8. Given the set
$$
T=\left\{n \mid n=5^{a}+5^{b}, 0 \leqslant a \leqslant b \leqslant 30, a, b \in \mathbf{Z}\right\} \text {. }
$$

If a number is randomly selected from set $T$, then the probability that this number is a multiple of 9 is $\qquad$ .
8. $\frac{5}{31}$.

It is easy to see that when $x=1,7,13,19,25$,
$$
\begin{array}{l}
5^{x} \equiv 5(\bmod 9) ; \\
x=4,10,16,22,28 \text { when, } \\
5^{x} \equiv 4(\bmod 9) ; \\
x=2,8,14,20,26 \text { when, } \\
5^{x} \equiv 7(\bmod 9) ; \\
x=5,11,17,23,29 \text { when, } \\
5^{x} \equiv 2(\bmod 9) ; \\
x=3,9,15,21,27 \text { when, } \\
5^{x} \equiv 8(\bmod 9) ; \\
x=0,6,12,18,24,30 \text { when, } \\
5^{x} \equiv 1(\bmod 9) ;
\end{array}
$$

The sample space contains $\mathrm{C}_{31}^{2}+31$ basic events, and there are $5 \times 5 \times 2+5 \times 6=80$ ways for $5^{a}+5^{b}$ to be divisible by 9.
Thus, $P=\frac{80}{31 \times 15+31}=\frac{5}{31}$.