3. A number $x$ is randomly chosen from $[0,10]$, and a number $y$ is randomly chosen from $[0,6]$. The probability that
$$
|x-5|+|y-3| \leqslant 4
$$

is ( ).
(A) $\frac{1}{5}$
(B) $\frac{1}{3}$
(C) $\frac{1}{2}$
(D) $\frac{3}{4}$
3. C.

As shown in Figure 2, the graph of $|x-5|+|y-3| \leqslant 4$ is a square $B N E M$ and its interior, centered at $G(5,3)$. Suppose $B M$ intersects $R Q$ at point $A$. It is easy to see that $A(4,6)$. Similarly, $F(6,6)$.
Thus, the required probability is
$$
p=\frac{S_{\text {hexagon } A B C D E F}}{S_{\text {rectangle } O P Q R}}=\frac{32-2}{60}=\frac{1}{2} .
$$