There are 7 seats in a row. Three persons take seats at random the probability that the middle seat is always occupied and no two persons are consecutive is

Options

A. `8/35`
B. `4/35`
C. `4/(.^(7)C_(3))`
D. `(.^(4)C_(1))/(.^(7)C_(3))`

Answer

To solve the problem, we need to find the probability that the middle seat (4th seat) is occupied and no two persons are seated consecutively when three persons take seats at random from a row of 7 seats.

### Step-by-Step Solution:

1. **Identify the total number of seats and the middle seat**:
- There are 7 seats in total, numbered from 1 to 7.
- The middle seat is seat number 4.

2. **Fix the middle seat**:
- Since the middle seat (seat 4) must be occupied, we place one person there. This leaves us with 6 remaining seats (1, 2, 3, 5, 6, 7) for the other two persons.

3. **Determine the available seats for the remaining persons**:
- After placing one person in seat 4, we need to ensure that the other two persons do not sit in consecutive seats. The remaining seats are 1, 2, 3, 5, 6, and 7.

4. **Count the arrangements**:
- We can visualize the remaining seats as follows:
- If we denote the occupied seats as "P" (for persons) and the unoccupied seats as "O", we can represent the arrangement as:
- O O O P O O O
- We need to place 2 persons (P) in the remaining 6 seats (1, 2, 3, 5, 6, 7) such that no two P's are adjacent.

5. **Calculate the number of ways to place the remaining persons**:
- To ensure that no two persons are seated next to each other, we can think of placing the persons in the available gaps created by the unoccupied seats.
- The arrangement of the remaining seats can be represented as:
- O O O O O O (6 gaps)
- We need to choose 2 gaps from these 6 to place the persons. The number of ways to choose 2 gaps from 6 is given by the combination formula:
\[ \text{Ways} = \binom{n}{r} = \binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15 \]

6. **Calculate the total arrangements of persons**:
- The total number of ways to arrange 3 persons in 7 seats without any restrictions is given by:
\[ \text{Total arrangements} = \binom{7}{3} \times 3! = \frac{7!}{3!(7-3)!} \times 6 = 35 \times 6 = 210 \]

7. **Calculate the probability**:
- The probability that the middle seat is occupied and no two persons are seated consecutively is given by the ratio of favorable outcomes to total outcomes:
\[ P = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{15}{210} = \frac{1}{14} \]

### Final Answer:
The probability that the middle seat is always occupied and no two persons are consecutive is \( \frac{1}{14} \).