An airplane has 4 rows of seats with 3 seats in each row. 8 passengers have boarded the plane and are distributed randomly among the seats. A married couple is next to board. What is the probability there will be 2 adjacent seats in the same row for the couple?

We are given that an airplane has 4 rows of seats, with 3 seats in each row, resulting in a total of \(4 \times 3 = 12\) seats. Out of these 12 seats, 8 passengers have already boarded, so there are \(12 - 8 = 4\) empty seats remaining.

We need to find the probability that the married couple will find 2 adjacent seats in the same row.

### Step 1: Total number of ways to seat the couple

The couple must occupy two of the remaining four empty seats. The total number of ways to choose 2 seats from 4 is given by the combination formula:
\[
\text{Total ways} = \binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6
\]

### Step 2: Number of favorable outcomes (adjacent seats in the same row)

The couple can only sit together in the same row if there are two adjacent seats available. Let's consider how many adjacent pairs are possible given the seating arrangement.

Each row has 3 seats, which gives 2 possible pairs of adjacent seats in each row (for example, seats 1 and 2, and seats 2 and 3). Since there are 4 rows, the total number of possible adjacent pairs in the airplane is:
\[
2 \times 4 = 8 \text{ adjacent pairs}.
\]

However, we are only interested in the adjacent pairs that are unoccupied. Since there are 4 empty seats and 8 seats occupied randomly, we will calculate how many adjacent pairs of seats are still unoccupied.

### Step 3: Probability calculation

We can break this problem down further based on which seats are still unoccupied and compute accordingly or move forward based on

Solve the probability problem where a married couple is looking for 2 adjacent seats in a partially filled airplane. Learn how combinatorics and basic probability are used to calculate the chances of finding adjacent seats out of 4 remaining empty seats. Suitable for Grades 10-12.

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