# For the experiment of tossing a six sided die twice, what is the probability that the sum is less than 11?

Jun 14, 2017

$P \left(\text{sum} < 11\right) = \frac{11}{12}$

#### Explanation:

Without loss of generality we could assume that one die is $\textcolor{red}{\text{red}}$ and the other is $\textcolor{g r e e n}{\text{green}}$

For each of the 6 $\textcolor{red}{\text{red}}$ outcomes there are 6 $\textcolor{g r e e n}{\text{green}}$ outcomes.
This means that there are a combined collection of $\textcolor{red}{6} \times \textcolor{g r e e n}{6} = \textcolor{b l u e}{36}$ possible outcomes.

Of the $\textcolor{b l u e}{36}$ possible outcomes only $\textcolor{m a \ge n t a}{3}$ result in totals greater than on equal to 11; namely
$\textcolor{w h i t e}{\text{XXX")< color(red)("red"),color(green)("green}} > \in \left\{< \textcolor{red}{5} , \textcolor{g r e e n}{6} > , < \textcolor{red}{6} , \textcolor{g r e e n}{5} > , < \textcolor{red}{6} , \textcolor{g r e e n}{6} >\right\}$

Therefore, there are $\textcolor{b l u e}{36} - \textcolor{m a \ge n t a}{3} = \textcolor{b r o w n}{33}$ possible outcomes for which the sum is less than 11.

Since each outcome is (assumed to be) equally probable,
$\textcolor{w h i t e}{\text{XXX")P("sum} < 11} = \frac{\textcolor{b r o w n}{33}}{\textcolor{b l u e}{36}} = \frac{11}{12}$