Consider the Cartesian plane `R^(2)` and let `X` denote the subset of points for which both coordinates are integer. A coin of diameter `1//2` is tossed randomly onto the plane. The probability `p` that the coin covers a point of `X`

Options

A. `0.2`
B. `0.8`
C. `1.2`
D. None of these

Explanation

To solve the problem, we need to find the probability \( p \) that a randomly tossed coin of diameter \( \frac{1}{2} \) covers a point in the set \( X \) of integer coordinates on the Cartesian plane \( \mathbb{R}^2 \).

### Step-by-Step Solution:

1. **Understanding the Coin's Coverage**:
The coin has a diameter of \( \frac{1}{2} \), which means its radius \( r \) is:
\[ r = \frac{1}{2} \div 2 = \frac{1}{4} \]

2. **Area of the Coin**:
The area \( A \) of the coin can be calculated using the formula for the area of a circle:
\[ A = \pi r^2 = \pi \left(\frac{1}{4}\right)^2 = \pi \cdot \frac{1}{16} = \frac{\pi}{16} \]

3. **Understanding the Grid**:
The integer points \( X \) are located at the intersections of a grid formed by the integer coordinates. Each square in this grid has a side length of \( 1 \), giving it an area of:
\[ \text{Area of one square} = 1 \times 1 = 1 \]

4. **Probability Calculation**:
The probability \( p \) that the coin covers at least one integer point is given by the ratio of the area of the coin to the area of the square grid (considering the effective area where the center of the coin can land):
\[ p = \frac{\text{Area of the coin}}{\text{Area of one square}} = \frac{\frac{\pi}{16}}{1} = \frac{\pi}{16} \]

5. **Numerical Approximation**:
To express this probability numerically, we can use the approximation \( \pi \approx 3.14 \):
\[ p \approx \frac{3.14}{16} \approx 0.19625 \]

6. **Final Result**:
Thus, the probability \( p \) that the coin covers a point of \( X \) is approximately:
\[ p \approx 0.19625 \quad \text{or} \quad p \approx 0.2 \]