11. A math competition problem: the probabilities of A, B, and C solving this problem independently are $\frac{1}{a}$, $\frac{1}{b}$, and $\frac{1}{c}$, respectively, where $a$, $b$, and $c$ are positive integers less than 10. Now A, B, and C are solving this problem independently. If the probability that exactly one of them solves the problem is $\frac{7}{15}$, then the probability that none of them solves the problem is
11. $\frac{4}{15}$.

Notice,
$P$ (exactly one person solves the problem)
$$
\begin{aligned}
= & P(\text { only Jia solves })+P(\text { only Yi solves })+P(\text { only Bing solves }) \\
= & \frac{(b-1)(c-1)}{a b c}+\frac{(a-1)(c-1)}{a b c}+ \\
& \frac{(a-1)(b-1)}{a b c} \\
= & \frac{7}{15} .
\end{aligned}
$$

Thus, $15|7 a b c \Rightarrow 15| a b c \Rightarrow 5 \mid a b c$
$\Rightarrow 5 \mid a$ or $5 \mid b$ or $5 \mid c$.
Without loss of generality, assume $5 \mid a$.
Since $0<a<10$, then, $a=5$.
Thus, equation (1) becomes
$$
\begin{array}{l}
\frac{b c-b-c+1+4 c-4+4 b-4}{b c}=\frac{7}{3} \\
\Rightarrow(4 b-9)(4 c-9)=-3 \\
\Rightarrow(4 b-9) \mid 3 \Rightarrow b=2 \text { or } 3 .
\end{array}
$$

When $b=2$, $c=3$; when $b=3$, $c=2$.
Therefore, $a, b, c$ are a permutation of $2, 3, 5$.
Hence, the required value is $\frac{a-1}{a} \cdot \frac{b-1}{b} \cdot \frac{c-1}{c}=\frac{4}{15}$.