If two unbaised six-faced dice are thrown simultaneously until a sum of either 7 or 11 occurs, then the porbability that 7 comes before 11 is 

two dice are thrown 

A=Sum is 7

B=Sum is 11

C=Sum neither 7 nor 11

$\begin{array}{rcl}\mathrm{P}\left(\mathrm{A}\right)& =& \frac{6}{36}=\frac{1}{6} \mathrm{P}\left(\mathrm{B}\right)=\frac{2}{36}=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$18$}\right.\\ \mathrm{P}\left(\mathrm{C}\right)& =& 1-\left(\frac{1}{6}+\frac{1}{18}\right)=1-\frac{2}{9}=\frac{7}{9}\end{array}$

$\begin{array}{rcl} & & \\ \mathrm{Probability} \mathrm{that} 7 \mathrm{comes} \mathrm{before} 11 & =& \frac{1}{6}+\frac{7}{9} \frac{1}{6}+{\left(\frac{7}{9}\right)}^2 \frac{1}{6} ......\\ & =& \frac{{\displaystyle \frac{1}{6}}}{1-{\displaystyle \frac{{\displaystyle 7}}{9}}}=\frac{9}{12}=\frac{3}{4}\end{array}$