The prime numbers between $2$ to $12$ are $2,3,5, 7,11 $ Case I When sum is $2$, total cases are $(1,1)$ ie, $1$ Case II When sum is $3$, total cases are $(1,2)$, $(2,1)$ ie, $2$ Case III When sum is $5$, total cases are $(1,4),(2,3),(4,1),(3,2)$ i e, $4$ Case IV When sum is $7$, total cases are $(1,6),(2,5),(3,4),(6,1),(5,2),(4,3)$ i e, $6$ Case V When sum is $11$, total cases are $(5,6)$, $(6,5)$ i e, $2$
$\therefore $ Required probability $=\frac{15}{36}=\frac{5}{12}$