5. $\frac{1}{140}$.

Analysis: The total number of groupings is $C_{9}^{3} \cdot C_{6}^{3} \cdot C_{3}^{3}=1680$, calculate the number of groupings that satisfy the condition.
Let the sums of the three groups be $a, b, c$, all of which are prime numbers, so we can assume $7 \leq a \leq b \leq c \leq 23$. The prime numbers between 7 and 23 are 6 in total: $7,11,13,17,19,23$, where $7,13,19 \equiv 1(\bmod 3), 11,17,23 \equiv 2(\bmod 3)$. Since $7+13+19 \neq 45,11+17+23 \neq 45$, it follows that $a, b, c$ must have two identical numbers. Therefore, the sets $\{a, b, c\}$ that satisfy the condition are $\{7,19,19\} ;\{11,11,23\} ;\{13,13,19\}$ or $\{11,17,17\}$, a total of four. Among them, $\{7,19,19\}$ can only be 124/379/568;
$\{11,11,23\}$ can only be $137 / 245 / 689 ; \quad\{13,13,19\}$ has a total of 5 groups: 139/256/478;
139/247/568; 148/256/379; 157/238/469; 157/346/289; $\{11,17,17\}$ also has a total of 5 groups: $128 / 359 / 467$;
$137 / 269 / 458 ; 146 / 278 / 359 ; 236 / 179 / 458 ; 245 / 179 / 368$, a total of 12 groups. Therefore, the required probability is $\frac{12}{1680}=\frac{1}{140}$.
5. Divide $1,2,3,4,5,6,7,8,9$ into 3 groups, each containing three numbers, then the probability that the sum of the numbers in each of these three groups is a prime number is