5. On the 4 surfaces of a regular tetrahedron, the numbers $1,2,3,4$ are written respectively. Four such uniform regular tetrahedrons are simultaneously thrown onto a table, and the product of the 4 numbers on the 4 surfaces that touch the table can be divided by 4. The probability of this is ( ).
(A) $\frac{1}{8}$
(B) $\frac{1}{16}$
(C) $\frac{13}{16}$
(D) $\frac{9}{64}$
5.C.

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
$$
P_{2}=\mathrm{C}_{4}^{3}\left(\frac{1}{2}\right)^{3} \times \frac{1}{4}=\frac{1}{8} \text {. }
$$

Therefore, the probability that the product of 4 numbers is divisible by 4 is
$$
P=1-P_{1}-P_{2}=\frac{13}{16} \text {. }
$$