4. There is a math competition problem, the probabilities of A, B, and C solving it alone are $\frac{1}{a} 、 \frac{1}{b} 、 \frac{1}{c}$, respectively, where $a 、 b 、 c$ are all positive integers in the range of 1 to 9. 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 solves the problem is $\qquad$
4. $\frac{4}{15}$.

According to the problem, we have
$$
\frac{1}{a} \cdot \frac{b-1}{b} \cdot \frac{c-1}{c}+\frac{a-1}{a} \cdot \frac{1}{b} \cdot \frac{c-1}{c}+\frac{a-1}{a} \cdot \frac{b-1}{b} \cdot \frac{1}{c}=\frac{7}{15},
$$

which simplifies to $15[(a-1)(b-1)+(b-1)(c-1)+(c-1)(a-1)]=7 a b c$.
Thus, $5 \mid a b c$.
Without loss of generality, assume $5 \mid c$, so $c=5$. Then,
$$
3[(a-1)(b-1)+4(b-1)+4(a-1)]=7 a b,
$$

which simplifies to $3(3 a+3 b-7)=4 a b$.
Thus, $3 \mid a b$. Without loss of generality, assume $3 \mid b$, so $b=3,6,9$.
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, the only solution is $a=2, b=3, c=5$.
Thus, the probability that none of the three people solve the problem is $\frac{4}{15}$.