4. Each experience of throwing the circle corresponds to a point $M-$ the position of the center of the circle on the vertical segment $[A ; B]$ of length 1.

![](https://cdn.mathpix.com/cropped/2024_05_06_8b002ac84843a2e42fafg-05.jpg?height=315&width=702&top_left_y=1016&top_left_x=382)

Let's introduce the following notations: $R$ - the radius of the circle, $O$ - the midpoint of the segment $[A ; B], \quad O M=x \in[0 ; 0.5]$ - a random variable uniformly distributed on this segment, $\alpha$ - the angle indicated in the figure, $A M=h=\frac{\cos \alpha}{2}$.

The area of the circular segment $S_{1}$ of a circle with radius $R$ is given by the formula:

$$
S_{1}=\left|\alpha R^{2}-\frac{1}{2} R^{2} \sin 2 \alpha\right|
$$

In our case, $R=\frac{1}{2}$, so,

$$
S_{1}=\frac{\alpha}{4}-\frac{\sin 2 \alpha}{8} .
$$

From the problem statement, it follows that

$$
\left\{\begin{array}{c}
S_{1}: S_{2}=(\pi-2):(3 \pi+2) \\
S_{1}+S_{2}=\frac{\pi}{4}
\end{array}\right.
$$

Then $S_{1}=k(\pi-2)$, and $S_{2}=k(3 \pi+2)$. Substituting this into the second equality, we get: $S_{1}+S_{2}=k \cdot 4 \pi=\frac{\pi}{4}$. Therefore, $k=\frac{1}{16}$. Then, $S_{1}=\frac{\alpha}{4}-\frac{\sin 2 \alpha}{8}=\frac{1}{16}(\pi-2)$. From this, we find $\alpha=\frac{\pi}{4}$. Thus, the conditions of the problem correspond to the central angle $P M Q$ equal to $\frac{\pi}{2}$. Then $h=\frac{\sqrt{2}}{4}$ and the favorable outcome of the experiment corresponds to points $M$ that are no more than $\frac{1}{2}-h=\frac{2-\sqrt{2}}{4}$ away from point $O$. Since the probability of the desired event is the ratio of the length of the segment of "favorable" outcomes, i.e., $\frac{2-\sqrt{2}}{2}$ to the length of the segment $[A ; B]$, i.e., to one, the desired probability $P(A)=\frac{2-\sqrt{2}}{2} \approx 0.29$
Answer: $P(A)=\frac{2-\sqrt{2}}{2} \approx 0.29$.