The probabilities of solving a problem by three students A, B, and C are respectively $\frac{3}{7}$, $\frac{5}{9}$, and $\frac{1}{5}$. What is the probability that the problem will be solved by at least one of them?

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**Solution:**

1. Given probabilities of students solving the problem:
   - $R(A) = \frac{3}{7}$
   - $R(B) = \frac{5}{9}$
   - $R(C) = \frac{1}{5}$

2. Calculate the probabilities of each student *not* solving the problem:
   - $R(A') = 1 - R(A) = 1 - \frac{3}{7} = \frac{4}{7}$
   - $R(B') = 1 - R(B) = 1 - \frac{5}{9} = \frac{4}{9}$
   - $R(C') = 1 - R(C) = 1 - \frac{1}{5} = \frac{4}{5}$

3. Since the events are independent, the probability that none of them solve the problem is:
   $$P(	ext{none}) = R(A') 	imes R(B') 	imes R(C') = \frac{4}{7} 	imes \frac{4}{9} 	imes \frac{4}{5} = \frac{64}{315}$$

4. The probability that the problem will be solved by at least one student is the complement:
   $$P(	ext{solved}) = 1 - P(	ext{none}) = 1 - \frac{64}{315} = \frac{315 - 64}{315} = \frac{251}{315}$$

**Answer:** $\frac{251}{315}$

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