A point is chosen at random inside a rectangle measuring 6 inches by 5 inches. What is the probability that the randomly selected point is at least one inch from the edge of the rectangle?

Options

A. `(2)/(3)`
B. `(1)/(3)`
C. `(1)/(4)`
D. `(2)/(5)`

Explanation

To solve the problem, we need to find the probability that a randomly selected point inside a rectangle measuring 6 inches by 5 inches is at least 1 inch away from the edge of the rectangle.

### Step-by-Step Solution:

1. **Calculate the Area of the Rectangle**:
- The area of the rectangle can be calculated using the formula:
\[ \text{Area} = \text{Length} \times \text{Breadth} \]
- For our rectangle:
\[ \text{Area} = 6 \, \text{inches} \times 5 \, \text{inches} = 30 \, \text{square inches} \]

2. **Determine the Dimensions of the Inner Rectangle**:
- To find the area of the region that is at least 1 inch away from the edges, we need to create a smaller rectangle inside the original rectangle.
- Since we need to stay 1 inch away from each edge, we subtract 2 inches from both the length and the breadth (1 inch from each side).
- The new dimensions will be:
- Length: \(6 - 2 = 4\) inches
- Breadth: \(5 - 2 = 3\) inches

3. **Calculate the Area of the Inner Rectangle**:
- Now we calculate the area of the inner rectangle:
\[ \text{Inner Area} = 4 \, \text{inches} \times 3 \, \text{inches} = 12 \, \text{square inches} \]

4. **Calculate the Probability**:
- The probability that a randomly selected point is at least 1 inch from the edge of the rectangle is given by the ratio of the area of the inner rectangle to the area of the outer rectangle:
\[ P = \frac{\text{Area of Inner Rectangle}}{\text{Area of Outer Rectangle}} = \frac{12}{30} \]
- Simplifying this fraction:
\[ P = \frac{2}{5} \]

5. **Final Answer**:
- Therefore, the probability that the randomly selected point is at least 1 inch from the edge of the rectangle is:
\[ \boxed{\frac{2}{5}} \]