4. On two concentric circles with radii of 3 and $\sqrt{2}$, respectively, a point $P$ and a point $Q$ are randomly taken. Then the probability that the distance $P Q$ between these two points is no greater than $\sqrt{17}$ is $\qquad$
4. $\frac{3}{4}$.

Assume the center of the concentric circles is at the origin of the coordinate system, and let
$$
P(3 \cos x, 3 \sin x), Q(\sqrt{2} \cos y, \sqrt{2} \sin y) \text {, }
$$

where $x, y \in [0, 2\pi]$.
$$
\begin{array}{l}
\text { Then } P Q \leqslant \sqrt{17} \Leftrightarrow P Q^{2} \leqslant 17 \\
\Leftrightarrow (3 \cos x - \sqrt{2} \cos y)^{2} + (3 \sin x - \sqrt{2} \sin y)^{2} \leqslant 17 \\
\Leftrightarrow \cos (x - y) \geqslant -\frac{\sqrt{2}}{2} \\
\Leftrightarrow |x - y| \leqslant \frac{3 \pi}{4} \text { or } \frac{5 \pi}{4} \leqslant |x - y| \leqslant 2 \pi .
\end{array}
$$

The set $\{(x, y) \mid x, y \in [0, 2\pi]\}$ is the square in Figure 2; the set
$$
\left\{(x, y) \mid |x - y| \leqslant \frac{3 \pi}{4} \text { or } \frac{5 \pi}{4} \leqslant |x - y| \leqslant 2 \pi \right\}
$$

is the shaded part in Figure 2. By symmetry, the required probability is
$$
\frac{S_{\triangle A F G} + S_{\text {shaded region ODEB }}}{S_{\triangle O A B}} = 1 - \frac{D F}{O A} = \frac{3}{4} \text {. }
$$