25. On the surface of a small pond, there are 11 lotus leaves, marked as $0,1, \cdots, 10$. A frog is on the 1st lotus leaf, and the probability of it jumping to the $N(0<N<10)$-marked lotus leaf is equal to the sum of the probability of it being on the $N-1$-marked lotus leaf multiplied by $\frac{N}{10}$ and the probability of it being on the $N+1$-marked lotus leaf multiplied by $\frac{10-N}{10}$. Assume that each jump of the frog is an independent event. If the frog jumps to the 0-marked lotus leaf, it will be eaten by a snake waiting patiently; if the frog jumps to the 10-marked lotus leaf, it can successfully jump out of the pond. What is the probability that the frog will jump out of the pond without being eaten by the snake? ( ).
(A) $\frac{32}{79}$
(B) $\frac{161}{384}$
(C) $\frac{63}{146}$
(D) $\frac{7}{16}$
(E) $\frac{1}{2}$
25. C.

Let the probability that the frog can escape when jumping from the 1st lily pad to the $k$th lily pad be $N_{k}(k=0,1, \cdots, 10)$.

By symmetry, when the frog jumps to the 5th lily pad, the probability of being eaten and the probability of escaping are both $\frac{1}{2}$.
From the problem, we have
$$
\left\{\begin{array}{l}
N_{0}=0, \\
N_{1}=\frac{1}{10} N_{0}+\frac{9}{10} N_{2}, \\
N_{2}=\frac{2}{10} N_{1}+\frac{8}{10} N_{3}, \\
N_{3}=\frac{3}{10} N_{2}+\frac{7}{10} N_{4}, \\
N_{4}=\frac{4}{10} N_{3}+\frac{6}{10} N_{5}, \\
N_{5}=\frac{1}{2} .
\end{array}\right.
$$

Solving these equations, we get $N_{1}=\frac{63}{146}$.
Therefore, the probability that the frog can escape from the 1st lily pad is
$$
N_{1}=\frac{63}{146} \text {. }
$$