To solve this, we need to find the probability that a person is sitting either in an odd-numbered row or an odd-numbered seat.

Step 1: Total number of seats

There are 20 rows, each with 15 seats, so the total number of seats is:

$\text{Total seats} = 20 \times 15 = 300$

Step 2: Counting the favorable outcomes

1. Odd-numbered rows:

• The odd-numbered rows are: 1, 3, 5, ..., 19.
• There are 10 odd-numbered rows (half of 20).
• Each row has 15 seats, so the total number of seats in odd-numbered rows is:

$10 \times 15 = 150$

2. Odd-numbered seats:

• The odd-numbered seats in each row are: 1, 3, 5, ..., 15.
• There are 8 odd-numbered seats in each row (since $1, 3, 5, \dots, 15$ is a set of 8 numbers).
• Across all 20 rows, the total number of odd-numbered seats is:

$20 \times 8 = 160$

3. Seats counted twice (odd-numbered rows and odd-numbered seats):

• For seats that are both in odd-numbered rows and odd-numbered seats, we need to count how many odd-numbered seats are in the odd-numbered rows.
• There are 8 odd-numbered seats in each odd-numbered row, and there are 10 odd-numbered rows. So, the number of seats that fall in both categories is:

$10 \times 8 = 80$

Step 3: Applying the inclusion-exclusion principle

To find the number of seats that are either in odd-numbered rows or odd-numbered seats, we apply the inclusion-exclusion principle:

$\text{Favorable outcomes} = (\text{Odd rows}) + (\text{Odd seats}) - (\text{Both odd rows and odd seats})$

$\text{Favorable outcomes} = 150 + 160 - 80 = 230$

Step 4: Probability calculation

The probability is the ratio of favorable outcomes to total outcomes:

$\text{Probability} = \frac{230}{300} = \frac{23}{30}$

Thus, the probability that a person chosen is