25. Given that the six vertices of a centrally symmetric hexagon in the complex plane form the complex number set
$$
V=\left\{ \pm \sqrt{2} \mathrm{i}, \frac{ \pm 1+\mathrm{i}}{\sqrt{8}}, \frac{ \pm 1-\mathrm{i}}{\sqrt{8}}\right\} .
$$

For each $j(1 \leqslant j \leqslant 12), z_{j}$ is an element randomly selected from the set $V$, and the selections are independent. Let $P=\prod_{j=1}^{12} z_{j}$, then the probability that $P=-1$ is ( ).
(A) $\frac{5 \times 11}{3^{10}}$
(B) $\frac{5^{2} \times 11}{2 \times 3^{10}}$
(C) $\frac{5 \times 11}{3^{9}}$
(D) $\frac{5 \times 7 \times 11}{2 \times 3^{10}}$
(E) $\frac{2^{2} \times 5 \times 11}{3^{10}}$
25. E.

Replace $z_{1}$ with $-z_{1}$, then $P=-1$ becomes $P=1$, and this mapping is a one-to-one correspondence. Therefore, calculate the number of ways for $P= \pm 1$, and then divide by 2.

Multiplying each vertex of the hexagon by $\mathrm{i}$ does not change the value of $P$. Replacing $z_{j}$ with $-z_{j}$ does not change the relationship $P= \pm 1$. Thus, the probability of selecting 12 elements such that the product is -1 is equal to selecting 12 elements from the following set:
$$
\left\{a=\sqrt{2}, b=\frac{1+\mathrm{i}}{\sqrt{8}}, c=\frac{1-\mathrm{i}}{\sqrt{8}}\right\} .
$$

Since $|a|=\sqrt{2}$ and $|b|=|c|=\frac{1}{2}$, $a$ must be chosen exactly eight times to make $|P|=1$.

To make $P$ a real number, $b$ and $c$ must each be chosen four times or each exactly twice. Therefore, the number of ways to satisfy $P= \pm 1$ is
$$
2 \mathrm{C}_{12}^{4}+\mathrm{C}_{12}^{2} \mathrm{C}_{10}^{2}=2^{3} \times 3^{2} \times 5 \times 11 .
$$

Thus, the required probability is
$$
\frac{1}{2} \times \frac{2^{3} \times 3^{2} \times 5 \times 11}{3^{12}}=\frac{2^{2} \times 5 \times 11}{3^{10}} .
$$