8. A math competition problem: the probabilities of A, B, and C solving it alone are $\frac{1}{a}$, $\frac{1}{b}$, and $\frac{1}{c}$, respectively, where $a$, $b$, and $c$ are all single-digit numbers. Now A, B, and C are solving this problem independently at the same time. If the probability that exactly one of them solves the problem is $\frac{7}{15}$, then the probability that none of them solve the problem is $\qquad$
8. $\frac{4}{15}$.

According to the problem, we have
$$
\begin{array}{l}
\frac{1}{a} \cdot \frac{b-1}{b} \cdot \frac{c-1}{c}+\frac{a-1}{a} \cdot \frac{1}{b} \cdot \frac{c-1}{c}+ \\
\frac{a-1}{a} \cdot \frac{b-1}{b} \cdot \frac{1}{c}=\frac{7}{15} \\
\Rightarrow 15[(a-1)(b-1)+(b-1)(c-1) \\
\quad(c-1)(a-1)]=7 a b c
\end{array}
$$
$\Rightarrow 5 \mid a b c$.
Assume without loss of generality that $5 \mid c$. Thus, $c=5$.
Then, $3[(a-1)(b-1)+4(b-1)+$ $4(a-1)]=7 a b$
$$
\Rightarrow 3(3 a+3 b-7)=4 a b \Rightarrow 3 \mid a b \text {. }
$$

Assume without loss of generality that $3 \mid b$. Thus, $b=3,6,9$.
When $b=3$, $a=2$;
When $b=6$, $3 a+11=8 a$, no integer solution;
When $b=9$, $3 a+20=12 a$, no integer solution.
Therefore, $a=2, b=3, c=5$.
Thus, the probability that none of the three people solve the problem is $\frac{4}{15}$.