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 for the three students are \( \frac{1}{2} \), \( \frac{1}{3} \), and \( \alpha \) respectively, where \( \alpha \) is in the interval \( (0, 1) \).

### Step-by-Step Solution:

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

2. **Calculate the probabilities of each student NOT solving the problem:**
- \( P(A') = 1 - P(A) = 1 - \frac{1}{2} = \frac{1}{2} \)
- \( P(B') = 1 - P(B) = 1 - \frac{1}{3} = \frac{2}{3} \)
- \( P(C') = 1 - P(C) = 1 - \alpha \)

3. **Calculate the probability that none of the students solve the problem:**
- Since the students solve the problem independently, the probability that none of them solve it is:
\[ P(S') = P(A') \times P(B') \times P(C') = \frac{1}{2} \times \frac{2}{3} \times (1 - \alpha) \]

4. **Simplify the expression for \( P(S') \):**
- Calculate \( P(S') \):
\[ P(S') = \frac{1}{2} \times \frac{2}{3} \times (1 - \alpha) = \frac{1}{3} \times (1 - \alpha) \]

5. **Calculate the probability that at least one student solves the problem:**
- The probability that at least one student solves the problem is:
\[ P(S) = 1 - P(S') = 1 - \left(\frac{1}{3} \times (1 - \alpha)\right) \]
- Simplifying this gives:
\[ P(S) = 1 - \frac{1 - \alpha}{3} = 1 - \frac{1}{3} + \frac{\alpha}{3} = \frac{2}{3} + \frac{\alpha}{3} \]
- Thus, we can express \( P(S) \) as:
\[ P(S) = \frac{2 + \alpha}{3} \]

6. **Determine the interval for \( P(S) \):**
- Since \( \alpha \) is in the interval \( (0, 1) \):
- When \( \alpha = 0 \):
\[ P(S) = \frac{2 + 0}{3} = \frac{2}{3} \]
- When \( \alpha = 1 \):
\[ P(S) = \frac{2 + 1}{3} = 1 \]
- Therefore, \( P(S) \) lies in the interval:
\[ \left( \frac{2}{3}, 1 \right) \]

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