Question

Two integers \( x \) and \( y \) are chosen with replacement from the set \( \{0, 1, 2, 3, \ldots, 10\} \). Then the probability that \( |x - y| > 5 \) is:

• A. \( \frac{30}{121} \)
• B. \( \frac{62}{121} \)
• C. \( \frac{60}{121} \)
• D. \( \frac{31}{121} \)

Answer 1

The Correct Option is A

The total number of outcomes when choosing \( x \) and \( y \) with replacement from the set \(\{0, 1, 2, \dots, 10\}\) is:
\[11 \times 11 = 121\]
To satisfy \(|x - y| > 5\), we need \(x - y > 5\) or \(x - y < -5\). We count the favorable pairs by analyzing each possible value of \( x \):
If \( x = 0 \), \( y \) can be 6, 7, 8, 9, 10 (5 values)
If \( x = 1 \), \( y \) can be 7, 8, 9, 10 (4 values)
If \( x = 2 \), \( y \) can be 8, 9, 10 (3 values)
If \( x = 3 \), \( y \) can be 9, 10 (2 values)
If \( x = 4 \), \( y \) can be 10 (1 value)
If \( x = 5 \), there are no possible values of \( y \)
If \( x = 6 \), \( y = 0 \) (1 value)
If \( x = 7 \), \( y = 0, 1 \) (2 values)
If \( x = 8 \), \( y = 0, 1, 2 \) (3 values)
If \( x = 9 \), \( y = 0, 1, 2, 3 \) (4 values)
If \( x = 10 \), \( y = 0, 1, 2, 3, 4 \) (5 values)
Adding these values, the total number of favorable outcomes is:
\[5 + 4 + 3 + 2 + 1 + 1 + 1 + 2 + 3 + 4 + 5 = 30\]
The required probability is:
\[\frac{30}{121}\]
Final Answer: \(\frac{30}{121}\)