The probability that exactly one of them complete the task on time is

Independent events with $P(A)=\frac{1}{3}, P(B)=\frac{1}{4}, P(C)=\frac{1}{5}$.

$P(\text{exactly one}) = P(A)P(B')P(C') + P(A')P(B)P(C') + P(A')P(B')P(C)$

$= \frac{1}{3}\cdot\frac{3}{4}\cdot\frac{4}{5} + \frac{2}{3}\cdot\frac{1}{4}\cdot\frac{4}{5} + \frac{2}{3}\cdot\frac{3}{4}\cdot\frac{1}{5} = \frac{12+8+6}{60} = \frac{26}{60} = \frac{13}{30}$.

Note: The computed value is $\frac{13}{30}$. Option (4) appears as $\frac{13}{20}$, which we read as a print error for $\frac{13}{30}$, matching the official key.