<Question>
A point $(x,y)$ is randomly selected such that $0 \le x \le 8$ and $0 \le y \le 4$. What is the probability that $x+y \le 4$? Express your answer as a common fraction.
</Question>

<think>
<Plan_1>
Okay, so I need to find the probability that x + y ≤ 4 when a point (x, y) is randomly selected from the region where 0 ≤ x ≤ 8 and 0 ≤ y ≤ 4. Let me start by visualizing the problem.

<prompt_1.1> Visualize the problem by identifying the total region where the point (x, y) can be selected and the area where x + y ≤ 4. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
First, the sample space is all the points (x, y) such that x is between 0 and 8, and y is between 0 and 4. So, this forms a rectangle in the coordinate plane. The total area of this rectangle would be the product of the lengths of the sides. The length along the x-axis is 8 - 0 = 8, and along the y-axis is 4 - 0 = 4. So, the total area is 8 * 4 = 32. That's straightforward.

Now, the event we're interested in is the set of points where x + y ≤ 4. This inequality describes a region in the plane. I need to find the area of the region where both 0 ≤ x ≤ 8, 0 ≤ y ≤ 4, and x + y ≤ 4. Then, the probability will be the ratio of this area to the total area (32).
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Let me sketch the region mentally. The inequality x + y ≤ 4 is a linear inequality. The line x + y = 4 is a straight line. To plot this line, I can find its intercepts.

<prompt_2.1> Determine the region defined by x + y ≤ 4 within the sample space and set up the problem to find the probability. </prompt_2.1>
-----

But let me check if there's another way to approach this problem, maybe using geometry or integration, just to confirm.

<prompt_2.2> Explore alternative methods using geometry or integration to confirm the probability. </prompt_2.2>

Alternatively, if I set up the double integral for the probability, it would be the integral over the region where x + y ≤4, with x from 0 to8 and y from0 to4, divided by the total area.

<prompt_2.3> Set up and evaluate a double integral to confirm the probability. </prompt_2.3>
</Plan_2>

<Execution_2>
<execution_2.1>
When x = 0, y = 4, so that's the point (0, 4). When y = 0, x = 4, so that's the point (4, 0). But wait, our x goes up to 8, but the line x + y = 4 only reaches x = 4 when y = 0. So, in the sample space rectangle (which goes up to x =8 and y=4), the line x + y =4 will start at (0,4) and go down to (4,0). Beyond x =4, the line x + y =4 would require y to be negative, which isn't allowed here since y ≥0. So, the region where x + y ≤4 is the area under the line from (0,4) to (4,0), and below that line within the rectangle.

But wait, since the rectangle extends to x=8 and y=4, but the line x + y =4 only exists in the rectangle from x=0 to x=4, because for x >4, even if y=0, x would already be 5,6,7,8 which are all greater than 4, so x + y would be greater than 4. Therefore, the region x + y ≤4 is a triangle with vertices at (0,0), (0,4), and (4,0).
</execution_2.1>

<execution_2.2>
Another way: Since the problem is about a uniform distribution over the rectangle [0,8]x[0,4], the joint probability density function is constant, specifically 1/(8*4) = 1/32. Therefore, the probability that (x,y) lies in a certain region is just the area of that region multiplied by 1/32. So, again, the probability we want is Area / 32.

So, the key is to compute the area where x + y ≤4, within the rectangle. As established, this area is a right triangle with legs of length 4, so area 8. Therefore, probability is 8/32 = 1/4.
</execution_2.2>

<execution_2.3>
But since the region is a rectangle, and the condition x + y ≤4 is a straight line, the area is a polygon, so the integral would just compute the area of the triangle, which is 1/2 *4*4=8. So 8/32=1/4. So same result.
</execution_2.3>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Wait, is that correct?

Wait, hold on. Let me check again.

<prompt_3.1> Verify the shape and area of the region defined by x + y ≤ 4. </prompt_3.1>
</Plan_3>

<Execution_3>
<execution_3.1>
If x is allowed to go up to 8, but the inequality x + y ≤4 would require y ≤4 -x. However, since y cannot be negative, for x >4, 4 -x would be negative, which is not allowed. Therefore, for x between 0 and 4, y can be between 0 and 4 -x, forming a triangular region. For x from 4 to8, there are no points satisfying x + y ≤4 because even the smallest y (which is 0) would make x + y =x ≥4, which is not ≤4. Therefore, the region is indeed a triangle with vertices at (0,0), (0,4), and (4,0).
</execution_3.1>
</Execution_3>

<Plan_4>
Based on execution_3.1:
But wait, let me confirm that. Suppose x is 0, then y can be up to 4, which is the point (0,4). If x is 1, then y can be up to 3.
So, the line x + y =4 goes from (0,4) to (4,0). So, yes, the region under this line in the first quadrant is a right triangle with base 4 and height 4.

But here's a confusion point: the original x goes up to 8, but the triangle only uses x from 0 to4. So, even though the rectangle is bigger, the region where x + y ≤4 is entirely within the left part of the rectangle, from x=0 to x=4. So, the area is 1/2 *4*4=8.
Therefore, the probability is 8/32=1/4.

But wait, that seems too straightforward. Let me double-check.

Wait, maybe I made a mistake here. Let me visualize again. The inequality x + y ≤4. If x can go up to 8, but for x >4, y would have to be negative to satisfy x + y ≤4, which isn't allowed. Therefore, the region is indeed the triangle in the corner from (0,0) to (0,4) to (4,0). So, area 8. So probability 8/32=1/4.

But let me think again. If the rectangle was from x=0 to x=4 and y=0 to y=4, then the area where x + y ≤4 would be the same triangle, area 8, and total area 16, so probability 1/2. But in this case, the rectangle is larger, x goes up to8, so the total area is 32, hence 8/32=1/4.

Alternatively, perhaps there's a different way to compute the area. Let me set up the integral.

<prompt_4.1> Set up and compute the integral to find the area where x + y ≤ 4. </prompt_4.1>

But let me check once more. Let's take a different approach. Suppose we fix x.

<prompt_4.2> Use a different approach to confirm the probability by fixing x and calculating the corresponding y values. </prompt_4.2>

Alternatively, if I graph the line x + y =4 in the rectangle [0,8]x[0,4], the line connects (0,4) to (4,0).

<prompt_4.3> Graph the line x + y = 4 in the rectangle [0,8]x[0,4] and confirm the area calculation. </prompt_4.3>

Therefore, the probability is 8/32=1/4.

But wait, I'm a bit confused because sometimes when regions are truncated, there might be something else, but in this case, since the line x + y =4 exits the rectangle at (4,0), and beyond that, x cannot be less than4 to satisfy x + y ≤4.

<prompt_4.4> Confirm the probability by considering the truncation of the region and finalize the answer. </prompt_4.4>
</Plan_4>

<Execution_4>
<execution_4.1>
The region where x + y ≤4 is a polygon with vertices at (0,0), (0,4), (4,0). So, the area can be calculated as the integral over x from 0 to4 of (4 -x) dx. Let's compute that:

Area = ∫ from x=0 to4 of (4 - x) dx

= [4x - (x^2)/2] evaluated from 0 to4

= (4*4 - (16)/2) - (0 -0)

= (16 -8) =8.

So, the area is 8, which matches the triangle area calculation. So, 8/32=1/4.
</execution_4.1>

<execution_4.2>
For each x between 0 and8, the possible y values are from0 to4. But the condition x + y ≤4 implies y ≤4 -x. However, if x >4, then 4 -x is negative, so there are no y in [0,4] that satisfy this. Therefore, for x in [0,4], the allowable y is from0 to4 -x, and for x in [4,8], there are no allowable y. Therefore, the area is the integral from x=0 to x=4 of (4 -x) dx.

Calculating that integral:

Area = ∫₀⁴ (4 - x) dx = [4x - (x²)/2] from 0 to4 = (4*4 - (16)/2) - (0 -0) = (16 -8) =8. So, the area is indeed8, which matches the triangle area. Therefore, probability is8/32=1/4.
</execution_4.2>

<execution_4.3>
For each x from 0 to4, the maximum y is 4 -x. For x from4 to8, there's no y that satisfies the inequality because y would have to be negative. Therefore, the area is the integral from x=0 to4 of (4 -x) dx. Let's compute that:

Integral from 0 to4 of (4 -x) dx = [4x - (x^2)/2] evaluated from0 to4 = (4*4 - (16)/2) - (0 -0) = (16 -8) =8. So, area is8. Therefore, probability is8/32=1/4.
</execution_4.3>

<execution_4.4>
So, the region is entirely within the rectangle, so the area is 8, so probability 1/4.

But let me check with another method. For example, suppose we consider x and y as random variables. X is uniformly distributed over [0,8], Y is uniformly distributed over [0,4], and they are independent. Then, the joint distribution is uniform over the rectangle. The probability that x + y ≤4 is the area of the region where x + y ≤4 divided by the total area. As we found, that area is 8, so 8/32=1/4. So, yes, 1/4.

Wait, but let me confirm with another perspective. Suppose I fix y. For a given y, what is the range of x such that x ≤4 - y. Since x must be ≥0. So, for each y, x ranges from0 to4 - y. But y must satisfy4 - y ≥0, so y ≤4. Which is always true here, since y is between0 and4. Therefore, the area can be computed as the integral from y=0 to y=4 of (4 - y) dy.

Calculating that:

Integral from0 to4 of (4 - y) dy = [4y - (1/2)y²] from0 to4 = (4*4 - (1/2)*16) - (0 -0) =16 -8=8. So, area is8. Therefore, probability 8/32=1/4.

Yes, that's another way to compute the area, and it confirms the same result. So, that seems solid.

But just to make sure, let's think of an example. Suppose x and y are selected such that x + y ≤4. If x is 0, y can be up to4. If x is1, y up to3, and so on until x=4, y=0. So, the area is a triangle with base4 and height4, area8. So, yes, 8/32=1/4.
</execution_4.4>
</Execution_4>

<Plan_5>
Based on execution_4.4:
**Final Answer**
The probability that \(x + y \leq 4\) is \boxed{\dfrac{1}{4}}.
</Plan_5>

<Execution_5>
No prompt to execute.
</Execution_5>

<Final_answer>
Therefore, the probability that $x + y \leq 4$ for a randomly selected point $(x,y)$ within the given region is $\boxed{\frac{1}{4}}$.
</Final_answer>
</think>