9. Given the sets:
$$
U=\{1,2,3,4,5\}, I=\{X \mid X \subseteq U\} .
$$

If two different elements $A$ and $B$ are randomly selected from set $I$, the probability that $A \cap B$ contains exactly three elements is $\qquad$
9. $\frac{5}{62}$.

When $A \cap B$ is determined, for example, $A \cap B=\{3,4,5\}$, we can set $A=A^{\prime} \cup\{3,4,5\}, B=B^{\prime} \cup\{3,4,5\}$, $A^{\prime} \cap B^{\prime}=\varnothing$.
Then the cases for $\left\{A^{\prime}, B^{\prime}\right\}$ are:
$$
\{\varnothing,\{1\}\},\{\varnothing,\{2\}\},\{\varnothing,\{1,2\}\},\{\{1\},\{2\}\},
$$

a total of 4 cases.
Therefore, the required probability is $\frac{\mathrm{C}_{5}^{3} \times 4}{\mathrm{C}_{32}^{2}}=\frac{5}{62}$.