416. In the circle $x^{2}+y^{2} \leqslant R^{2}$, the two-dimensional probability density is $f(x, y)=C\left(R-\sqrt{x^{2}+y^{2}}\right)$; outside the circle $f(x, y)=0$. Find: a) the constant $C$; b) the probability that a random point $(X, Y)$ falls into the circle of radius $r=1$ centered at the origin, if $R=2$.
Solution. a) Using the second property of the two-dimensional probability density:

$$
\iint_{(D)} C\left(R-\sqrt{x^{2}+y^{2}}\right) \mathrm{d} x \mathrm{~d} y=1
$$

From this,

$$
C=1 / \int_{(D)}\left(R-\sqrt{\left.x^{2}+y^{2}\right)} \mathrm{d} x \mathrm{~d} y\right.
$$

By switching to polar coordinates, we get

$$
C=1 / \int_{0}^{2 \pi} d \varphi \int_{0}^{R}(R-\rho) \rho \mathrm{d} \rho=3 /\left(\pi R^{2}\right)
$$

b) According to the condition, $R=2$; therefore, $C=3 /(8 \pi)$ and

$$
f(x, y)=(3 / 8 \pi)\left(2-\sqrt{x^{2}+y^{2}}\right)
$$

The probability of a random point $(X, Y)$ falling into a circle of radius $r=1$ centered at the origin (region $D_{1}$)

$$
P\left[(X, Y) \subset D_{1}\right]=(3 / 8 \pi) \iint_{\left(D_{1}\right)}\left(2-\sqrt{x^{2}+y^{2}}\right) d x d y
$$

By switching to polar coordinates, we finally get the desired probability:

$$
P=(3 / 8 \pi) \int_{0}^{2 \pi} d \varphi \int_{0}^{1}(2-\rho) \rho d \rho=1 / 2
$$