Key Concepts and Formulas

• Classical Probability: The probability of an event $E$ is defined as the ratio of the number of outcomes favorable to the event to the total number of equally likely possible outcomes in the sample space $S$. $P(E) = \frac{\text{Number of Favorable Outcomes (n(E))}}{\text{Total Number of Possible Outcomes (n(S))}}$
• Fundamental Principle of Counting: If an event consists of a sequence of $k$ independent choices, where the first choice has $n_1$ options, the second has $n_2$ options, and so on, then the total number of possible outcomes for the event is $n_1 \times n_2 \times \ldots \times n_k$.
• Absolute Value Inequality: The inequality $|A| > k$ (where $k > 0$) is equivalent to the disjunction ($A > k$ or $A < -k$). This property is essential for correctly interpreting the condition given in the problem.

Step-by-Step Solution

Step 1: Determine the Total Number of Possible Outcomes (Sample Space)

Our first step is to establish the complete set of all possible pairs $(x,y)$ that can be chosen. This forms our sample space, $S$.

1. Identify the Source Set: The integers $x$ and $y$ are chosen from the set $S_{base} = \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$.
2. Count Elements in the Source Set: To find the number of distinct integers in $S_{base}$, we use the formula (last element - first element + 1). $\text{Number of elements in } S_{base} = 10 - 0 + 1 = 11$
3. Account for "With Replacement": The problem states that $x$ and $y$ are chosen with replacement. This means that the choice of $x$ does not affect the possible choices for $y$, and $x$ and $y$ can be the same value.
   • Since there are 11 choices for $x$, $x$ can take any value from $\{0, 1, \ldots, 10\}$.
   • Similarly, there are 11 choices for $y$, and $y$ can take any value from $\{0, 1, \ldots, 10\}$.
4. Calculate Total Outcomes: Using the Fundamental Principle of Counting, the total number of distinct ordered pairs $(x,y)$ is the product of the number of choices for $x$ and $y$. $\text{Total Number of Possible Outcomes (n(S))} = 11 \times 11 = 121$

Step 2: Determine the Number of Favorable Outcomes

We are interested in the event $E$ where the absolute difference between $x$ and $y$ is greater than 5. This is expressed as $|x-y| > 5$.

Using the property of absolute value inequalities, $|x-y| > 5$ can be broken down into two distinct and mutually exclusive conditions:

1. $x-y > 5$
2. $x-y < -5$ (which is equivalent to $y-x > 5$)

We will count the number of pairs $(x,y)$ for each condition separately.

Sub-step 2.1: Count outcomes for the condition $x-y > 5$

This condition can be rewritten as $x \ge y+6$. We need to find all pairs $(x,y)$ from $S_{base} \times S_{base}$ that satisfy this. Let's systematically list the possibilities by iterating through possible values of $y$ from $0$ to $10$.

• If $y=0$: $x$ must be $\ge 0+6=6$. Possible $x$ values from $S_{base}$ are $\{6, 7, 8, 9, 10\}$. This gives 5 pairs: $(6,0), (7,0), (8,0), (9,0), (10,0)$.
• If $y=1$: $x$ must be $\ge 1+6=7$. Possible $x$ values from $S_{base}$ are $\{7, 8, 9, 10\}$. This gives 4 pairs: $(7,1), (8,1), (9,1), (10,1)$.
• If $y=2$: $x$ must be $\ge 2+6=8$. Possible $x$ values from $S_{base}$ are $\{8, 9, 10\}$. This gives 3 pairs: $(8,2), (9,2), (10,2)$.
• If $y=3$: $x$ must be $\ge 3+6=9$. Possible $x$ values from $S_{base}$ are $\{9, 10\}$. This gives 2 pairs: $(9,3), (10,3)$.
• If $y=4$: $x$ must be $\ge 4+6=10$. Possible $x$ values from $S_{base}$ are $\{10\}$. This gives 1 pair: $(10,4)$.
• If $y=5$: $x$ must be $\ge 5+6=11$. There are no values $\ge 11$ in $S_{base}=\{0, \ldots, 10\}$. This gives 0 pairs.
• For any $y > 5$, the required value for $x$ would be even larger (e.g., if $y=6$, $x \ge 12$), so there will be no more pairs.

The number of favorable outcomes for $x-y > 5$ is the sum of these counts: $\text{Number of pairs for } (x-y > 5) = 5 + 4 + 3 + 2 + 1 = 15$

Sub-step 2.2: Count outcomes for the condition $y-x > 5$

This condition can be rewritten as $y \ge x+6$. This case is perfectly symmetric to the previous one, with the roles of $x$ and $y$ simply swapped. We can leverage this symmetry to quickly determine the count.

• If $x=0$: $y$ must be $\ge 0+6=6$. Possible $y$ values from $S_{base}$ are $\{6, 7, 8, 9, 10\}$. This gives 5 pairs: $(0,6), (0,7), (0,8), (0,9), (0,10)$.
• If $x=1$: $y$ must be $\ge 1+6=7$. Possible $y$ values from $S_{base}$ are $\{7, 8, 9, 10\}$. This gives 4 pairs: $(1,7), (1,8), (1,9), (1,10)$.
• If $x=2$: $y$ must be $\ge 2+6=8$. Possible $y$ values from $S_{base}$ are $\{8, 9, 10\}$. This gives 3 pairs: $(2,8), (2,9), (2,10)$.
• If $x=3$: $y$ must be $\ge 3+6=9$. Possible $y$ values from $S_{base}$ are $\{9, 10\}$. This gives 2 pairs: $(3,9), (3,10)$.
• If $x=4$: $y$ must be $\ge 4+6=10$. Possible $y$ values from $S_{base}$ are $\{10\}$. This gives 1 pair: $(4,10)$.
• For any $x \ge 5$, the required value for $y$ would be $\ge 11$, which is not in $S_{base}$. This gives 0 pairs.

The number of favorable outcomes for $y-x > 5$ is: $\text{Number of pairs for } (y-x > 5) = 5 + 4 + 3 + 2 + 1 = 15$

Sub-step 2.3: Combine all Favorable Outcomes

Since the two conditions ($x-y>5$ and $y-x>5$) are mutually exclusive (an ordered difference cannot be both greater than 5 and less than -5 simultaneously), we simply add the counts from both cases to get the total number of favorable outcomes for event $E$. $\text{Total Number of Favorable Outcomes (n(E))} = 15 + 15 = 30$

Step 3: Calculate the Probability

Finally, we apply the classical probability formula using the total number of outcomes from Step 1 and the total number of favorable outcomes from Step 2. $P(|x-y|>5) = \frac{\text{Number of Favorable Outcomes (n(E))}}{\text{Total Number of Possible Outcomes (n(S))}}$ $P(|x-y|>5) = \frac{30}{121}$

Common Mistakes & Tips

• Counting the Set Size: Always be careful when counting elements in a set that includes 0. The set $\{0, 1, \ldots, n\}$ has $n+1$ elements, not $n$. Here, $\{0, \ldots, 10\}$ has $10-0+1=11$ elements.
• Understanding "With Replacement": This phrase is crucial. If choices were "without replacement," the total number of outcomes would be $11 \times 10 = 110$, as $x$ and $y$ could not be the same.
• Absolute Value Interpretation: A common error is to only consider $x-y > 5$ and forget the $x-y < -5$ (or $y-x > 5$) case. This would lead to half the correct number of favorable outcomes.
• Leveraging Symmetry: For conditions involving absolute differences like $|x-y|$, there is often symmetry. Once you've counted outcomes for $x-y > k$, the count for $y-x > k$ is usually the same. This saves time and provides a good cross-check.

Summary

This problem is a fundamental application of classical probability. The solution involved two main parts: first, correctly determining the total number of possible ordered pairs $(x,y)$ by understanding the "with replacement" condition and accurately counting the elements in the given set. Second, we meticulously counted the number of favorable outcomes by carefully breaking down the absolute value inequality $|x-y|>5$ into two mutually exclusive conditions ($x-y>5$ and $y-x>5$). We systematically listed and summed the pairs for each condition, then combined them. Finally, the probability was calculated using the ratio of favorable outcomes to total outcomes.

The final answer is $\boxed{\frac{30}{121}}$, which corresponds to option (A).