Calculating Probability: |x - y| > 5

The problem asks for the probability that the absolute difference between two integers, x and y, is greater than 5 when they are chosen with replacement from the set S = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.

Understanding the Sample Space

The set S contains 11 integers (from 0 to 10, inclusive). Since x and y are chosen with replacement, the choice of x does not affect the choice of y, and both can be any value from the set S.

The total number of possible ordered pairs (x, y) is the number of elements in S multiplied by itself:

\(\text{Total Outcomes} = \text{|S|} \times \text{|S|} = 11 \times 11 = 121\).

Identifying Favorable Outcomes

We are interested in the event where \(\vert x - y \vert > 5\). The absolute difference \(\vert x - y \vert\) being greater than 5 means the difference between x and y is at least 6 in magnitude. This condition is satisfied if either \(x - y > 5\) or \(y - x > 5\).

Since x and y are integers, \(x - y > 5\) is equivalent to \(x - y \ge 6\), and \(y - x > 5\) is equivalent to \(y - x \ge 6\).

Let's find the pairs (x, y) that satisfy these conditions:

Case 1: \(x - y \ge 6\)

In this case, x must be at least 6 greater than y. We can list the possible pairs by fixing y and finding the possible values of x:

• If \(y = 0\), then \(x \ge 6\). Possible x values from S are {6, 7, 8, 9, 10}. This gives 5 pairs: (6,0), (7,0), (8,0), (9,0), (10,0).
• If \(y = 1\), then \(x \ge 7\). Possible x values from S are {7, 8, 9, 10}. This gives 4 pairs: (7,1), (8,1), (9,1), (10,1).
• If \(y = 2\), then \(x \ge 8\). Possible x values from S are {8, 9, 10}. This gives 3 pairs: (8,2), (9,2), (10,2).
• If \(y = 3\), then \(x \ge 9\). Possible x values from S are {9, 10}. This gives 2 pairs: (9,3), (10,3).
• If \(y = 4\), then \(x \ge 10\). Possible x values from S are {10}. This gives 1 pair: (10,4).
• If \(y \ge 5\), there are no possible x values in S such that \(x \ge y + 6\), since the maximum value for x is 10 (e.g., if \(y=5\), \(x \ge 11\), which is not in S).

The total number of favorable outcomes for \(x - y \ge 6\) is \(5 + 4 + 3 + 2 + 1 = 15\).

Case 2: \(y - x \ge 6\)

In this case, y must be at least 6 greater than x. This situation is symmetric to Case 1. We can list the possible pairs by fixing x and finding the possible values of y:

• If \(x = 0\), then \(y \ge 6\). Possible y values from S are {6, 7, 8, 9, 10}. This gives 5 pairs: (0,6), (0,7), (0,8), (0,9), (0,10).
• If \(x = 1\), then \(y \ge 7\). Possible y values from S are {7, 8, 9, 10}. This gives 4 pairs: (1,7), (1,8), (1,9), (1,10).
• If \(x = 2\), then \(y \ge 8\). Possible y values from S are {8, 9, 10}. This gives 3 pairs: (2,8), (2,9), (2,10).
• If \(x = 3\), then \(y \ge 9\). Possible y values from S are {9, 10}. This gives 2 pairs: (3,9), (3,10).
• If \(x = 4\), then \(y \ge 10\). Possible y values from S are {10}. This gives 1 pair: (4,10).
• If \(x \ge 5\), there are no possible y values in S such that \(y \ge x + 6\), since the maximum value for y is 10 (e.g., if \(x=5\), \(y \ge 11\), which is not in S).

The total number of favorable outcomes for \(y - x \ge 6\) is \(5 + 4 + 3 + 2 + 1 = 15\).

The two conditions, \(x - y \ge 6\) and \(y - x \ge 6\), represent mutually exclusive events (a pair like (8,1) satisfies the first but not the second, and (1,8) satisfies the second but not the first). Therefore, the total number of favorable outcomes for \(\vert x - y \vert > 5\) is the sum of the outcomes from Case 1 and Case 2.

\(\text{Total Favorable Outcomes} = 15 (\text{for } x - y \ge 6) + 15 (\text{for } y - x \ge 6) = 30\).

Calculating the Probability

The probability of the event \(\vert x - y \vert > 5\) is calculated by dividing the number of favorable outcomes by the total number of possible outcomes:

\(\text{Probability} = \frac{\text{Total Favorable Outcomes}}{\text{Total Outcomes}}\)

\(\text{Probability} = \frac{30}{121}\)

Thus, the probability that \(\vert x - y \vert > 5\) when x and y are chosen with replacement from {0, 1, ..., 10} is \(\frac{30}{121}\).

Revision Table: Probability Concepts

Additional Information: Discrete Uniform Probability

This problem is an example of discrete uniform probability because every possible outcome (each pair (x,y)) in the sample space is equally likely. The set of possible outcomes is discrete (a finite set of integer pairs).

In general, for a discrete uniform probability space, if there are \(N\) total possible outcomes and \(M\) favorable outcomes for an event E, the probability of event E is given by:

\(P(E) = \frac{M}{N}\)

For this question:

• The sample space consists of all ordered pairs \((x, y)\) where \(x, y \in \{0, 1, \dots, 10\}\).
• Total outcomes \(N = 11 \times 11 = 121\).
• The event E is \(\vert x - y \vert > 5\), which is equivalent to \((x - y \ge 6) \text{ or } (y - x \ge 6)\).
• The number of favorable outcomes \(M = 30\), calculated by summing the counts for \(x - y \ge 6\) and \(y - x \ge 6\).
• The probability is \(P(E) = \frac{30}{121}\).

This method of listing favorable pairs based on the condition is a direct way to solve such discrete probability problems.