A problem is given to three students A, B and C whose probabilities of solving the problem are `(1)/(2),(3)/(4)` and `(1)/(4)` respectively. What is the probability that the problem will be solved if they all slove the problem independently?

Options

A. `(29)/(32)`
B. `(27)/(32)`
C. `(25)/(32)`
D. `(23)/(32)`

Explanation

To solve the problem, we need to find the probability that at least one of the students A, B, or C solves the problem. We can use the complementary probability approach, which involves calculating the probability that none of the students solve the problem and then subtracting that from 1.

### Step-by-Step Solution:

1. **Identify the probabilities of each student solving the problem:**
- Probability that student A solves the problem, \( P(A) = \frac{1}{2} \)
- Probability that student B solves the problem, \( P(B) = \frac{3}{4} \)
- Probability that student C solves the problem, \( P(C) = \frac{1}{4} \)

2. **Calculate the probabilities of each student not solving the problem:**
- Probability that student A does not solve the problem, \( P(A') = 1 - P(A) = 1 - \frac{1}{2} = \frac{1}{2} \)
- Probability that student B does not solve the problem, \( P(B') = 1 - P(B) = 1 - \frac{3}{4} = \frac{1}{4} \)
- Probability that student C does not solve the problem, \( P(C') = 1 - P(C) = 1 - \frac{1}{4} = \frac{3}{4} \)

3. **Calculate the probability that none of the students solve the problem:**
- Since the students solve the problem independently, we can multiply their individual probabilities of not solving the problem:
\[ P(A' \cap B' \cap C') = P(A') \times P(B') \times P(C') = \frac{1}{2} \times \frac{1}{4} \times \frac{3}{4} \]
- Performing the multiplication:
\[ P(A' \cap B' \cap C') = \frac{1}{2} \times \frac{1}{4} \times \frac{3}{4} = \frac{3}{32} \]

4. **Calculate the probability that at least one student solves the problem:**
- Now, we can find the probability that at least one student solves the problem by subtracting the probability that none solve it from 1:
\[ P(\text{at least one solves}) = 1 - P(A' \cap B' \cap C') = 1 - \frac{3}{32} \]
- Performing the subtraction:
\[ P(\text{at least one solves}) = \frac{32}{32} - \frac{3}{32} = \frac{29}{32} \]

5. **Conclusion:**
- The probability that the problem will be solved by at least one of the students is \( \frac{29}{32} \).

### Final Answer:
The probability that the problem will be solved is \( \frac{29}{32} \).