8. A positive integer $x \in$ $\{1,2, \cdots, 2016\}$ is generated with equal probability. Then the probability that the sum of the digits of $x$ in binary is no more than 8 is $\qquad$
8. $\frac{655}{672}$.

Let $x \in\{1,2, \cdots, 2016\}$ be represented in binary as $\left(x_{10} x_{9} \cdots x_{1} x_{0}\right)_{2}$,
i.e., $x=\sum_{i=0}^{10} x_{i} 2^{i}\left(x_{i} \in\{0,1\}\right)$.
In particular, the binary representation of 2016 is $(11111100000)_{2}$.
Next, consider the number of $x$ such that $\sum_{i=0}^{10} x_{i}>8$, which is equivalent to
$$
p=\sum_{i=5}^{10} x_{i} \leqslant 5 \text {, and } q=\sum_{i=0}^{4} x_{i} \geqslant 9-p \text {. }
$$

Thus, the number of such $x$ is
$$
\sum_{\substack{p \leqslant 5 \\ q \geqslant 9-p}} \mathrm{C}_{6}^{p} \mathrm{C}_{5}^{q}=\mathrm{C}_{6}^{4} \mathrm{C}_{5}^{5}+\mathrm{C}_{6}^{5}\left(\mathrm{C}_{5}^{4}+\mathrm{C}_{5}^{5}\right)=51 .
$$

Therefore, the number of $x$ whose binary digits sum to no more than 8 is $2016-51=1965$.
Hence, the required probability is $\frac{1965}{2016}=\frac{655}{672}$.