7. Zheng rolls four dice. The probability that the product of the four numbers is divisible by 6 is $\qquad$
7. $\frac{325}{432}$.

Let the product of the four numbers be $m$, i.e., find $P(61 m)$.
$$
\begin{array}{l}
\text { Then } P(2 \nmid m)=\left(\frac{1}{2}\right)^{4}=\frac{1}{16} \\
\Rightarrow P(2 \mid m)=\frac{15}{16}, \\
P(3 \nmid m)=\left(\frac{2}{3}\right)^{4}=\frac{16}{81} \Rightarrow P(3 \mid m)=\frac{65}{81}, \\
P(2 \nmid m \text { and } 3 \nmid m)=\left(\frac{1}{3}\right)^{4}=\frac{1}{81} \\
\Rightarrow P(2 \mid m \text { or } 3 \mid m)=\frac{80}{81} .
\end{array}
$$

Therefore, $P(6 \mid m)=P(2 \mid m$ and $3 \mid m)$ $=P(2 \mid m)+P(3 \mid m)-P(2 \mid m$ or $31 m)$ $=\frac{15}{16}+\frac{65}{81}-\frac{80}{81}=\frac{325}{432}$.