Okay, so I need to find the probability that a randomly selected point (x, y) within the rectangle defined by 0 ≤ x ≤ 8 and 0 ≤ y ≤ 4 satisfies the inequality x + y ≤ 4. Let me start by visualizing the problem. 

First, the sample space is the entire area where the point can be chosen. Since x ranges from 0 to 8 and y from 0 to 4, the area of the sample space is 8 multiplied by 4, which is 32. So, the total area is 32. 

Now, the event we're interested in is all the points (x, y) where x + y ≤ 4. This inequality represents a region in the coordinate plane. To find the area of this region, I need to graph the line x + y = 4 and determine the area below it within the given rectangle.

Let me sketch the line x + y = 4. When x is 0, y is 4, and when y is 0, x is 4. However, since the rectangle only goes up to x = 8 and y = 4, the line x + y = 4 will intersect the rectangle at two points. Wait, let's check that. If x can be up to 8, but the line x + y = 4 would have x-intercept at (4, 0) and y-intercept at (0, 4). But the rectangle's boundaries are x from 0 to 8 and y from 0 to 4. So, the line x + y = 4 will intersect the rectangle at (0, 4) and (4, 0). 

But wait, the rectangle's height is up to y = 4, so the line x + y = 4 will start at (0, 4) and go down to (4, 0). However, since the rectangle's x goes up to 8, but the line x + y = 4 only reaches x = 4 when y = 0. So, the region x + y ≤ 4 within the rectangle is a polygon with vertices at (0, 0), (0, 4), (4, 0), and back to (0, 0). Wait, actually, no. Let me think again. 

If we have the line x + y = 4, then for x between 0 and 4, y can be from 0 up to 4 - x. For x from 4 to 8, the inequality x + y ≤ 4 would require y ≤ 4 - x. But since x is already greater than 4, 4 - x would be negative, and since y is non-negative, the inequality y ≤ 4 - x would only be satisfied if y = 0. But since y is between 0 and 4, for x from 4 to 8, the only points that satisfy x + y ≤ 4 would be along the x-axis from x = 4 to x = 8 with y = 0. But since the line x + y = 4 at x = 4 is at y = 0, so beyond x = 4, the inequality x + y ≤ 4 would not hold because y can't be negative. Therefore, the region x + y ≤ 4 is a triangle with vertices at (0, 4), (0, 0), and (4, 0). 

Wait, but if I consider the rectangle from x = 0 to 8 and y = 0 to 4, then the line x + y = 4 cuts through the rectangle at (0, 4) and (4, 0). So the region x + y ≤ 4 within the rectangle is a triangle with vertices at (0, 0), (0, 4), and (4, 0). That makes sense because for x from 0 to 4, y can go from 0 up to 4 - x, which is the area under the line. For x from 4 to 8, as I said earlier, y would have to be less than or equal to a negative number, which isn't possible since y is non-negative. So, the area where x + y ≤ 4 is indeed a triangle.

The area of this triangle can be calculated as half the base times the height. The base is from (0, 4) to (4, 0), which is a diagonal, but in terms of base and height for the triangle, the base can be considered as 4 units along the x-axis from (0, 0) to (4, 0), and the height is 4 units along the y-axis from (0, 0) to (0, 4). So, the area is (1/2)*base*height = (1/2)*4*4 = 8. 

Therefore, the area of the event region is 8. The total area of the sample space is 32, so the probability is 8/32, which simplifies to 1/4. 

Wait, but let me verify this again. If the triangle has vertices at (0, 0), (0, 4), and (4, 0), then yes, that's a right triangle with legs of length 4 each. The area should indeed be (1/2)*4*4 = 8. So the probability is 8/32 = 1/4. 

But wait, let me check if there's another way to calculate this area. Maybe using integration? For x from 0 to 4, y goes from 0 to 4 - x. So integrating over x from 0 to 4, the integral of (4 - x) dx. Let's compute that:

∫ from 0 to 4 of (4 - x) dx = [4x - (1/2)x²] from 0 to 4 = (16 - 8) - (0 - 0) = 8. 

So that's another way to get the area as 8. So that confirms the area is 8. 

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

But wait, another thought: the rectangle is 8 units in x and 4 units in y. The line x + y = 4 cuts through the rectangle at (0,4) and (4,0). The region x + y ≤ 4 is a triangle with area 8. But what if the line x + y = 4 was outside the rectangle somewhere else? For example, if the rectangle was larger in one dimension, but in this case, since the rectangle's x goes up to 8 and y up to 4, the line x + y = 4 only intersects at (0,4) and (4,0), so the region is entirely within the rectangle. 

Therefore, the area is 8, sample area is 32, probability is 1/4. 

But let me check once more with a different approach. Suppose we consider the entire rectangle. The region x + y ≤ 4 is a polygon. Since x can go up to 8, but beyond x=4, the inequality x + y ≤4 can't be satisfied because y would have to be negative. So, for x between 0 and 4, y can be from 0 to 4 - x. For x between 4 and 8, there's no y in [0,4] that satisfies x + y ≤4. Therefore, the area is indeed the integral from x=0 to x=4 of (4 - x) dx, which is 8. 

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

So, I think that's the answer. But just to make sure I didn't make any miscalculations. Let me compute the integral again:

∫₀⁴ (4 - x) dx = [4x - (x²)/2] from 0 to 4. Plugging in 4: 4*4 - (16)/2 = 16 - 8 = 8. Plugging in 0: 0 - 0 = 0. So the integral is 8. 

Yes, that's correct. So, the area is 8. Total area is 8*4=32. 8/32=1/4. 

Alternatively, if I use geometric probability, the region is a triangle with base 4 and height 4, area 8, so probability 8/32=1/4. 

Therefore, the probability should be 1/4. 

I don't see any mistakes in this reasoning. So, I think that's the answer.
Final answer: \boxed{\dfrac{1}{4}}
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