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: \(\frac{30}{121}\)

Answer 2

To solve this problem, we need to find the probability that the absolute difference between two integers \(x\) and \(y\) picked with replacement from the set \(\{0, 1, 2, \ldots, 10\}\) is greater than 5. The total number of possible outcomes when picking two integers from this set with replacement is 121, because there are 11 choices for \(x\) and 11 choices for \(y\), giving us \(11 \times 11 = 121\) total outcomes.

Now, we need to find the number of favorable outcomes where \(|x - y| > 5\). This inequality can be written as two conditions:

• \(x - y > 5\)
• \(y - x > 5\), which is equivalent to \(x - y < -5\)

Let's analyze these conditions:

1. For \(x - y > 5\):
   This means \(x > y + 5\). Let's count the possibilities for each value of \(y\):

y	Possible x values (x > y + 5)	Count
0	6, 7, 8, 9, 10	5
1	7, 8, 9, 10	4
2	8, 9, 10	3
3	9, 10	2
4	10	1

Total count for this condition: \(5 + 4 + 3 + 2 + 1 = 15\).

1. For \(x - y < -5\):
   This means \(y > x + 5\). Let’s count the possibilities for each value of \(x\):

x	Possible y values (y > x + 5)	Count
0	6, 7, 8, 9, 10	5
1	7, 8, 9, 10	4
2	8, 9, 10	3
3	9, 10	2
4	10	1

Total count for this condition: \(5 + 4 + 3 + 2 + 1 = 15\).

Adding the counts for both conditions, we have \(15 + 15 = 30\) favorable outcomes.

The probability is thus given by: \(\frac{30}{121}\)

The correct option is \( \frac{30}{121} \).