Let A be the event that the absolute difference between two randomly chosen real numbers in the samp

Q: Why can't we just use integration to find the area?

You absolutely can! Integration is a powerful tool. However, in geometric probability problems involving linear constraints, the region is often a simple polygon. Using geometry (area of triangles) is significantly faster and less prone to calculation errors than setting up double integrals. Both methods will yield the same result, but geometry builds better intuition.

Q: Why is the area of the excluded triangles $\frac{1}{2}(60 - a)^2$?

Great question. The line $y = x + a$ intersects the $y$-axis at $(0, a)$. The top edge of the square is $y = 60$. The vertical height of the triangle is $60 - a$. Since the line has a slope of $1$, the horizontal base is also $60 - a$. Thus, the area is $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2}(60 - a)^2$.

Q: What if $a$ was larger than $60$?

If $a \geq 60$, the condition $|x - y| \leq a$ would cover the entire square because the maximum possible difference between any two numbers in $[0, 60]$ is $60$. In that case, the probability would be $1$. Our calculation assumes $a < 60$, which is consistent with the given probability $\frac{11}{36}$.

Question

Let $A$ be the event that the absolute difference between two randomly chosen real numbers in the sample space $[0, 60]$ is less than or equal to $a$. If $P(A) = \frac{11}{36}$, then $a$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, square field. This field is defined by two coordinates, $x$ and $y$, both ranging from $0$ to $60$.

The total area of this square is:

$$3600 = 60 	imes 60$$

This represents our universe of possibilities.

### The Strip of Truth

We introduce a constraint: the absolute difference between these two numbers must be small, specifically $|x - y| \leq a$.

This inequality implies that $x - y$ is trapped between $-a$ and $a$. Rearranging this, we obtain two parallel lines:

$$y \leq x + a$$
$$y \geq x - a$$

These lines slice through our square, creating a diagonal strip. Any point $(x, y)$ that falls within this strip satisfies our condition and represents a favorable outcome.

### The Complementary Shortcut

Calculating the area of this strip directly can be cumbersome. Instead, we look at the corners of the square that are *not* covered by this strip.

These regions form two identical, right-angled triangles. By calculating the area of these two triangles and subtracting them from the total area, we find the favorable region efficiently.

### Unveiling the Dimensions

Consider the top-left triangle. The line $y = x + a$ intersects the $y$-axis at $a$. Since the top boundary of our square is $y = 60$, the vertical side of this triangle is $60 - a$.

Because the line has a slope of $1$, the horizontal side is also $60 - a$. The area of one such triangle is:

$$	ext{Area} = \frac{1}{2}(60 - a)^2$$

Since we have two such triangles (one at the top-left and one at the bottom-right), the total excluded area is:

$$2 	imes \frac{1}{2}(60 - a)^2 = (60 - a)^2$$

### The Final Algebraic Dance

The probability $P(A)$ is the ratio of the favorable area to the total area. Our favorable area is $3600 - (60 - a)^2$.

The probability equation is:

$$P(A) = \frac{3600 - (60 - a)^2}{3600}$$

Given that $P(A) = \frac{11}{36}$, we set up the following equality:

$$1 - \frac{(60 - a)^2}{3600} = \frac{11}{36}$$

Subtracting $1$ from both sides and rearranging, we find:

$$\frac{(60 - a)^2}{3600} = 1 - \frac{11}{36} = \frac{25}{36}$$

Multiplying by $3600$, we obtain:

$$(60 - a)^2 = 2500$$

Taking the square root, we get $60 - a = 50$. This leads us directly to the final result:

**$a = 10$**

Animated Solution for Mathematics - Probability: Let $A$ be the event that the absolute difference between two randomly chosen real numbers in the sample space $[0, 60]$ is less than or equal to $a$ . If $P (A) = \frac{11}{36}$ , then $a$ is equal to

Enter Numerical Value:

Visualized Solution

Defining the Sample Space

The Event Condition

Visualizing the Event Region

The Complementary Approach

Dimensions of the Triangles

Area of Excluded Triangles

Formulating the Probability

Equating to Given Probability

Simplifying the Equation

Rearranging the Terms

Solving for the Squared Term

Finding the Value of $a$

The Sigma Insight: Classical Definition of Probability

Analyzing the Setup

The Strip of Truth

The Complementary Shortcut

Unveiling the Dimensions

The Final Algebraic Dance

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