Consider $A$ and $B$ be the events when the sum is obtained as $7$
and $11$ respectively.

Number of times sum$7$ and $11$ occur are,

$n\left(A\right)=\left\{\right(2,5),(5,2),(3,4),(4,3),(1,6),(6,1\left)\right\}=6$

$n\left(B\right)=\left\{\right(5,6),(6,5\left)\right\}=2$

Now, the probability is,

$P\left(A\right)=\frac{6}{36}=\frac{1}{6}$

$P\left(B\right)=\frac{2}{36}=\frac{1}{18}$

The probability when neither the sum of$7$or $11$ occurs,

$P\left(C\right)=\frac{36-(6+2)}{36}=\frac{7}{9}$

The probability that$7$ comes before $11$,

$P=\frac{1}{6}+\frac{1}{6}\times \frac{7}{9}+{\left(\frac{7}{9}\right)}^2\left(\frac{1}{6}\right)+{\left(\frac{7}{9}\right)}^3\left(\frac{1}{6}\right)\dots$

$=\frac{1}{6}\left(\frac{1}{1-{\displaystyle \frac{7}{9}}}\right)$

$=\frac{3}{4}$