Let there are 4 sections of 25 students each in a coaching class. Now, out of 150 students 100 are to be selected randomly and enrolled in these sections. Then, the probability that the students A and B (both present in 150 students) are selected and placed in the same section, is

Options

A. `(10)/(13)`
B. `(16)/(149)`
C. `(2)/(33)`
D. `(5)/(149)`

Answer

To solve the problem, we need to find the probability that two specific students, A and B, are selected from a total of 150 students and placed in the same section of 25 students out of 4 sections.

### Step-by-Step Solution:

1. **Total Students and Sections**: We have 150 students in total and 4 sections, each containing 25 students.

2. **Choosing Students**: We need to select 100 students out of the 150. However, since we are specifically interested in A and B being in the same section, we will first consider the scenario where A and B are already selected.

3. **Selecting A and B**: If A and B are selected, we have already chosen 2 students. Therefore, we need to select 23 more students to fill one section of 25 students.

4. **Remaining Students**: After selecting A and B, there are 148 students left (150 - 2 = 148).

5. **Choosing Remaining Students**: The number of ways to choose the remaining 23 students from the 148 students is given by the combination formula:
\[ \binom{148}{23} \]

6. **Total Ways to Choose 25 Students**: The total number of ways to choose any 25 students from the original 150 is:
\[ \binom{150}{25} \]

7. **Calculating Probability**: The probability that A and B are selected and placed in the same section can be expressed as:
\[ P(A \text{ and } B \text{ in same section}) = \frac{\binom{148}{23}}{\binom{150}{25}} \]

8. **Simplifying the Probability**:
- We can express the combinations in terms of factorials:
\[ \binom{148}{23} = \frac{148!}{23! \cdot (148-23)!} = \frac{148!}{23! \cdot 125!} \]
\[ \binom{150}{25} = \frac{150!}{25! \cdot (150-25)!} = \frac{150!}{25! \cdot 125!} \]

- Thus, the probability becomes:
\[ P(A \text{ and } B \text{ in same section}) = \frac{\frac{148!}{23! \cdot 125!}}{\frac{150!}{25! \cdot 125!}} = \frac{148! \cdot 25!}{150! \cdot 23!} \]

9. **Cancelling Factorials**: Simplifying further:
- We can cancel \(148!\) and \(125!\):
\[ P(A \text{ and } B \text{ in same section}) = \frac{25!}{150 \times 149 \times 148!} = \frac{25 \times 24}{150 \times 149} \]

10. **Calculating Final Probability**:
- This simplifies to:
\[ P(A \text{ and } B \text{ in same section}) = \frac{4}{149} \]

11. **Considering All Sections**: Since there are 4 sections, we multiply the probability by 4:
\[ P(\text{A and B in any section}) = 4 \times \frac{4}{149} = \frac{16}{149} \]

### Final Answer:
Thus, the probability that students A and B are selected and placed in the same section is:
\[ \frac{16}{149} \]