Set S is the set of all points (x, y) in the coordinate plane such that x and y both integers with absolute value less than 4. If one of these points is chosen at random, what is the probability that this point will be 2 units or less from the origin ?

Options

A. `0.189`
B. `0.227`
C. `0.265`
D. `0.314`

Explanation

To solve the problem, we need to follow these steps:

### Step 1: Define the Set S
Set S consists of all points (x, y) in the coordinate plane where both x and y are integers with absolute values less than 4. This means:
- x can take values from -3 to 3 (i.e., -3, -2, -1, 0, 1, 2, 3)
- y can also take values from -3 to 3.

### Step 2: Count the Total Points in Set S
Since both x and y can take 7 different integer values (-3, -2, -1, 0, 1, 2, 3), the total number of points in set S is:
\[ \text{Total points} = 7 \times 7 = 49 \]

### Step 3: Determine Points Within 2 Units from the Origin
We need to find the points (x, y) that are 2 units or less from the origin (0, 0). The distance from the origin is given by the formula:
\[ \sqrt{x^2 + y^2} \leq 2 \]
Squaring both sides, we get:
\[ x^2 + y^2 \leq 4 \]

### Step 4: Identify Integer Points Satisfying the Condition
We will check integer combinations of (x, y) that satisfy \( x^2 + y^2 \leq 4 \):
- For \( x = 0 \): \( y^2 \leq 4 \) → y can be -2, -1, 0, 1, 2 (5 points)
- For \( x = 1 \): \( 1 + y^2 \leq 4 \) → \( y^2 \leq 3 \) → y can be -1, 0, 1 (3 points)
- For \( x = -1 \): Same as above (3 points)
- For \( x = 2 \): \( 4 + y^2 \leq 4 \) → \( y^2 \leq 0 \) → y can only be 0 (1 point)
- For \( x = -2 \): Same as above (1 point)
- For \( x = 3 \) or \( x = -3 \): \( 9 + y^2 \leq 4 \) → No valid points.

### Step 5: Count the Valid Points
Now, we can count the valid points:
- From \( x = 0 \): 5 points
- From \( x = 1 \): 3 points
- From \( x = -1 \): 3 points
- From \( x = 2 \): 1 point
- From \( x = -2 \): 1 point

Adding these gives:
\[ 5 + 3 + 3 + 1 + 1 = 13 \text{ points} \]

### Step 6: Calculate the Probability
The probability that a randomly chosen point from set S is 2 units or less from the origin is given by:
\[ \text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{13}{49} \]

### Final Answer
Thus, the probability that a randomly chosen point will be 2 units or less from the origin is:
\[ \frac{13}{49} \]

---