7. Let the elements in set $T$ be integers between 1 and $2^{30}$, and their binary representations contain exactly two 1s. If a number is randomly selected from set $T$, then the probability that this number is divisible by 9 is $\qquad$.
7. $\frac{5}{29}$.

Notice that, the elements in set $T$ are of the form $2^{a}+2^{b}$, where $0 \leqslant a<b \leqslant 29$.
Thus, $|T|=\mathrm{C}_{30}^{2}=435$.
By $9\left|\left(2^{a}+2^{b}\right) \Leftrightarrow 9\right| 2^{a}\left(2^{b-a}+1\right)$
$\Leftrightarrow 9 \mid\left(2^{b-a}+1\right)$
$\Leftrightarrow b-a=6 k-3\left(k \in \mathbf{Z}_{+}\right)$.
This implies that,
$0 \leqslant a \leqslant 29-(6 k-3)$.
Thus, $30-(6 k-3)=33-6 k$ for the ordered integer pairs $(a, b)$, satisfying $b-a=6 k-3$.
Since $6 k-3 \leqslant 29$, we have $1 \leqslant k \leqslant 5$.

Therefore, the number of multiples of 9 in set $T$ is $\sum_{k=1}^{5}(33-6 k)=75$.
Hence, the required probability is $\frac{75}{435}=\frac{5}{29}$.