1. The 4 faces of a regular tetrahedron are labeled with $1,2,3,4$. Four such uniform regular tetrahedrons are simultaneously 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. $\frac{13}{16}$
1. D Event “4 numbers are all odd” has a probability $P_{1}=\left(\frac{1}{2}\right)^{4}=\frac{1}{16}$; Event “3 are odd, 1 is 2” has a probability $P_{2}=C 1 \cdot \frac{1}{4} \cdot\left(\frac{1}{2}\right)^{3}=\frac{1}{8}$. Therefore, the probability that the product of 4 numbers is divisible by 4 is $P=1-P_{1}-P_{2}=\frac{13}{16}$.