The probability that $A, B$ and $C$ can solve a problem are $\displaystyle \frac{4}{5}, \frac{2}{3}$ and $\dfrac{3}{7}$ respectively. The probability of the problem being solved by $A$ and $B$ is $\dfrac{8}{15}$, $B$ and $C$ is $\dfrac{2}{7}$, $A$ and $C$ is $\dfrac{12}{35}$. The probability of the problem being solved by all the three is $\dfrac{8}{35}$.

Find the probability that the problem can be solved by at least one of them.

$\begin{array}{l}P(A)=4 / 5, P(B)=213, P(C)=3 / 7, \\ P(A \cap B)=8 / 15, P(B \cap C)=217, \\ P(A \cap C)=12 / 35, P(A \cap B \cap C)=8 / 35\end{array}$

$\begin{aligned} P(\text { at least one })= & P(A \cup B \cup C) \\ = & P(A)+P(B)+P(C)-P(A \cap B) -P(B \cap C)-P(A \cap C)+P(A \cap B \cap C) \\ = & \frac{4}{5}+\frac{2}{3}+\frac{3}{7}-\frac{8}{15}-\frac{2}{7}-\frac{12}{35} +\frac{8}{35}\end{aligned}$

$\begin{array}{l}=\frac{84+70+45-56-30-36+24}{105} \\ =\frac{101}{105}\end{array}$