410. The distribution function of a two-dimensional random variable is given by

$$
F(x, y)=\left\{\begin{array}{c}
\sin x \cdot \sin y \text { for } 0 \leqslant x \leqslant \pi / 2,0 \leqslant y \leqslant \pi / 2 \\
0 \quad \text { for } x<0 \text { or } y<0
\end{array}\right.
$$

Find the probability that a random point $(X, Y)$ falls within the rectangle bounded by the lines $x=0, x=\pi / 4$, $\boldsymbol{y}=\pi / 6, y=\pi / 3$.
Solution. We use the formula

$$
\begin{gathered}
P\left(x_{1}<X<x_{2}, y_{1}<Y<y_{2}\right)=\left[F\left(x_{2}, y_{2}\right)-F\left(x_{1}, y_{2}\right)\right]- \\
-\left[F\left(x_{2}, y_{1}\right)-F\left(x_{1}, y_{1}\right)\right] .
\end{gathered}
$$

Setting $x_{1}=0, x_{2}=\pi / 4, y_{1}=\pi / 6, y_{2}=\pi / 3$, we get

$$
P=[\sin (\pi / 4) \cdot \sin (\pi / 3)-\sin 0 \cdot \sin (\pi / 3)]-
$$

$-[\sin (\pi ; 4) \cdot \sin (\pi / 6)-\sin 0 \cdot \sin (\pi ; 6)]=(\sqrt{6}-\sqrt{2}) / 4=0.26$.