9. Given the set $U=\{1,2,3,4,5\}, I=\{X \mid X \subseteq U\}$, if two different elements $A, B$ are randomly selected from set $I$, then the probability that $A \cap B$ contains exactly 3 elements is $\qquad$ .
The total number of subsets of $U$ is $2^{5}=32$, the number of 3-element subsets is $C_{5}^{3}=10$, and for each 3-element subset, it can form the corresponding $(A, B)$ pairs, a total of 8. Therefore, the required probability is $P=\frac{80}{A_{32}^{2}}=\frac{5}{62}$.