$.
For example, the probability of obtaining $(D1,D2)=(1,1)$ is $\left(\frac{1}{6}\right)^2 = \left(\frac{1}{36}\right)$
Every individual outcome in the table is obtained with probability $\frac{1}{36}$ as each result is equally likely.
Since there are 36 (6 $\times$ 6 table) total outcomes, the probability will be $\frac{x}{36}$ or an equivalent fraction.
Try to find $x$ via a simple counting method (count how many numbers in red there are).
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