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:
\[11 \times 11 = 121\]
The condition \(|x - y| > 5\) implies \(x - y > 5\) or \(x - y < -5\). We enumerate the favorable pairs based on the value of \( x \):
For \( x = 0 \), \( y \in \{6, 7, 8, 9, 10\} \) (5 pairs)
For \( x = 1 \), \( y \in \{7, 8, 9, 10\} \) (4 pairs)
For \( x = 2 \), \( y \in \{8, 9, 10\} \) (3 pairs)
For \( x = 3 \), \( y \in \{9, 10\} \) (2 pairs)
For \( x = 4 \), \( y \in \{10\} \) (1 pair)
For \( x = 5 \), no \( y \) satisfies the condition.
For \( x = 6 \), \( y = 0 \) (1 pair)
For \( x = 7 \), \( y \in \{0, 1\} \) (2 pairs)
For \( x = 8 \), \( y \in \{0, 1, 2\} \) (3 pairs)
For \( x = 9 \), \( y \in \{0, 1, 2, 3\} \) (4 pairs)
For \( x = 10 \), \( y \in \{0, 1, 2, 3, 4\} \) (5 pairs)
The sum of these favorable outcomes is:
\[5 + 4 + 3 + 2 + 1 + 1 + 1 + 2 + 3 + 4 + 5 = 30\]
The probability is:
\[\frac{30}{121}\]
Final Answer: \(\frac{30}{121}\)