The probability of a problem being solved by 3 students independently are `(1)/(2), (1)/(3)` and `alpha` respectively. If the probability that the problem is solved in P(S), then P(S) lies in the interval (where, `alpha in (0, 1)`)

Options

A. `(0, (1)/(2))`
B. `((1)/(3),(1)/(2))`
C. `((2)/(3),1)`
D. `((1)/(3),(2)/(3))`

Answer

To solve the problem, we need to find the probability \( P(S) \) that a problem is solved by at least one of the three students, given their individual probabilities of solving the problem. The probabilities of the three students solving the problem are \( \frac{1}{2} \), \( \frac{1}{3} \), and \( \alpha \) respectively.

### Step-by-Step Solution:

1. **Identify the probabilities of not solving the problem**:
- For the first student: The probability of not solving the problem is \( 1 - \frac{1}{2} = \frac{1}{2} \).
- For the second student: The probability of not solving the problem is \( 1 - \frac{1}{3} = \frac{2}{3} \).
- For the third student: The probability of not solving the problem is \( 1 - \alpha \).

2. **Calculate the probability that none of the students solve the problem**:
The probability that none of the students solve the problem is the product of their individual probabilities of not solving it:
\[ P(\text{not solved}) = \left(1 - \frac{1}{2}\right) \times \left(1 - \frac{1}{3}\right) \times (1 - \alpha) = \frac{1}{2} \times \frac{2}{3} \times (1 - \alpha) \]

3. **Simplify the expression**:
\[ P(\text{not solved}) = \frac{1}{2} \times \frac{2}{3} \times (1 - \alpha) = \frac{1}{3} (1 - \alpha) \]

4. **Find the probability that the problem is solved**:
The probability that the problem is solved is the complement of the probability that it is not solved:
\[ P(S) = 1 - P(\text{not solved}) = 1 - \frac{1}{3}(1 - \alpha) \]
\[ P(S) = 1 - \frac{1}{3} + \frac{\alpha}{3} = \frac{2}{3} + \frac{\alpha}{3} \]

5. **Determine the range of \( P(S) \)**:
Since \( \alpha \) lies in the interval \( (0, 1) \):
- When \( \alpha = 0 \):
\[ P(S) = \frac{2}{3} + \frac{0}{3} = \frac{2}{3} \]
- When \( \alpha = 1 \):
\[ P(S) = \frac{2}{3} + \frac{1}{3} = 1 \]

Thus, the probability \( P(S) \) lies in the interval:
\[ \left(\frac{2}{3}, 1\right) \]

### Final Answer:
The probability \( P(S) \) lies in the interval \( \left(\frac{2}{3}, 1\right) \).