Here's a clear, educational, and well-structured solution to the problem.

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1. Key Concepts and Formulas

• Geometric Probability: For continuous random variables, probability can be found by comparing geometric measures (lengths, areas, or volumes). $P(\text{Event}) = \frac{\text{Measure of Favorable Region}}{\text{Measure of Sample Space}}$
• Absolute Value Inequality: An inequality of the form $|Z| \le K$ is equivalent to $-K \le Z \le K$. This is crucial for defining the region of interest.
• Area of a Square and Triangle: Standard formulas for calculating the areas of these basic geometric shapes will be used.

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2. Step-by-Step Solution

Step 1: Define the Sample Space ($\mathcal{S}$)

• What we are doing: We are choosing two real numbers, $x$ and $y$, independently and uniformly from the interval $[0, 60]$. We represent these pairs as points $(x,y)$ in a 2D Cartesian plane.
• Why this approach: This allows us to visualize all possible outcomes as a geometric region.
• Defining the boundaries: Since $x \in [0, 60]$ and $y \in [0, 60]$, the possible values are constrained by: $0 \le x \le 60$ $0 \le y \le 60$
• Geometric representation: These inequalities define a square region in the $xy$-plane with vertices at $(0,0)$, $(60,0)$, $(60,60)$, and $(0,60)$.
• Calculating the measure (Area): The area of this square represents the total measure of our sample space. $\text{Area}(\mathcal{S}) = (\text{side length})^2 = (60)^2 = 3600 \text{ square units}$ This value will be the denominator in our probability calculation.

Step 2: Define the Event Space ($\mathcal{E}$) for Event A

• What we are doing: Event A is defined by the condition that the absolute difference between the two chosen numbers, $x$ and $y$, is less than or equal to $a$. We need to translate this condition into inequalities that can be graphed.
• Why breaking down absolute value: The absolute value inequality is best handled by converting it into two linear inequalities, which are easier to work with geometrically.
• Translating the condition: The condition for Event A is $|x - y| \le a$. Using the property of absolute value inequalities, this is equivalent to: $-a \le x - y \le a$ This gives us two separate inequalities:
  1. $x - y \le a \implies y \ge x - a$
  2. $x - y \ge -a \implies y \le x + a$
• Combining with sample space boundaries: The event space $\mathcal{E}$ consists of all points $(x, y)$ that satisfy all of the following conditions:
  • $0 \le x \le 60$
  • $0 \le y \le 60$
  • $y \ge x - a$
  • $y \le x + a$

Step 3: Visualize and Calculate the Area of the Event Space

• What we are doing: We will find the area of the region defined by Event A within the sample space square. A common strategy for this type of problem is to calculate the area of the complementary event (where the condition is not met) and subtract it from the total sample space area.
• Why use the complementary event: The region defined by $|x-y| \le a$ is a band around the line $y=x$. Calculating its area directly can be done by considering it as a hexagon, but subtracting the areas of the two corner triangles (the complementary event) is often simpler.
• Complementary Event ($\mathcal{E}^c$): The complementary event occurs when $|x - y| > a$, which means either $x - y > a$ or $x - y < -a$.
  • $y < x - a$ (Region below the line $y = x - a$)
  • $y > x + a$ (Region above the line $y = x + a$)
• Condition for 'a': For the probability to be between 0 and 1 (exclusive), $a$ must be greater than 0 and less than 60. If $a=0$, $P(A)=0$. If $a \ge 60$, $P(A)=1$. So, we assume $0 < a < 60$.
• Area of the complementary region:
  1. Region 1 ($y > x + a$): This region, bounded by $y=x+a$, $x=0$, and $y=60$, forms a right-angled triangle in the upper-left corner of the sample space.
     • The line $y = x + a$ intersects $x=0$ at $(0, a)$ and $y=60$ at $(60-a, 60)$.
     • The vertices of this triangle are $(0, a)$, $(0, 60)$, and $(60-a, 60)$.
     • Its base length (along $x=0$) is $60-a$. Its height (along $y=60$) is $60-a$.
     • Area of this triangle $= \frac{1}{2} \times (60-a) \times (60-a) = \frac{1}{2}(60-a)^2$.
  2. Region 2 ($y < x - a$): This region, bounded by $y=x-a$, $y=0$, and $x=60$, forms another right-angled triangle in the lower-right corner.
     • The line $y = x - a$ intersects $y=0$ at $(a, 0)$ and $x=60$ at $(60, 60-a)$.
     • The vertices of this triangle are $(a, 0)$, $(60, 0)$, and $(60, 60-a)$.
     • Its base length (along $y=0$) is $60-a$. Its height (along $x=60$) is $60-a$.
     • Area of this triangle $= \frac{1}{2} \times (60-a) \times (60-a) = \frac{1}{2}(60-a)^2$.
  • Total Area of Complementary Event: The sum of the areas of these two triangles is: $\text{Area}(\mathcal{E}^c) = \frac{1}{2}(60-a)^2 + \frac{1}{2}(60-a)^2 = (60-a)^2$
• Area of the Favorable Region: Now, we subtract the unfavorable area from the total sample space area: $\text{Area}(\mathcal{E}) = \text{Area}(\mathcal{S}) - \text{Area}(\mathcal{E}^c)$ $\text{Area}(\mathcal{E}) = 3600 - (60-a)^2$

Step 4: Calculate the Probability and Solve for 'a'

• What we are doing: We are given $P(A) = \frac{11}{36}$. We will use the geometric probability formula to set up an equation and solve for $a$.
• Why this setup: This directly links the given probability to our geometric model, allowing us to find the unknown parameter 'a'. Using the geometric probability formula: $P(A) = \frac{\text{Area}(\mathcal{E})}{\text{Area}(\mathcal{S})}$ Substitute the calculated areas and the given probability: $\frac{11}{36} = \frac{3600 - (60-a)^2}{3600}$ To solve for $a$, multiply both sides by 3600: $11 \times \frac{3600}{36} = 3600 - (60-a)^2$ $11 \times 100 = 3600 - (60-a)^2$ $1100 = 3600 - (60-a)^2$ Rearrange the equation to isolate $(60-a)^2$: $(60-a)^2 = 3600 - 1100$ $(60-a)^2 = 2500$ Take the square root of both sides. Remember that $\sqrt{X^2} = |X|$: $|60-a| = \sqrt{2500}$ $|60-a| = 50$ This absolute value equation gives two possible scenarios:
  1. $60 - a = 50$ $a = 60 - 50$ $a = 10$
  2. $60 - a = -50$ $a = 60 + 50$ $a = 110$
• Why check the solutions: The parameter $a$ represents a maximum absolute difference between two numbers chosen from $[0,60]$. The maximum possible difference is $60-0=60$. Therefore, $a$ must be in the range $0 \le a \le 60$.
  • $a = 10$ is within the valid range.
  • $a = 110$ is outside the valid range ($110 > 60$). This solution is not physically meaningful for this problem.

Therefore, the only valid value for $a$ is $10$.

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3. Common Mistakes & Tips

• Incorrect Absolute Value Expansion: A common mistake is to incorrectly expand $|x-y| \le a$ as just $x-y \le a$. Always remember it implies two inequalities: $-a \le x-y \le a$.
• Overlooking the Range of 'a': Always check if your calculated value of 'a' makes sense in the context of the problem. For differences like $|x-y| \le a$ within an interval $[0, L]$, $a$ must be between $0$ and $L$.
• Calculation Errors: Be careful with arithmetic, especially when squaring or taking square roots. Double-check your algebra.

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4. Summary

We used geometric probability to solve this problem. The sample space was represented by a $60 \times 60$ square with an area of $3600$. The event space, defined by $|x-y| \le a$, was found by subtracting the area of its complementary event (two right-angled triangles) from the total sample space area, resulting in $3600 - (60-a)^2$. By equating the ratio of the event space area to the sample space area with the given probability $\frac{11}{36}$, we set up an equation. Solving this equation yielded two possible values for $a$, $10$ and $110$. Considering the valid range for $a$ (i.e., $0 \le a \le 60$), we concluded that $a=10$.

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The final answer is $\boxed{10}$.