3. On the 4 faces of a regular tetrahedron, the numbers $1, 2, 3,$ and $4$ are written. Four such uniform regular tetrahedra are thrown onto a table, and the product of the 4 numbers on the 4 faces that touch the table is divisible by 4. The probability of this happening is ( ).
(A) $\frac{1}{8}$
(B) $\frac{9}{64}$
(C) $\frac{1}{16}$
(D) $\frac{13}{16}$
3.D.

The probability of the event "all 4 numbers are odd" is
$$
P_{1}=\left(\frac{1}{2}\right)^{4}=\frac{1}{16} \text {; }
$$

The probability of the event "3 numbers are odd, 1 number is 2" is
$$
\begin{array}{l}
P_{2}=\mathrm{C}_{4}^{1} \cdot \frac{1}{4} \cdot\left(\frac{1}{2}\right)^{3}=\frac{1}{8} . \\
\text { Therefore, } P=1-P_{1}-P_{2}=\frac{13}{16} .
\end{array}
$$