Total number of sample space,
$n(S)=6 \times 6=36$
Let $E=$ Event of getting maximum sum of two dices is $5$
$=\{(1,1),(1,2),(2,1),(1,3),(3,1),(2,2) (1,3),(3,1),(1,4),(4,1),(2,3),(3,2)\}$
$\therefore n(E)=12 $
$\therefore $ Probability (the maximum sum of two dices is $5$ )
$=\frac{n(E)}{n(S)}=\frac{12}{36}=\frac{5}{3}$
$\therefore $ Probability (the sum of two dices is more than $5$ )
$=1-$ Probability (the sum of two dices is $5$ )
$=1-\frac{1}{3}=\frac{2}{3}$