Solution. First, we compute the area of the region bounded by the ellipse $x^{2} / a^{2}+y^{2} / b^{2}=1$, by writing its parametric equations: $x=a \cos t, y=b \sin t$. Since the ellipse is symmetric with respect to the coordinate axes, it is sufficient

![](https://cdn.mathpix.com/cropped/2024_05_22_720b3cbdc89c57e2b629g-029.jpg?height=415&width=713&top_left_y=1159&top_left_x=379)

Fig. 1.6 to compute the area
of one quarter of the specified region:

$$
\begin{aligned}
& S=4 \int_{0}^{a} y d x=4 \int_{\pi / 2}^{0} b \sin t(-a \sin t) d t=-4 a b \int_{\pi / 2}^{0} \sin ^{2} t d t= \\
& =4 a b \int_{0}^{\pi / 2} \frac{1-\cos 2 t}{2} d t=2 a b \int_{0}^{\pi / 2} d t-2 a b \int_{0}^{\pi / 2} \cos 2 t d t=\pi a b
\end{aligned}
$$

Thus, the area $S$ of the region bounded by the ellipse $x^{2} / a^{2}+y^{2} / b^{2}=1$ is given by the formula $S=\pi a b$, where $a$ and $b$ are the semi-axes of the ellipse.

In this case, $S_{C}=\pi \cdot 7 \cdot 4=28 \pi, S_{g}=\pi \cdot 5 \cdot 4=20 \pi$. Using formula (1.5.1), we find the desired probability

$$
P=\frac{S_{g}}{S_{G}}=\frac{20 \pi}{28 \pi}=\frac{20}{28}=\frac{5}{7} \approx 0.714
$$
Example 5. On the plane, the region $G$ is bounded by the ellipse $x^{2} / 49 + y^{2} / 16 = 1$, and the region $g$ - by the ellipse $x^{2} / 25 + y^{2} / 9 = 1$ (Fig. I.6). A point is thrown into the region $G$. What is the probability that the point will land in the region $g?$