If `x, in {0, 1, 2, 3,....10}` then the probability that `|x -y| > 5` is

Options

A. `frac{30}{121}`
B. `frac{31}{121}`
C. `frac{60}{121}`
D. `frac{62}{121}`

Answer

To solve the problem of finding the probability that \(|x - y| > 5\) where \(x\) and \(y\) can take values from the set \(\{0, 1, 2, 3, \ldots, 10\}\), we can follow these steps:

### Step 1: Determine the total number of outcomes
Since both \(x\) and \(y\) can take any of the 11 values (from 0 to 10), the total number of possible pairs \((x, y)\) is:
\[ \text{Total outcomes} = 11 \times 11 = 121 \]

**Hint:** Count the number of choices for \(x\) and \(y\) independently, then multiply to find the total combinations.

### Step 2: Identify the pairs where \(|x - y| > 5\)
To satisfy the condition \(|x - y| > 5\), we can break it down into two cases:
1. \(x - y > 5\)
2. \(y - x > 5\)

#### Case 1: \(x - y > 5\)
This implies:
\[ x > y + 5 \]
For each value of \(x\), we can find the corresponding values of \(y\):
- If \(x = 6\), then \(y\) can be \(0\) (1 way)
- If \(x = 7\), then \(y\) can be \(0, 1\) (2 ways)
- If \(x = 8\), then \(y\) can be \(0, 1, 2\) (3 ways)
- If \(x = 9\), then \(y\) can be \(0, 1, 2, 3\) (4 ways)
- If \(x = 10\), then \(y\) can be \(0, 1, 2, 3, 4\) (5 ways)

Adding these gives:
\[ 1 + 2 + 3 + 4 + 5 = 15 \text{ ways} \]

#### Case 2: \(y - x > 5\)
This implies:
\[ y > x + 5 \]
For each value of \(x\), we can find the corresponding values of \(y\):
- If \(x = 0\), then \(y\) can be \(6, 7, 8, 9, 10\) (5 ways)
- If \(x = 1\), then \(y\) can be \(7, 8, 9, 10\) (4 ways)
- If \(x = 2\), then \(y\) can be \(8, 9, 10\) (3 ways)
- If \(x = 3\), then \(y\) can be \(9, 10\) (2 ways)
- If \(x = 4\), then \(y\) can be \(10\) (1 way)
- If \(x = 5\), then \(y\) cannot be greater than \(10\) (0 ways)

Adding these gives:
\[ 5 + 4 + 3 + 2 + 1 = 15 \text{ ways} \]

### Step 3: Total favorable outcomes
Now, we sum the favorable outcomes from both cases:
\[ \text{Total favorable outcomes} = 15 + 15 = 30 \]

### Step 4: Calculate the probability
The probability that \(|x - y| > 5\) is given by the ratio of favorable outcomes to total outcomes:
\[ P(|x - y| > 5) = \frac{\text{Total favorable outcomes}}{\text{Total outcomes}} = \frac{30}{121} \]

### Final Answer
Thus, the probability that \(|x - y| > 5\) is:
\[ \frac{30}{121} \]

---