Two integers x and y are chosen with replacement from the set {0, 1, 2, 3, ..., 10}. Then the probab

Q: Wait, why is the total number of outcomes $11 \times 11$ and not $11 \times 10$?

That is a great question! The phrase 'with replacement' is the key here. It means that after we pick the first integer $x$, we put it back into the set. So, for the second pick $y$, we still have all $11$ options available. If it were 'without replacement', then yes, we would have $11 \times 10$ outcomes.

Q: Why do we exclude the case where $|x - y| = 5$?

The problem explicitly asks for the probability that $|x - y| > 5$. This is a strict inequality. If the distance between $x$ and $y$ is exactly $5$, it does not satisfy the condition of being *greater than* $5$. Always pay close attention to whether the inequality is strict ($>$) or non-strict ($\ge$).

Q: Can I just use the formula for the number of solutions to an inequality?

You certainly can! For $x - y > 5$, you are looking for integer solutions to $x - y \ge 6$. You could use stars and bars or combinatorial methods, but for a small set like this, systematic counting is often faster and less prone to off-by-one errors. The geometric approach we used is very robust for these types of problems.

Question

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

Options:

(A) $\frac{30}{121}$

(B) $\frac{62}{121}$

(C) $\frac{60}{121}$

(D) $\frac{31}{121}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a vast, empty grid. On both the horizontal and vertical axes, we have the integers from zero to ten.

This is our sample space, a collection of $11 	imes 11$ points, each representing a unique pair $(x, y)$. Because we are choosing with replacement, every single point on this $121$-point grid is equally likely to be our outcome.

The total number of outcomes is simply:
$$11 	imes 11 = 121$$

Now, we introduce a constraint: $|x - y| > 5$. This is not just an algebraic expression; it is a geometric boundary. It carves out a 'forbidden zone' where the difference is small, and two 'favorable zones' where the difference is large.

### Breaking the Modulus

The absolute value $|x - y| > 5$ splits our world into two distinct realities: either $x - y > 5$ or $y - x > 5$. Let's focus on the first case: $x > y + 5$.

If we visualize this on our grid, it represents the points lying strictly below the line $y = x - 5$. We can count these systematically:

*   If $y = 0$, then $x > 5$, meaning $x \in \{6, 7, 8, 9, 10\}$. That is $5$ points.
*   If $y = 1$, then $x > 6$, meaning $x \in \{7, 8, 9, 10\}$. That is $4$ points.
*   If $y = 2$, then $x \in \{8, 9, 10\}$. That is $3$ points.
*   If $y = 3$, then $x \in \{9, 10\}$. That is $2$ points.
*   If $y = 4$, then $x = 10$. That is $1$ point.

If $y$ were $5$, $x$ would have to be greater than $10$, which is impossible in our set. Summing these up, we get:
$$5 + 4 + 3 + 2 + 1 = 15$$

### The Elegance of Symmetry

Now, we could repeat this entire process for the second case, $y > x + 5$. But here is where we can be clever.

The condition $y > x + 5$ is just a reflection of the first case across the diagonal line $y = x$. Because our set is symmetric, the number of favorable outcomes in this upper region must be exactly the same as in the lower region.

We do not need to count again; we simply recognize the symmetry. Thus, the second case also yields $15$ favorable outcomes.

### The Final Synthesis

We have $15$ outcomes from the first case and $15$ from the second. Since these two regions do not overlap—they are mutually exclusive—we simply add them together to get $30$ favorable outcomes.

The probability is the ratio of favorable outcomes to the total sample space:
$$P = \frac{30}{121}$$

It is a beautiful result, born from the simple act of visualizing the grid and respecting the symmetry of the numbers. You have navigated the constraints, avoided the traps, and arrived at the **final answer of $\frac{30}{121}$** with mathematical precision.

Animated Solution for Mathematics - Probability: Two integers $x$ and $y$ are chosen with replacement from the set ${0, 1, 2, 3, \dots, 10}$ . Then the probability that $∣ x - y ∣ > 5$ is :

Select Answer:

Visualized Solution

Defining the Sample Space

Total Outcomes

Analyzing the Condition

Splitting the Modulus

Counting Case 1 ( $y = 0$ )

Counting Case 1 ( $y = 1, 2$ )

Counting Case 1 ( $y = 3, 4$ )

Using Symmetry for Case 2

Total Favorable Outcomes

Final Probability Calculation

The Sigma Insight: Classical Definition of Probability

Analyzing the Setup

Breaking the Modulus

The Elegance of Symmetry

The Final Synthesis

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Animated Solution for Mathematics - Probability: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Two integers x and y are chosen with replacement from the set {0,1,2,3,…,10}. Then the probability that ∣x−y∣>5 is :