Explanation:
Okay, so I need to find the probability that three randomly chosen points from a 3x3 grid lie on the same straight line. Let me start by understanding the problem.

First, the grid is 3x3, which means there are 9 points in total. The question says a set of three points is chosen randomly, and each set has the same probability. So, the total number of possible three-point sets is the combination of 9 points taken 3 at a time. I remember that combinations are calculated using the formula C(n, k) = n! / (k! * (n - k)!).

Let me calculate that first. So n is 9, and k is 3. Plugging in:

C(9, 3) = 9! / (3! * 6!) = (9 × 8 × 7) / (3 × 2 × 1) = 84. So there are 84 possible ways to choose three points. That's the denominator of my probability.

Now, I need to find the number of favorable outcomes, which are the sets of three points that lie on a straight line. Then, the probability will be favorable / total, which is favorable / 84.

So, how do I count the number of such lines?

First, I should consider all possible lines that can be formed by three points on the grid. The grid is 3x3, so there are horizontal lines, vertical lines, and diagonal lines. Also, maybe there are other lines with different slopes that pass through three points? Let me check.

Starting with horizontal lines. Each row of the grid is a horizontal line. Since there are 3 rows, each containing 3 points, so there are 3 horizontal lines. Each of these lines has exactly 3 points, so each row contributes 1 set of three colinear points. Therefore, horizontal lines contribute 3 favorable sets.

Similarly, vertical lines. Each column is a vertical line. Again, there are 3 columns, each with three points. So another 3 favorable sets from vertical lines.

Then, the main diagonals. There are two main diagonals: from top-left to bottom-right and from top-right to bottom-left. Each of these diagonals has exactly three points. So that's 2 more favorable sets.

So so far, horizontal (3) + vertical (3) + diagonal (2) = 8. But wait, are there other lines? For example, lines with different slopes?

Let me think. In a 3x3 grid, are there any lines with slopes other than 0, infinity, 1, -1 that pass through three points?

Let me try to visualize the grid. Let me label the points with coordinates. Let me assume the grid goes from (0,0) to (2,2), with each coordinate being integers from 0 to 2. So the points are:

(0,0), (0,1), (0,2),

(1,0), (1,1), (1,2),

(2,0), (2,1), (2,2).

Now, let me check for other lines.

For example, consider slope 2. Let's see if there is a line with slope 2 that passes through three points. Let me check points (0,0), (1,2). The next point would be (2,4), but that's outside the grid. So only two points on that line.

How about slope 1/2? Starting at (0,0), going to (2,1). Then the next point would be (4,2), which is outside. So again only two points.

What about slope -2? Let's see. Starting at (0,2), going to (1,0). Next point would be (2, -2), which is outside. Similarly, slope -1/2: starting at (0,2), going to (2,1). Next point would be (4,0), outside. So again only two points.

How about other slopes? Let me think. What about slope 1/1, which is already considered in the main diagonals. Slope -1 is also considered. Is there a line with slope 2/1? Let me check. Starting at (0,0), going to (1,2), then (2,4) which is outside. Not possible.

How about slope -1/1? That's the other diagonal already considered.

What about lines that are not aligned with the grid or diagonals? For example, maybe some other lines?

Wait, maybe there are lines that pass through three points with different slopes. Let me try to check.

For example, consider the points (0,1), (1,0), (2,-1). But (2,-1) is not in the grid. Similarly, points (0,1), (1,1), (2,1)? No, that's horizontal. Wait, maybe points (0,0), (1,1), (2,2) which is the main diagonal. Already counted.

How about points (0,2), (1,1), (2,0)? That's the other main diagonal, already counted.

What about points (0,0), (1, 1), (2, 2)? Already counted. How about points (0,1), (1, 2), (2, 3)? Again, (2,3) is outside. Not in grid. Similarly, any other points?

Wait, maybe there are lines that pass through three points but aren't horizontal, vertical, or main diagonals. Let me think again. For example, consider points (0,0), (1, 2), and (2, 0). Wait, (0,0), (1,2), (2,0). Do these lie on a straight line?

Let me check. The slope between (0,0) and (1,2) is (2 - 0)/(1 - 0) = 2. The slope between (1,2) and (2,0) is (0 - 2)/(2 - 1) = -2. So different slopes, so they are not colinear. So that's not a line.

How about points (0,1), (1,1), (2,1)? That's horizontal, already counted.

How about points (1,0), (1,1), (1,2)? Vertical, already counted.

How about points (0,0), (1,1), (2,2)? Diagonal, already counted.

How about points (0,2), (1,1), (2,0)? Diagonal, already counted.

What about points (0,0), (2,1), (something)? Let me check if there is another point on that line. The line from (0,0) to (2,1) has slope 1/2. The next point would be (4, 2), which is outside. So only two points.

Similarly, other lines. How about points (0,1), (1,0), (2,-1)? Again, outside the grid.

Alternatively, maybe lines that are not straight? No, the question is about straight lines.

Wait, is there any line that passes through three points with different slopes?

Wait, let me check another possibility. Let me think of the points (0,0), (1, 1), (2, 2). That's slope 1. Similarly, (0, 0), (2, 1). Slope 1/2. Not three points.

Alternatively, maybe some other lines. Let me take another approach. For a 3x3 grid, how many lines are there that contain exactly three points?

We know horizontal, vertical, and two main diagonals. But maybe there are more lines that pass through exactly three points?

Wait, actually, in some grids, there are more lines. For example, in a 3x3 grid, there are lines with slopes 2, 1/2, -2, -1/2, but do they pass through three points?

Earlier, I thought they don't, but let me check again.

Take slope 2. Starting from (0,0): next point would be (1,2), then (2,4). But (2,4) is outside the grid. So only two points.

Starting from (0,1): next point would be (1,3), which is outside. Starting from (1,0): next point would be (2,2), so that's two points: (1,0), (2,2). Similarly, starting from (1,1): next point would be (2,3), outside. Similarly for others. So only two points on those lines.

Same with slope -2: starting from (0,2), next point (1,0), then (2,-2). So only two points. Similarly for other slopes. Therefore, there are no lines with slopes other than 0, infinity, 1, -1 that pass through three points.

Therefore, the total number of lines with three points is 3 horizontal + 3 vertical + 2 diagonal = 8 lines. Therefore, each of these lines has exactly one set of three points. Therefore, the number of favorable outcomes is 8.

Wait, but hold on, is that correct? Because each line has exactly one combination of three points. So if there are 8 such lines, then there are 8 favorable three-point sets?

But wait, let me think again. For example, each horizontal line has three points. The number of ways to choose three points on that line is 1, since all three are on the line. Similarly for vertical and diagonal. So yes, each line contributes one favorable set. So 3 horizontal, 3 vertical, 2 diagonal: 8 total. So 8 favorable sets.

Therefore, the probability is 8 / 84. Simplifying that, divide numerator and denominator by 4: 2 / 21. Wait, but wait, that seems low. Let me check again.

Wait, but maybe I missed some lines? Because sometimes people forget that there are more lines in the grid. Let me think again.

For example, in a 3x3 grid, there are lines that are not horizontal, vertical, or main diagonals but still contain three points. Wait, let me try to find such a line.

Wait, how about the line that goes through (0, 1), (1, 1), (2, 1)? That's horizontal. Already counted. How about (1, 0), (1, 1), (1, 2)? Vertical, already counted. How about (0,0), (1,1), (2,2)? Diagonal, counted. (0,2), (1,1), (2,0)? Diagonal, counted. What about (0,0), (0,1), (0,2)? Horizontal, counted.

Wait, is there a line that isn't aligned with grid or diagonals but still passes through three points? Hmm. Let me think. Let me take points (0,0), (1,2), and (2,1). Wait, are these colinear?

Let me check. The slope between (0,0) and (1,2) is (2 - 0)/(1 - 0) = 2. The slope between (1,2) and (2,1) is (1 - 2)/(2 - 1) = -1. Not the same, so not colinear.

How about (0,1), (1, 2), (2, 3)? But (2,3) is outside. No. How about (0,1), (1, 0), (2, -1)? Also outside. How about (1,0), (1,1), (1,2)? Vertical. Already counted. (0,0), (1,0), (2,0)? Horizontal. Already counted. (0,2), (1,2), (2,2)? Horizontal. Counted.

Wait, maybe another diagonal? For example, if you shift the diagonal? But in a 3x3 grid, the diagonals are only the main ones. Wait, what about a line that goes through (0, 1), (1, 0), (2, -1)? But that's going out of the grid. Similarly, the other way.

Wait, perhaps there's a line that passes through three points with a different slope? Wait, let me check all possible triples.

Alternatively, maybe I can think about how many lines there are in total. For a 3x3 grid, the number of lines that contain at least three points. But I thought it was 8, but maybe I'm missing some?

Wait, let me check online... Wait, no, I can't do that. I need to think through.

Alternatively, think about all possible lines that can be formed. For a 3x3 grid, the lines can be categorized by their slopes.

Slope 0 (horizontal): 3 lines.

Slope infinity (vertical): 3 lines.

Slope 1: 2 lines (main diagonals).

Slope -1: 2 lines? Wait, no, slope -1 is two lines? Wait, slope -1 is only two main diagonals? Wait, no, in a 3x3 grid, there are only two main diagonals: one from top-left to bottom-right (slope 1) and from top-right to bottom-left (slope -1). Wait, but for slope -1, there are only two points on that diagonal? Wait no, in the 3x3 grid, the main diagonals have three points each. For example, (0,2), (1,1), (2,0) is slope -1, and that has three points. Similarly, (0,0), (1,1), (2,2) is slope 1. So slope 1 and -1 each have one line with three points. Wait, but slope -1? There is only one line with slope -1 that has three points. Wait, no. Wait, in a 3x3 grid, how many diagonals with slope -1 have three points? Just the main one. Similarly for slope 1. So that gives 2 lines.

So total lines with three points: horizontal (3) + vertical (3) + slope 1 (1) + slope -1 (1) = 8 lines. So that's 8 lines, each contributing one set of three colinear points, so 8 favorable combinations.

But wait, hold on. Maybe there are more lines with three points? For example, if you have lines that are not main diagonals but still pass through three points.

Wait, for example, take the line y = x + 1. Let's see which points lie on that. When x=0, y=1; x=1, y=2; x=2, y=3. So (0,1), (1,2), but (2,3) is outside. So only two points. Similarly, y = x -1: (1,0), (2,1). Again two points. Similarly for other offsets.

How about lines with slope 1/2? For example, starting at (0,0), then (2,1). Then (4, 2). Outside. So only two points.

How about lines with slope 2? Starting at (0,1), going to (1,3). Outside. Not helpful.

How about lines with slope -1/2? Starting at (0,2), going to (2,1). Then (4, 0), which is outside. So only two points.

Same with other slopes. So, seems like there are no lines with three points other than the ones already considered.

Wait, but I remember that in some cases, like the 3x3 grid, there are more lines. Maybe I need to think about all possible lines, not just those with integral coordinates?

Wait, but the grid only has points with integer coordinates from (0,0) to (2,2). So any line passing through three of those points must have slope that is a rational number, but even so, only the ones we considered have three points. For example, if a line passes through three points, then the slope must be such that between any two points, the slope is same. But given that, and the grid size, I think we have already considered all.

So, going back, 8 lines. So 8 favorable combinations. Therefore, probability is 8/84, which simplifies to 2/21. But let me check once more to be sure.

Wait, but maybe there are more lines? Let me try to think differently. How many lines are there in total? For each pair of points, there is a line. But how many of these lines contain at least three points?

But that approach might be complicated. Alternatively, perhaps I can think of all possible lines that can be formed.

But maybe I can think of the number of lines with exactly three points. Each of the 8 lines as I thought before. So 8.

Alternatively, maybe I can think of how many lines there are with exactly three points. For example, in addition to the 3 horizontal, 3 vertical, 2 diagonals, are there others?

Wait, suppose I take the line that passes through (0,0), (1,1), (2,2) – that's one.

Then (0,0), (0,1), (0,2) – horizontal.

Then (0,2), (1,1), (2,0) – another diagonal.

Then (1,0), (1,1), (1,2) – vertical.

Similarly for others. So I don't think there are any others.

Alternatively, here's a thought: maybe some lines that aren't aligned with the grid can pass through three points? For example, take points (0,1), (1,2), (2, something). Wait, (0,1), (1,2). The next point would be (2,3), which is outside. Similarly, (0,1), (1,0), (2, -1) – outside. Similarly, (1,0), (2,1), (3,2) – outside. So no.

Alternatively, maybe non-integer slopes? Wait, no, because the points have integer coordinates, so any line passing through two points will have a slope that is a rational number. But even then, unless the slope is such that it passes through a third point.

But given that the grid is 3x3, the only way a line can pass through three points is if it's horizontal, vertical, or the main diagonals. Because otherwise, the spacing is too small to get three points. So I think my initial count of 8 is correct.

Therefore, the number of favorable outcomes is 8. Therefore, probability is 8/84.

But wait, let me check again. Let me actually list all possible lines with three points.

Horizontal lines:

1. (0,0), (0,1), (0,2)

2. (1,0), (1,1), (1,2)

3. (2,0), (2,1), (2,2)

Vertical lines:

4. (0,0), (1,0), (2,0)

5. (0,1), (1,1), (2,1)

6. (0,2), (1,2), (2,2)

Diagonals:

7. (0,0), (1,1), (2,2)

8. (0,2), (1,1), (2,0)

Yes, that's 8 lines. So 8 sets of three points. Therefore, 8 favorable combinations.

Thus, probability is 8/84. Simplify:

Divide numerator and denominator by 4: 2/21. So the probability is 2/21.

But wait, now I'm confused because I recall that in some problems similar to this, the answer is different. Let me verify once more.

Wait, perhaps I made a mistake in counting? Let me check the total number of lines again.

Alternatively, maybe there are more lines. Wait, for example, the line that goes through (0,1), (1,0), (2,-1) – but that's outside. Similarly, (0,1), (1,2), (2,3) – outside. So no.

Alternatively, maybe the line that goes through (1,0), (1,1), (1,2) – vertical, already counted.

Wait, perhaps lines that aren't aligned in the obvious ways? Wait, suppose you have a line that goes through (0,0), (1,2), and (2, something). But as we saw, (0,0) to (1,2) has slope 2. Next point would be (2,4), which is outside. Similarly for others.

Alternatively, (0,1), (1, 0), (2, -1) – outside. So no.

Alternatively, (1,0), (2,1), (something). Slope is 1, so next would be (3, 2), outside.

So again, nothing.

Wait, but here's another thought. What about the line that goes through (0, 1), (1, 2), and (2, 1)? Wait, no, those three points. Let me check if they are colinear.

Slope from (0,1) to (1,2) is 1. Slope from (1,2) to (2,1) is -1. Not same. Not colinear.

How about (0,1), (1,1), (2,1)? That's horizontal, already counted.

How about (0, 0), (1, 1), (2, 2)? Diagonal, counted.

How about (0, 0), (2,1), (something)? As before, no.

Wait, perhaps there are lines with three points that are not axis-aligned or main diagonals? For example, let me check if there are lines with three points with different slopes. Let me take points (0,0), (1, 1), (2, 2) – already there. How about (0, 0), (1, 0), (2, 0) – horizontal. How about (0, 0), (0, 1), (0, 2) – vertical. How about (0, 0), (1, 2), (2, something)? As before, no. How about (0, 1), (1, 0), (2, -1)? No.

Alternatively, (1, 0), (1, 1), (1, 2) – vertical. (0, 0), (1, 1), (2, 2) – diagonal. So I think all lines with three points have been considered.

Therefore, 8 lines. Therefore, 8 favorable combinations.

Wait, but let me check another way. Let me calculate the total number of lines in a 3x3 grid. Wait, but that's more complicated.

Alternatively, think of all possible triples of points and see how many are colinear. But since there are 84 total triples, that would be tedious, but maybe I can check some numbers.

Alternatively, think of how many lines there are. For a 3x3 grid, the number of lines is 3 horizontal + 3 vertical + 2 diagonals + ...? If I think that there are more.

Wait, here's a different approach. In general, for an n x n grid, the number of lines that contain exactly k points is... Well, for our case n=3. The formula isn't straightforward, but maybe I can recall that in a 3x3 grid, there are 8 lines with three points. Wait, but maybe that's not accurate.

Wait, let me think of the number of lines with exactly three points. For horizontal lines: 3. Vertical: 3. Diagonals: 2. So 8. But maybe there are more?

Wait, suppose I consider lines that are not axis-aligned or diagonals but still have three points. Wait, for example, in a 3x3 grid, there are also lines that go through the centers? Wait, for example, the center point (1,1). But how?

Wait, for example, take the line that goes through (0,1), (1,1), (2,1). That's horizontal. Already counted.

How about the line through (1,0), (1,1), (1,2). Vertical. Counted.

How about the line through (0,0), (1,1), (2,2). Diagonal. Counted.

How about (0,2), (1,1), (2,0). Diagonal. Counted.

Is there a line that passes through three points not in these categories?

Wait, maybe the line that goes through (0,1), (1, 0.5), (2, 0)? But (1, 0.5) is not a grid point. Similarly, any line that isn't passing through grid points can't be considered.

Therefore, there are no such lines. Therefore, the total number of lines is indeed 8.

Hence, the number of favorable outcomes is 8, leading to probability 8/84 = 2/21 ≈ 0.095.

Wait, but I feel like this might be wrong. Let me check with another method.

Another way: total number of ways to choose 3 points is C(9,3) = 84. Now, number of lines with three points is 8, each contributing 1 set, so 8. Therefore probability is 8/84 = 2/21 ≈ 0.095.

But I have a feeling that the answer might be higher. Maybe I missed some lines?

Wait, let me think again about diagonals. In addition to the main diagonals, are there other diagonals with three points? For example, in a 3x3 grid, if you consider diagonals with different slopes?

Wait, suppose you take the points (0,0), (1,1), (2,2) – that's one diagonal. Similarly, (0,1), (1,2) – but that's only two points. Similarly for others. How about (1,0), (1,1), (1,2)? Vertical. How about (0,0), (1, 0.5), (2,1)? Again, not grid points. So no.

Wait, maybe there is a line that goes through (0, 1), (1, 2), (2, 3)? But (2,3) is outside. So no. Similarly, (0, 1), (1, 0), (2, -1). Also outside.

Wait, here's a different idea: maybe there are lines that pass through three points where two are on one row and one on another?

No, because if they are colinear, they have to be on a straight line.

Wait, another way: think of all possible lines in the grid. For example, in a 3x3 grid, how many different lines are there?

Well, for each pair of points, there is a line. But many lines are repeated. But how many unique lines?

But this seems complicated, but let me try.

Total number of pairs: C(9,2)=36. Each line is defined by two points. But multiple pairs can define the same line.

For example, the horizontal line y=0 has three points: (0,0), (1,0), (2,0). This line is defined by C(3,2)=3 pairs. Similarly for other lines.

Similarly, the main diagonal (0,0), (1,1), (2,2) is defined by C(3,2)=3 pairs.

So if I can compute how many unique lines there are, considering that lines with three points have been counted multiple times.

But perhaps this is getting too complex. However, if I proceed:

Total number of unique lines can be calculated, but maybe we can use this to find the number of lines with exactly three points.

But since we know that lines with exactly three points are the ones we have already considered (horizontal, vertical, diagonals), and others have less.

But let me try. Let me calculate the number of lines with exactly two points, and exactly three points.

But perhaps not necessary. Wait, since we are interested only in lines with exactly three points, which would be the ones that can have three points selected. So, if we can find how many lines have at least three points, which would be the ones with exactly three points, since there are no lines with four or more points in a 3x3 grid.

So for a 3x3 grid, the maximum number of points on a line is three.

Therefore, the number of lines that have exactly three points is the number of lines we can have. Which, in our case, we said was 8.

Therefore, the number of favorable outcomes is 8, so probability is 8/84.

Wait, but now I recall that in some other similar problems, for example, the probability that three randomly chosen points form a right triangle or something, the answer might differ. But maybe not.

Alternatively, let me think of specific examples. For example, how many lines with three points are there?

Let me count again:

Horizontal lines: 3 (each row)

Vertical lines: 3 (each column)

Diagonals: 2 (main diagonals)

So 3 + 3 + 2 = 8. Yes.

Therefore, 8 lines. So 8 favorable outcomes. Therefore, probability is 8/84 = 2/21 ≈ 0.095. So 2/21 is the probability.

But wait, I found a source once where the answer was 8/84, which reduces to 2/21. But I want to make sure.

Alternatively, maybe there are more lines. Let me think again. For example, take the line connecting (0,0), (1, 2), and (2, 0). Wait, no, these three points are not colinear. As we checked before, the slopes differ.

Another idea: what about the line connecting (0, 1), (1, 2), and (2, 1)? No, that's not colinear. Slope between first two is 1, between last two is -1. Not same.

Another one: (0, 0), (1, 1), (2, 2) – yes, counted.

How about (0, 0), (0, 1), (0, 2) – horizontal. Counted.

Is there a line that goes through (0, 1), (1, 2), (2, 3)? No, beyond the grid.

Wait, here's a different thought. What about lines that are not straight? No, the question is about straight lines.

Wait, so maybe I really have only 8 lines. So the answer is 8/84.

But let me check with actual counting.

Let me list all possible lines with three points:

1. Row 0: (0,0), (0,1), (0,2)

2. Row 1: (1,0), (1,1), (1,2)

3. Row 2: (2,0), (2,1), (2,2)

4. Column 0: (0,0), (1,0), (2,0)

5. Column 1: (0,1), (1,1), (2,1)

6. Column 2: (0,2), (1,2), (2,2)

7. Diagonal 1: (0,0), (1,1), (2,2)

8. Diagonal 2: (0,2), (1,1), (2,0)

Yes, that's 8 lines. Therefore, only 8 sets of three colinear points.

Therefore, probability is 8 / 84.

Simplify: divide numerator and denominator by 4: 2 / 21.

But I'm still hesitant because sometimes problems like this have more lines. Wait, let me check an alternative resource in my mind. For example, in a 3x3 grid, how many lines are there with three points?

I think that the answer is indeed 8. For example, in the game of Tic-Tac-Toe, there are 8 winning lines: 3 rows, 3 columns, 2 diagonals. Therefore, if you think of the 3x3 grid as a tic-tac-toe board, the number of ways to have three in a row is 8. Therefore, the number of favorable outcomes is 8. So probability is 8 / 84 = 2/21.

Therefore, I think that is correct. Therefore, the answer is 2/21.

But just to be absolutely thorough, let me check if there are other lines.

Suppose someone says, what about the line that goes through (0, 0), (1, 1), (2, 2)? That's already counted.

How about the line through (0, 0), (2, 1)? That's a line with two points. Not enough.

How about the line through (1, 0), (2, 1)? Also two points.

How about (1, 0), (2, 2)? Slope is (2 - 0)/(2 - 1) = 2. So line from (1, 0) to (2, 2). The third point would be (0, -2), which is outside. So no.

Similarly for others.

Therefore, I think my initial count is correct. Therefore, the probability is 2/21.

But let me just check one more thing. Suppose I miscalculated the total number of combinations. Wait, C(9,3) is 84. Yes. 9*8*7 / 6 = 84. Correct.

So 8/84 = 2/21. Simplify 8 divided by 84: divide numerator and denominator by 4: 2/21. Yes.

Alternatively, maybe there are more lines? Wait, for example, in the 3x3 grid, is there a line that passes through (0, 1), (1, 0), (2, 1)? Wait, let me check if they are colinear.

The slope between (0,1) and (1,0) is (0 - 1)/(1 - 0) = -1. The slope between (1,0) and (2,1) is (1 - 0)/(2 - 1) = 1. Not same, so not colinear.

Another example: (0, 1), (1, 2), (2, 1). Slope between first two is 1, between last two is -1. Not same.

How about (0, 1), (1, 1), (2, 1)? That's horizontal, already counted.

How about (1, 0), (1, 1), (1, 2)? Vertical, already counted.

So no new lines here.

Therefore, I think I can confidently say that the number of favorable outcomes is 8. Hence, the probability is 2/21.

But just to be 100% sure, I can think of all the lines again:

Rows:

- Top row: (0,0), (0,1), (0,2)

- Middle row: (1,0), (1,1), (1,2)

- Bottom row: (2,0), (2,1), (2,2)

Columns:

- Left column: (0,0), (1,0), (2,0)

- Middle column: (0,1), (1,1), (2,1)

- Right column: (0,2), (1,2), (2,2)

Diagonals:

- Main diagonal: (0,0), (1,1), (2,2)

- Anti-diagonal: (0,2), (1,1), (2,0)

That's 3 + 3 + 2 = 8 lines. Therefore, 8 favorable sets. Therefore, probability is 8/84 = 2/21.

Thus, the answer should be \boxed{\dfrac{2}{21}}.

**Final Answer**
The probability is \boxed{\dfrac{2}{21}}.

To solve the problem of determining the probability that three randomly selected points from a 3×3 grid lie on a straight line, we proceed in the following structured way:

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### **Step 1: Total Number of Ways to Choose Three Points**

The grid consists of 9 points arranged in a 3×3 grid. We are to select any 3 of these points, so the total number of possible three-point sets is calculated using combinations:

$$
\text{Total combinations} = \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84
$$

So, there are **84** possible sets of three points.

---

### **Step 2: Count the Number of Favorable Outcomes**

We are now interested in counting the number of 3-point sets that lie **on a straight line**.

In a 3×3 grid, lines that pass through three points can be of the following types:

#### **Horizontal Lines:**
- Top row: (0,0), (0,1), (0,2)
- Middle row: (1,0), (1,1), (1,2)
- Bottom row: (2,0), (2,1), (2,2)

👉 3 lines

#### **Vertical Lines:**
- Left column: (0,0), (1,0), (2,0)
- Middle column: (0,1), (1,1), (2,1)
- Right column: (0,2), (1,2), (2,2)

👉 3 lines

#### **Diagonal Lines:**
- Main diagonal (top-left to bottom-right): (0,0), (1,1), (2,2)
- Anti-diagonal (top-right to bottom-left): (0,2), (1,1), (2,0)

👉 2 lines

**Total lines with exactly 3 points:**
$$
3 \text{ (horizontal)} + 3 \text{ (vertical)} + 2 \text{ (diagonal)} = 8 \text{ lines}
$$

Each of these lines contains exactly **three distinct grid points**, and selecting the entire line (i.e., all three points) results in one valid **favorable outcome**.

Therefore, there are **8 favorable sets** of 3 points that lie on a straight line.

---

### **Step 3: Compute the Probability**

$$
\text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{8}{84}
$$

Simplify the fraction:

$$
\frac{8}{84} = \frac{2}{21}
$$

---

### **Final Answer**

$$
\boxed{\dfrac{2}{21}}
$$

[Assistant]
Explanation:
Okay, so I need to find the probability that three randomly chosen points from a 3x3 grid lie on the same straight line. Let me start by understanding the problem.

First, the grid is 3x3, which means there are 9 points in total. The question says a set of three points is chosen randomly, and each set has the same probability. So, the total number of possible three-point sets is the combination of 9 points taken 3 at a time. I remember that combinations are calculated using the formula C(n, k) = n! / (k! * (n - k)!).

Let me calculate that first. So n is 9, and k is 3. Plugging in:

C(9, 3) = 9! / (3! * 6!) = (9 × 8 × 7) / (3 × 2 × 1) = 84. So there are 84 possible ways to choose three points. That's the denominator of my probability.

Now, I need to find the number of favorable outcomes, which are the sets of three points that lie on a straight line. Then, the probability will be favorable / total, which is favorable / 84.

So, how do I count the number of such lines?

First, I should consider all possible lines that can be formed by three points on the grid. The grid is 3x3, so there are horizontal lines, vertical lines, and diagonal lines. Also, maybe there are other lines with different slopes that pass through three points? Let me check.

Starting with horizontal lines. Each row of the grid is a horizontal line. Since there are 3 rows, each containing 3 points, so there are 3 horizontal lines. Each of these lines has exactly 3 points, so each row contributes 1 set of three colinear points. Therefore, horizontal lines contribute 3 favorable sets.

Similarly, vertical lines. Each column is a vertical line. Again, there are 3 columns, each with three points. So another 3 favorable sets from vertical lines.

Then, the main diagonals. There are two main diagonals: from top-left to bottom-right and from top-right to bottom-left. Each of these diagonals has exactly three points. So that's 2 more favorable sets.

So so far, horizontal (3) + vertical (3) + diagonal (2) = 8. But wait, are there other lines? For example, lines with different slopes?

Let me think. In a 3x3 grid, are there any lines with slopes other than 0, infinity, 1, -1 that pass through three points?

Let me try to visualize the grid. Let me label the points with coordinates. Let me assume the grid goes from (0,0) to (2,2), with each coordinate being integers from 0 to 2. So the points are:

(0,0), (0,1), (0,2),

(1,0), (1,1), (1,2),

(2,0), (2,1), (2,2).

Now, let me check for other lines.

For example, consider slope 2. Let's see if there is a line with slope 2 that passes through three points. Let me check points (0,0), (1,2). The next point would be (2,4), but that's outside the grid. So only two points on that line.

How about slope 1/2? Starting at (0,0), going to (2,1). Then the next point would be (4,2), which is outside. So again only two points.

What about slope -2? Let's see. Starting at (0,2), going to (1,0). Next point would be (2, -2), which is outside. Similarly, slope -1/2: starting at (0,2), going to (2,1). Next point would be (4,0), outside. So again only two points.

How about other slopes? Let me think. What about slope 1/1, which is already considered in the main diagonals. Slope -1 is also considered. Is there a line with slope 2/1? Let me check. Starting at (0,0), going to (1,2), then (2,4) which is outside. Not possible.

How about slope -1/1? That's the other diagonal already considered.

What about lines that are not aligned with the grid or diagonals? For example, maybe some other lines?

Wait, maybe there are lines that pass through three points with different slopes. Let me try to check.

For example, consider the points (0,1), (1,0), (2,-1). But (2,-1) is not in the grid. Similarly, points (0,1), (1,1), (2,1)? No, that's horizontal. Wait, maybe points (0,0), (1,1), (2,2) which is the main diagonal. Already counted.

How about points (0,2), (1,1), (2,0)? That's the other main diagonal, already counted.

What about points (0,0), (1, 1), (2, 2)? Already counted. How about points (0,1), (1, 2), (2, 3)? Again, (2,3) is outside. Not in grid. Similarly, any other points?

Wait, maybe there are lines that pass through three points but aren't horizontal, vertical, or main diagonals. Let me think again. For example, consider points (0,0), (1, 2), and (2, 0). Wait, (0,0), (1,2), (2,0). Do these lie on a straight line?

Let me check. The slope between (0,0) and (1,2) is (2 - 0)/(1 - 0) = 2. The slope between (1,2) and (2,0) is (0 - 2)/(2 - 1) = -2. So different slopes, so they are not colinear. So that's not a line.

How about points (0,1), (1,1), (2,1)? That's horizontal, already counted.

How about points (1,0), (1,1), (1,2)? Vertical, already counted.

How about points (0,0), (1,1), (2,2)? Diagonal, already counted.

How about points (0,2), (1,1), (2,0)? Diagonal, already counted.

What about points (0,0), (2,1), (something)? Let me check if there is another point on that line. The line from (0,0) to (2,1) has slope 1/2. The next point would be (4, 2), which is outside. So only two points.

Similarly, other lines. How about points (0,1), (1,0), (2,-1)? Again, outside the grid.

Alternatively, maybe lines that are not straight? No, the question is about straight lines.

Wait, is there any line that passes through three points with different slopes?

Wait, let me check another possibility. Let me think of the points (0,0), (1, 1), (2, 2). That's slope 1. Similarly, (0, 0), (2, 1). Slope 1/2. Not three points.

Alternatively, maybe some other lines. Let me take another approach. For a 3x3 grid, how many lines are there that contain exactly three points?

We know horizontal, vertical, and two main diagonals. But maybe there are more lines that pass through exactly three points?

Wait, actually, in some grids, there are more lines. For example, in a 3x3 grid, there are lines with slopes 2, 1/2, -2, -1/2, but do they pass through three points?

Earlier, I thought they don't, but let me check again.

Take slope 2. Starting from (0,0): next point would be (1,2), then (2,4). But (2,4) is outside the grid. So only two points.

Starting from (0,1): next point would be (1,3), which is outside. Starting from (1,0): next point would be (2,2), so that's two points: (1,0), (2,2). Similarly, starting from (1,1): next point would be (2,3), outside. Similarly for others. So only two points on those lines.

Same with slope -2: starting from (0,2), next point (1,0), then (2,-2). So only two points. Similarly for other slopes. Therefore, there are no lines with slopes other than 0, infinity, 1, -1 that pass through three points.

Therefore, the total number of lines with three points is 3 horizontal + 3 vertical + 2 diagonal = 8 lines. Therefore, each of these lines has exactly one set of three points. Therefore, the number of favorable outcomes is 8.

Wait, but hold on, is that correct? Because each line has exactly one combination of three points. So if there are 8 such lines, then there are 8 favorable three-point sets?

But wait, let me think again. For example, each horizontal line has three points. The number of ways to choose three points on that line is 1, since all three are on the line. Similarly for vertical and diagonal. So yes, each line contributes one favorable set. So 3 horizontal, 3 vertical, 2 diagonal: 8 total. So 8 favorable sets.

Therefore, the probability is 8 / 84. Simplifying that, divide numerator and denominator by 4: 2 / 21. Wait, but wait, that seems low. Let me check again.

Wait, but maybe I missed some lines? Because sometimes people forget that there are more lines in the grid. Let me think again.

For example, in a 3x3 grid, there are lines that are not horizontal, vertical, or main diagonals but still contain three points. Wait, let me try to find such a line.

Wait, how about the line that goes through (0, 1), (1, 1), (2, 1)? That's horizontal. Already counted. How about (1, 0), (1, 1), (1, 2)? Vertical, already counted. How about (0,0), (1,1), (2,2)? Diagonal, counted. (0,2), (1,1), (2,0)? Diagonal, counted. What about (0,0), (0,1), (0,2)? Horizontal, counted.

Wait, is there a line that isn't aligned with grid or diagonals but still passes through three points? Hmm. Let me think. Let me take points (0,0), (1,2), and (2,1). Wait, are these colinear?

Let me check. The slope between (0,0) and (1,2) is (2 - 0)/(1 - 0) = 2. The slope between (1,2) and (2,1) is (1 - 2)/(2 - 1) = -1. Not the same, so not colinear.

How about (0,1), (1, 2), (2, 3)? But (2,3) is outside. No. How about (0,1), (1, 0), (2, -1)? Also outside. How about (1,0), (1,1), (1,2)? Vertical. Already counted. (0,0), (1,0), (2,0)? Horizontal. Already counted. (0,2), (1,2), (2,2)? Horizontal. Counted.

Wait, maybe another diagonal? For example, if you shift the diagonal? But in a 3x3 grid, the diagonals are only the main ones. Wait, what about a line that goes through (0, 1), (1, 0), (2, -1)? But that's going out of the grid. Similarly, the other way.

Wait, perhaps there's a line that passes through three points with a different slope? Wait, let me check all possible triples.

Alternatively, maybe I can think about how many lines there are in total. For a 3x3 grid, the number of lines that contain at least three points. But I thought it was 8, but maybe I'm missing some?

Wait, let me check online... Wait, no, I can't do that. I need to think through.

Alternatively, think about all possible lines that can be formed. For a 3x3 grid, the lines can be categorized by their slopes.

Slope 0 (horizontal): 3 lines.

Slope infinity (vertical): 3 lines.

Slope 1: 2 lines (main diagonals).

Slope -1: 2 lines? Wait, no, slope -1 is two lines? Wait, slope -1 is only two main diagonals? Wait, no, in a 3x3 grid, there are only two main diagonals: one from top-left to bottom-right (slope 1) and from top-right to bottom-left (slope -1). Wait, but for slope -1, there are only two points on that diagonal? Wait no, in the 3x3 grid, the main diagonals have three points each. For example, (0,2), (1,1), (2,0) is slope -1, and that has three points. Similarly, (0,0), (1,1), (2,2) is slope 1. So slope 1 and -1 each have one line with three points. Wait, but slope -1? There is only one line with slope -1 that has three points. Wait, no. Wait, in a 3x3 grid, how many diagonals with slope -1 have three points? Just the main one. Similarly for slope 1. So that gives 2 lines.

So total lines with three points: horizontal (3) + vertical (3) + slope 1 (1) + slope -1 (1) = 8 lines. So that's 8 lines, each contributing one set of three colinear points, so 8 favorable combinations.

But wait, hold on. Maybe there are more lines with three points? For example, if you have lines that are not main diagonals but still pass through three points.

Wait, for example, take the line y = x + 1. Let's see which points lie on that. When x=0, y=1; x=1, y=2; x=2, y=3. So (0,1), (1,2), but (2,3) is outside. So only two points. Similarly, y = x -1: (1,0), (2,1). Again two points. Similarly for other offsets.

How about lines with slope 1/2? For example, starting at (0,0), then (2,1). Then (4, 2). Outside. So only two points.

How about lines with slope 2? Starting at (0,1), going to (1,3). Outside. Not helpful.

How about lines with slope -1/2? Starting at (0,2), going to (2,1). Then (4, 0), which is outside. So only two points.

Same with other slopes. So, seems like there are no lines with three points other than the ones already considered.

Wait, but I remember that in some cases, like the 3x3 grid, there are more lines. Maybe I need to think about all possible lines, not just those with integral coordinates?

Wait, but the grid only has points with integer coordinates from (0,0) to (2,2). So any line passing through three of those points must have slope that is a rational number, but even so, only the ones we considered have three points. For example, if a line passes through three points, then the slope must be such that between any two points, the slope is same. But given that, and the grid size, I think we have already considered all.

So, going back, 8 lines. So 8 favorable combinations. Therefore, probability is 8/84, which simplifies to 2/21. But let me check once more to be sure.

Wait, but maybe there are more lines? Let me try to think differently. How many lines are there in total? For each pair of points, there is a line. But how many of these lines contain at least three points?

But that approach might be complicated. Alternatively, perhaps I can think of all possible lines that can be formed.

But maybe I can think of the number of lines with exactly three points. Each of the 8 lines as I thought before. So 8.

Alternatively, maybe I can think of how many lines there are with exactly three points. For example, in addition to the 3 horizontal, 3 vertical, 2 diagonals, are there others?

Wait, suppose I take the line that passes through (0,0), (1,1), (2,2) – that's one.

Then (0,0), (0,1), (0,2) – horizontal.

Then (0,2), (1,1), (2,0) – another diagonal.

Then (1,0), (1,1), (1,2) – vertical.

Similarly for others. So I don't think there are any others.

Alternatively, here's a thought: maybe some lines that aren't aligned with the grid can pass through three points? For example, take points (0,1), (1,2), (2, something). Wait, (0,1), (1,2). The next point would be (2,3), which is outside. Similarly, (0,1), (1,0), (2, -1) – outside. Similarly, (1,0), (2,1), (3,2) – outside. So no.

Alternatively, maybe non-integer slopes? Wait, no, because the points have integer coordinates, so any line passing through two points will have a slope that is a rational number. But even then, unless the slope is such that it passes through a third point.

But given that the grid is 3x3, the only way a line can pass through three points is if it's horizontal, vertical, or the main diagonals. Because otherwise, the spacing is too small to get three points. So I think my initial count of 8 is correct.

Therefore, the number of favorable outcomes is 8. Therefore, probability is 8/84.

But wait, let me check again. Let me actually list all possible lines with three points.

Horizontal lines:

1. (0,0), (0,1), (0,2)

2. (1,0), (1,1), (1,2)

3. (2,0), (2,1), (2,2)

Vertical lines:

4. (0,0), (1,0), (2,0)

5. (0,1), (1,1), (2,1)

6. (0,2), (1,2), (2,2)

Diagonals:

7. (0,0), (1,1), (2,2)

8. (0,2), (1,1), (2,0)

Yes, that's 8 lines. So 8 sets of three points. Therefore, 8 favorable combinations.

Thus, probability is 8/84. Simplify:

Divide numerator and denominator by 4: 2/21. So the probability is 2/21.

But wait, now I'm confused because I recall that in some problems similar to this, the answer is different. Let me verify once more.

Wait, perhaps I made a mistake in counting? Let me check the total number of lines again.

Alternatively, maybe there are more lines. Wait, for example, the line that goes through (0,1), (1,0), (2,-1) – but that's outside. Similarly, (0,1), (1,2), (2,3) – outside. So no.

Alternatively, maybe the line that goes through (1,0), (1,1), (1,2) – vertical, already counted.

Wait, perhaps lines that aren't aligned in the obvious ways? Wait, suppose you have a line that goes through (0,0), (1,2), and (2, something). But as we saw, (0,0) to (1,2) has slope 2. Next point would be (2,4), which is outside. Similarly for others.

Alternatively, (0,1), (1, 0), (2, -1) – outside. So no.

Alternatively, (1,0), (2,1), (something). Slope is 1, so next would be (3, 2), outside.

So again, nothing.

Wait, but here's another thought. What about the line that goes through (0, 1), (1, 2), and (2, 1)? Wait, no, those three points. Let me check if they are colinear.

Slope from (0,1) to (1,2) is 1. Slope from (1,2) to (2,1) is -1. Not same. Not colinear.

How about (0,1), (1,1), (2,1)? That's horizontal, already counted.

How about (0, 0), (1, 1), (2, 2)? Diagonal, counted.

How about (0, 0), (2,1), (something)? As before, no.

Wait, perhaps there are lines with three points that are not axis-aligned or main diagonals? For example, let me check if there are lines with three points with different slopes. Let me take points (0,0), (1, 1), (2, 2) – already there. How about (0, 0), (1, 0), (2, 0) – horizontal. How about (0, 0), (0, 1), (0, 2) – vertical. How about (0, 0), (1, 2), (2, something)? As before, no. How about (0, 1), (1, 0), (2, -1)? No.

Alternatively, (1, 0), (1, 1), (1, 2) – vertical. (0, 0), (1, 1), (2, 2) – diagonal. So I think all lines with three points have been considered.

Therefore, 8 lines. Therefore, 8 favorable combinations.

Wait, but let me check another way. Let me calculate the total number of lines in a 3x3 grid. Wait, but that's more complicated.

Alternatively, think of all possible triples of points and see how many are colinear. But since there are 84 total triples, that would be tedious, but maybe I can check some numbers.

Alternatively, think of how many lines there are. For a 3x3 grid, the number of lines is 3 horizontal + 3 vertical + 2 diagonals + ...? If I think that there are more.

Wait, here's a different approach. In general, for an n x n grid, the number of lines that contain exactly k points is... Well, for our case n=3. The formula isn't straightforward, but maybe I can recall that in a 3x3 grid, there are 8 lines with three points. Wait, but maybe that's not accurate.

Wait, let me think of the number of lines with exactly three points. For horizontal lines: 3. Vertical: 3. Diagonals: 2. So 8. But maybe there are more?

Wait, suppose I consider lines that are not axis-aligned or diagonals but still have three points. Wait, for example, in a 3x3 grid, there are also lines that go through the centers? Wait, for example, the center point (1,1). But how?

Wait, for example, take the line that goes through (0,1), (1,1), (2,1). That's horizontal. Already counted.

How about the line through (1,0), (1,1), (1,2). Vertical. Counted.

How about the line through (0,0), (1,1), (2,2). Diagonal. Counted.

How about (0,2), (1,1), (2,0). Diagonal. Counted.

Is there a line that passes through three points not in these categories?

Wait, maybe the line that goes through (0,1), (1, 0.5), (2, 0)? But (1, 0.5) is not a grid point. Similarly, any line that isn't passing through grid points can't be considered.

Therefore, there are no such lines. Therefore, the total number of lines is indeed 8.

Hence, the number of favorable outcomes is 8, leading to probability 8/84 = 2/21 ≈ 0.095.

Wait, but I feel like this might be wrong. Let me check with another method.

Another way: total number of ways to choose 3 points is C(9,3) = 84. Now, number of lines with three points is 8, each contributing 1 set, so 8. Therefore probability is 8/84 = 2/21 ≈ 0.095.

But I have a feeling that the answer might be higher. Maybe I missed some lines?

Wait, let me think again about diagonals. In addition to the main diagonals, are there other diagonals with three points? For example, in a 3x3 grid, if you consider diagonals with different slopes?

Wait, suppose you take the points (0,0), (1,1), (2,2) – that's one diagonal. Similarly, (0,1), (1,2) – but that's only two points. Similarly for others. How about (1,0), (1,1), (1,2)? Vertical. How about (0,0), (1, 0.5), (2,1)? Again, not grid points. So no.

Wait, maybe there is a line that goes through (0, 1), (1, 2), (2, 3)? But (2,3) is outside. So no. Similarly, (0, 1), (1, 0), (2, -1). Also outside.

Wait, here's a different idea: maybe there are lines that pass through three points where two are on one row and one on another?

No, because if they are colinear, they have to be on a straight line.

Wait, another way: think of all possible lines in the grid. For example, in a 3x3 grid, how many different lines are there?

Well, for each pair of points, there is a line. But many lines are repeated. But how many unique lines?

But this seems complicated, but let me try.

Total number of pairs: C(9,2)=36. Each line is defined by two points. But multiple pairs can define the same line.

For example, the horizontal line y=0 has three points: (0,0), (1,0), (2,0). This line is defined by C(3,2)=3 pairs. Similarly for other lines.

Similarly, the main diagonal (0,0), (1,1), (2,2) is defined by C(3,2)=3 pairs.

So if I can compute how many unique lines there are, considering that lines with three points have been counted multiple times.

But perhaps this is getting too complex. However, if I proceed:

Total number of unique lines can be calculated, but maybe we can use this to find the number of lines with exactly three points.

But since we know that lines with exactly three points are the ones we have already considered (horizontal, vertical, diagonals), and others have less.

But let me try. Let me calculate the number of lines with exactly two points, and exactly three points.

But perhaps not necessary. Wait, since we are interested only in lines with exactly three points, which would be the ones that can have three points selected. So, if we can find how many lines have at least three points, which would be the ones with exactly three points, since there are no lines with four or more points in a 3x3 grid.

So for a 3x3 grid, the maximum number of points on a line is three.

Therefore, the number of lines that have exactly three points is the number of lines we can have. Which, in our case, we said was 8.

Therefore, the number of favorable outcomes is 8, so probability is 8/84.

Wait, but now I recall that in some other similar problems, for example, the probability that three randomly chosen points form a right triangle or something, the answer might differ. But maybe not.

Alternatively, let me think of specific examples. For example, how many lines with three points are there?

Let me count again:

Horizontal lines: 3 (each row)

Vertical lines: 3 (each column)

Diagonals: 2 (main diagonals)

So 3 + 3 + 2 = 8. Yes.

Therefore, 8 lines. So 8 favorable outcomes. Therefore, probability is 8/84 = 2/21 ≈ 0.095. So 2/21 is the probability.

But wait, I found a source once where the answer was 8/84, which reduces to 2/21. But I want to make sure.

Alternatively, maybe there are more lines. Let me think again. For example, take the line connecting (0,0), (1, 2), and (2, 0). Wait, no, these three points are not colinear. As we checked before, the slopes differ.

Another idea: what about the line connecting (0, 1), (1, 2), and (2, 1)? No, that's not colinear. Slope between first two is 1, between last two is -1. Not same.

Another one: (0, 0), (1, 1), (2, 2) – yes, counted.

How about (0, 0), (0, 1), (0, 2) – horizontal. Counted.

Is there a line that goes through (0, 1), (1, 2), (2, 3)? No, beyond the grid.

Wait, here's a different thought. What about lines that are not straight? No, the question is about straight lines.

Wait, so maybe I really have only 8 lines. So the answer is 8/84.

But let me check with actual counting.

Let me list all possible lines with three points:

1. Row 0: (0,0), (0,1), (0,2)

2. Row 1: (1,0), (1,1), (1,2)

3. Row 2: (2,0), (2,1), (2,2)

4. Column 0: (0,0), (1,0), (2,0)

5. Column 1: (0,1), (1,1), (2,1)

6. Column 2: (0,2), (1,2), (2,2)

7. Diagonal 1: (0,0), (1,1), (2,2)

8. Diagonal 2: (0,2), (1,1), (2,0)

Yes, that's 8 lines. Therefore, only 8 sets of three colinear points.

Therefore, probability is 8 / 84.

Simplify: divide numerator and denominator by 4: 2 / 21.

But I'm still hesitant because sometimes problems like this have more lines. Wait, let me check an alternative resource in my mind. For example, in a 3x3 grid, how many lines are there with three points?

I think that the answer is indeed 8. For example, in the game of Tic-Tac-Toe, there are 8 winning lines: 3 rows, 3 columns, 2 diagonals. Therefore, if you think of the 3x3 grid as a tic-tac-toe board, the number of ways to have three in a row is 8. Therefore, the number of favorable outcomes is 8. So probability is 8 / 84 = 2/21.

Therefore, I think that is correct. Therefore, the answer is 2/21.

But just to be absolutely thorough, let me check if there are other lines.

Suppose someone says, what about the line that goes through (0, 0), (1, 1), (2, 2)? That's already counted.

How about the line through (0, 0), (2, 1)? That's a line with two points. Not enough.

How about the line through (1, 0), (2, 1)? Also two points.

How about (1, 0), (2, 2)? Slope is (2 - 0)/(2 - 1) = 2. So line from (1, 0) to (2, 2). The third point would be (0, -2), which is outside. So no.

Similarly for others.

Therefore, I think my initial count is correct. Therefore, the probability is 2/21.

But let me just check one more thing. Suppose I miscalculated the total number of combinations. Wait, C(9,3) is 84. Yes. 9*8*7 / 6 = 84. Correct.

So 8/84 = 2/21. Simplify 8 divided by 84: divide numerator and denominator by 4: 2/21. Yes.

Alternatively, maybe there are more lines? Wait, for example, in the 3x3 grid, is there a line that passes through (0, 1), (1, 0), (2, 1)? Wait, let me check if they are colinear.

The slope between (0,1) and (1,0) is (0 - 1)/(1 - 0) = -1. The slope between (1,0) and (2,1) is (1 - 0)/(2 - 1) = 1. Not same, so not colinear.

Another example: (0, 1), (1, 2), (2, 1). Slope between first two is 1, between last two is -1. Not same.

How about (0, 1), (1, 1), (2, 1)? That's horizontal, already counted.

How about (1, 0), (1, 1), (1, 2)? Vertical, already counted.

So no new lines here.

Therefore, I think I can confidently say that the number of favorable outcomes is 8. Hence, the probability is 2/21.

But just to be 100% sure, I can think of all the lines again:

Rows:

- Top row: (0,0), (0,1), (0,2)

- Middle row: (1,0), (1,1), (1,2)

- Bottom row: (2,0), (2,1), (2,2)

Columns:

- Left column: (0,0), (1,0), (2,0)

- Middle column: (0,1), (1,1), (2,1)

- Right column: (0,2), (1,2), (2,2)

Diagonals:

- Main diagonal: (0,0), (1,1), (2,2)

- Anti-diagonal: (0,2), (1,1), (2,0)

That's 3 + 3 + 2 = 8 lines. Therefore, 8 favorable sets. Therefore, probability is 8/84 = 2/21.

Thus, the answer should be \boxed{\dfrac{2}{21}}.

**Final Answer**
The probability is \boxed{\dfrac{2}{21}}.

To solve the problem of determining the probability that three randomly selected points from a 3×3 grid lie on a straight line, we proceed in the following structured way:

---

### **Step 1: Total Number of Ways to Choose Three Points**

The grid consists of 9 points arranged in a 3×3 grid. We are to select any 3 of these points, so the total number of possible three-point sets is calculated using combinations:

$$
\text{Total combinations} = \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84
$$

So, there are **84** possible sets of three points.

---

### **Step 2: Count the Number of Favorable Outcomes**

We are now interested in counting the number of 3-point sets that lie **on a straight line**.

In a 3×3 grid, lines that pass through three points can be of the following types:

#### **Horizontal Lines:**
- Top row: (0,0), (0,1), (0,2)
- Middle row: (1,0), (1,1), (1,2)
- Bottom row: (2,0), (2,1), (2,2)

👉 3 lines

#### **Vertical Lines:**
- Left column: (0,0), (1,0), (2,0)
- Middle column: (0,1), (1,1), (2,1)
- Right column: (0,2), (1,2), (2,2)

👉 3 lines

#### **Diagonal Lines:**
- Main diagonal (top-left to bottom-right): (0,0), (1,1), (2,2)
- Anti-diagonal (top-right to bottom-left): (0,2), (1,1), (2,0)

👉 2 lines

**Total lines with exactly 3 points:**
$$
3 \text{ (horizontal)} + 3 \text{ (vertical)} + 2 \text{ (diagonal)} = 8 \text{ lines}
$$

Each of these lines contains exactly **three distinct grid points**, and selecting the entire line (i.e., all three points) results in one valid **favorable outcome**.

Therefore, there are **8 favorable sets** of 3 points that lie on a straight line.

---

### **Step 3: Compute the Probability**

$$
\text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{8}{84}
$$

Simplify the fraction:

$$
\frac{8}{84} = \frac{2}{21}
$$

---

### **Final Answer**

$$
\boxed{\dfrac{2}{21}}
$$

Solution:
The number of three-point sets that can be chosen from the nine grid points is \[
\binom{9}{3} = \frac{9!}{3!\cdot 6!} = 84.
\]Eight of these sets consist of three collinear points: 3 sets of points lie on vertical lines, 3 on horizontal lines, and 2 on diagonals. Hence the probability is $8/84 = \boxed{\frac{2}{21}}$.


Explanation:
Okay, so I need to find the probability that a randomly selected point (x, y) in the region 0 ≤ x ≤ 8 and 0 ≤ y ≤ 4 satisfies the condition x + y ≤ 4. Hmm, let me think about how to approach this.

First, probability in this context is usually the area of the region where the condition holds divided by the total area of the possible region. So I should probably start by figuring out the total area of the region where the point can be selected. The problem says 0 ≤ x ≤ 8 and 0 ≤ y ≤ 4. So that's a rectangle, right? The length along the x-axis is 8 units, and along the y-axis is 4 units. Therefore, the total area should be 8 multiplied by 4, which is 32. Let me write that down:

Total area = 8 * 4 = 32.

Okay, now I need to find the area of the region where x + y ≤ 4. This is a linear inequality, so the region should be a polygon. Since we're dealing with x and y both being non-negative, the region defined by x + y ≤ 4 should be a triangle or a trapezoid, maybe? Let me sketch it mentally.

If I consider the first quadrant (since x and y are non-negative), the line x + y = 4 is a straight line starting from (0, 4) to (4, 0). So the area where x + y ≤ 4 would be the area under this line within the rectangle defined by 0 ≤ x ≤ 8 and 0 ≤ y ≤ 4. Wait, but if x goes up to 8, but the line x + y = 4 only goes up to x=4. So beyond x=4, the line x + y = 4 would have y negative, which isn't in our region. Therefore, the region x + y ≤ 4 within the rectangle is actually a triangle with vertices at (0, 0), (0, 4), and (4, 0). Because beyond x=4, even if y is 0, x=4 is the maximum for x + y ≤ 4. Wait, let me check that.

Let me take some example points. For example, (0,0): 0 + 0 = 0 ≤ 4, which is true. (0,4): 0 + 4 = 4 ≤ 4, which is true. (4, 0): 4 + 0 = 4 ≤ 4, which is true. Now, if I take a point like (5, 0): 5 + 0 = 5 > 4, which is not allowed. Similarly, (0, 5) is outside the original region because y can only go up to 4. So the region where x + y ≤ 4 is indeed a triangle with vertices at (0, 0), (0, 4), and (4, 0). Therefore, the area of this triangle is (base * height)/2.

Wait, base and height? Let me think. If we consider the triangle with vertices at (0,0), (0,4), and (4,0), the base can be along the y-axis from (0,0) to (0,4), which is 4 units. The height would be along the x-axis from (0,0) to (4,0), which is 4 units. But since it's a right triangle, the area is (base * height)/2 = (4 * 4)/2 = 8. Is that correct? Let me confirm with another approach.

Alternatively, the area under the line x + y = 4 from x=0 to x=4. For each x between 0 and 4, y can go from 0 up to 4 - x. Therefore, integrating with respect to x, the area would be the integral from 0 to 4 of (4 - x) dx. Let me compute that integral:

Integral of 4 dx from 0 to 4 is 4x evaluated from 0 to 4, which is 4*4 - 4*0 = 16.

Integral of x dx from 0 to 4 is (1/2)x² evaluated from 0 to 4, which is (1/2)(16) - 0 = 8.

Therefore, the integral of (4 - x) dx from 0 to 4 is 16 - 8 = 8. Yep, that matches the area of the triangle. So that area is 8.

Therefore, the probability is area of favorable region divided by total area, so 8 / 32. Simplifying that gives 1/4. Wait, is that correct? Wait, 8 divided by 32 is indeed 1/4. But wait, let me double-check everything because sometimes there might be a mistake in the area of the region.

Wait, the total area is 8*4=32, correct. The region where x + y ≤ 4 is a triangle with area 8, so the probability is 8/32=1/4. But let me make sure that there isn't any other area beyond that triangle where x + y ≤ 4. Wait, since x can go up to 8, but if x is greater than 4, say x=5, then y would have to be negative to satisfy x + y ≤4, but since y is non-negative, there are no such points. So indeed, the entire region where x + y ≤4 is that triangle. Therefore, area is 8, so probability is 1/4. Therefore, is the answer 1/4? Wait, but I need to make sure.

Alternatively, maybe I can think of the coordinates. Let me visualize the rectangle with x from 0 to 8 and y from 0 to 4. The line x + y = 4 cuts through the rectangle. At x=0, y=4; at y=0, x=4. So that line is only within the rectangle from (0,4) to (4,0). So the area below that line is a triangle, as previously thought. So yes, area is 8. Hence, probability is 8/32=1/4.

Wait, but just to be thorough, let me consider if there are other regions. For example, if the line x + y = 4 extended beyond the rectangle? Well, no. Since in the rectangle, the maximum x is 8 and maximum y is 4. The line x + y =4 would not extend beyond (4,0) in the x-direction because if you go beyond x=4, say x=5, then y would have to be -1, which is outside the rectangle. Similarly, if y=5, which is outside the rectangle, x would have to be -1. Therefore, there's no other regions in the rectangle that satisfy x + y ≤4 except that triangle.

Therefore, probability is 1/4. Wait, but just to be 100% sure, maybe I can compute the area another way. Suppose I think about the region x ≥0, y ≥0, x ≤8, y ≤4, and x + y ≤4. So this is the intersection of all these regions. Since x and y are positive and their sum is less than or equal to 4, so x can go from 0 to 4, and for each x, y goes from 0 to 4 -x. Therefore, the area is the integral from x=0 to 4 of (4 -x) dx, which we already did and got 8. Therefore, same answer. So the area is 8. Then the total area is 32. So 8/32=1/4.

But wait, let me check if the rectangle is from 0 to 8 in x and 0 to 4 in y. So the area is 8*4=32. So the probability is 1/4. So answer is 1/4? Wait, but maybe I made a mistake here. Maybe I need to think again.

Alternatively, let me think about the coordinates. For example, if I have a square (but it's actually a rectangle) with sides 8 and 4. The line x + y =4 divides the rectangle. The area where x + y ≤4 is a triangle with area 8 as calculated before. But let me think again: if I consider the rectangle from (0,0) to (8,4). The line x + y =4 intersects the rectangle at (0,4) and (4,0). So the area below the line is the triangle with vertices at (0,0), (0,4), (4,0). So yes, as before. Therefore, area is 8. Hence, probability is 8/32=1/4.

But let me think if there's a different interpretation. For instance, maybe the problem is considering continuous probability, so it's uniform distribution over the rectangle. Therefore, probability is the area ratio. So I think that is correct. Therefore, maybe the answer is 1/4. But wait, let me check with another method.

Alternatively, maybe I can use geometry. The region x + y ≤4 in the first quadrant is a triangle with legs of length 4. Therefore, area is (4*4)/2=8. Then divide by the total area of the rectangle, which is 8*4=32, so 8/32=1/4. So seems consistent. Therefore, the probability is 1/4.

Wait, but I remember sometimes problems like this can have different answers if the line intersects the rectangle in a different way. Let me check if the line x + y =4 actually intersects the rectangle. At x=0, y=4 which is on the top edge of the rectangle. At y=0, x=4 which is on the right edge of the rectangle (since x goes up to 8). So yes, the line goes from the top left corner (0,4) to the middle of the right edge (4,0). Therefore, the area under the line is the triangle. Therefore, yes, area is 8.

Alternatively, maybe I can use coordinates to calculate the area. Let's think of coordinates. The triangle has coordinates (0,0), (0,4), (4,0). Using the formula for area of a triangle with coordinates, which is |(x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2))/2|. Let me plug in the coordinates:

x1 = 0, y1 = 0; x2 = 0, y2 = 4; x3 = 4, y3 = 0.

So area is |0*(4 - 0) + 0*(0 - 0) + 4*(0 - 4)| / 2 = |0 + 0 + 4*(-4)| / 2 = | -16 | / 2 = 16 / 2 = 8. So same result. Therefore, area is 8. So probability is 8/32 = 1/4. Therefore, the probability is 1/4.

But now, just to be absolutely thorough, let me consider if there's a possibility that I have misread the problem. The problem says "a point (x, y) is randomly selected such that 0 ≤ x ≤ 8 and 0 ≤ y ≤ 4". So it's the rectangle as I considered. The question is the probability that x + y ≤ 4. So, yes, the area where x + y ≤4 is 8, total area 32. Therefore, answer is 1/4. So maybe the answer is 1/4? Wait, but wait, let me check with some other approach.

Alternatively, think of it as a uniform distribution over the rectangle. So the probability density function is 1/(area) = 1/32. The probability that x + y ≤4 is the integral over x from 0 to 4, and for each x, y from 0 to 4 - x. So integrating 1/32 over that area. Which is the same as area times 1/32, so 8*(1/32) = 1/4. Yep, same answer.

Alternatively, maybe think of it as a ratio. Since the region is a triangle with area 8, and total area is 32, so ratio is 1/4. Seems straightforward.

But now I'm thinking, maybe the problem is from a math competition or something, and maybe the answer is different? Wait, maybe I need to check with specific values. For example, if I take x from 0 to 4, and y from 0 to 4 - x. Let me compute the area again.

For x between 0 and 4, y goes from 0 to 4 - x. Therefore, the area is ∫₀⁴ (4 - x) dx = [4x - (x²)/2]₀⁴ = (16 - 8) - 0 = 8. Yep, same answer. So that seems right.

Alternatively, if I consider that for each x between 0 and 4, the length in y is decreasing from 4 to 0. So average length is (4 + 0)/2 = 2. Therefore, area is average length multiplied by the range in x, which is 4. Therefore, 2 * 4 = 8. Therefore, same answer. So that's another way to see it.

Therefore, I think this is solid. The probability is 1/4. Therefore, the answer should be \boxed{\dfrac{1}{4}}.

Wait, but just to be 100% sure, let me check with a different approach. Imagine if the rectangle was a square, say 4x4. Then the area where x + y ≤4 would be 8, and total area 16, so probability 1/2. But in our case, the rectangle is 8x4, so it's twice as wide. However, the line x + y =4 is still only intersecting at x=4. So the area under the line is same as in the 4x4 square? Wait, no. Wait, if the rectangle was 4x4, then the area where x + y ≤4 would be the same triangle, area 8. But total area is 16, so probability 1/2. But in our problem, the rectangle is 8x4, so the same triangle area is 8, total area 32, so probability 1/4. So that seems to make sense. It's like the rectangle is twice as wide, so the probability is half of the previous case. So that seems logical.

Alternatively, if I think of scaling. If you have a rectangle that's twice as wide but same height, then the area where x + y ≤4 would not scale in the same way. But since the line x + y =4 is fixed, the area under it would be same as in the smaller rectangle. So in our case, the area is 8, total area is 32. So 1/4. So yes.

Alternatively, if I have a rectangle that's 8x4, and I want to find the area where x + y ≤4. Since x can be up to 8, but the condition x + y ≤4 restricts x to be at most 4. So the x ranges from 0 to 4, and y from 0 to 4 -x. So that's the same as if we had a smaller rectangle 4x4. So the area is same as 4x4 square's area where x + y ≤4, which is 8, but since the total area is 32 instead of 16, the probability is 8/32=1/4. So that makes sense.

Therefore, I think I have verified this multiple ways, so I can be confident that the answer is 1/4.

**Final Answer**
The probability is \boxed{\dfrac{1}{4}}.

To solve this problem, we are asked to find the probability that a randomly selected point $(x, y)$ in the region defined by $0 \le x \le 8$ and $0 \le y \le 4$ satisfies the condition:

$$
x + y \le 4
$$

---

### **Step 1: Understand the Total Region**

We are selecting points from a rectangular region in the $xy$-plane with:

- $x$ ranging from 0 to 8
- $y$ ranging from 0 to 4

This defines a **rectangle** with:

$$
\text{Length along } x\text{-axis} = 8, \quad \text{Length along } y\text{-axis} = 4
$$

So the **total area** of this rectangle is:

$$
\text{Total Area} = 8 \times 4 = 32
$$

---

### **Step 2: Understand the Favorable Region**

We now want to determine the **area of the region** where the condition $x + y \le 4$ is satisfied, given the same domain $0 \le x \le 8$, $0 \le y \le 4$.

This inequality describes the area **below** the line $x + y = 4$ in the first quadrant.

Let's graphically consider the bounds of this region:

- The line $x + y = 4$ intersects the $x$-axis at $(4, 0)$ and the $y$-axis at $(0, 4)$
- The region $x + y \le 4$ in the positive quadrant is the **triangle** with vertices at:

  $$
  (0, 0), \quad (0, 4), \quad (4, 0)
  $$

So, this is a **right triangle** with base = 4 and height = 4 (along the axes), and its area is:

$$
\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 4 = 8
$$

---

### **Step 3: Calculate the Probability**

Now that we have both the **favorable area** (where $x + y \le 4$) and the **total area** (of the rectangle), the desired probability is:

$$
\text{Probability} = \frac{\text{Favorable Area}}{\text{Total Area}} = \frac{8}{32} = \frac{1}{4}
$$

---

### **Final Answer**

$$
\boxed{\dfrac{1}{4}}
$$

[Assistant]
Explanation:
Okay, so I need to find the probability that a randomly selected point (x, y) in the region 0 ≤ x ≤ 8 and 0 ≤ y ≤ 4 satisfies the condition x + y ≤ 4. Hmm, let me think about how to approach this.

First, probability in this context is usually the area of the region where the condition holds divided by the total area of the possible region. So I should probably start by figuring out the total area of the region where the point can be selected. The problem says 0 ≤ x ≤ 8 and 0 ≤ y ≤ 4. So that's a rectangle, right? The length along the x-axis is 8 units, and along the y-axis is 4 units. Therefore, the total area should be 8 multiplied by 4, which is 32. Let me write that down:

Total area = 8 * 4 = 32.

Okay, now I need to find the area of the region where x + y ≤ 4. This is a linear inequality, so the region should be a polygon. Since we're dealing with x and y both being non-negative, the region defined by x + y ≤ 4 should be a triangle or a trapezoid, maybe? Let me sketch it mentally.

If I consider the first quadrant (since x and y are non-negative), the line x + y = 4 is a straight line starting from (0, 4) to (4, 0). So the area where x + y ≤ 4 would be the area under this line within the rectangle defined by 0 ≤ x ≤ 8 and 0 ≤ y ≤ 4. Wait, but if x goes up to 8, but the line x + y = 4 only goes up to x=4. So beyond x=4, the line x + y = 4 would have y negative, which isn't in our region. Therefore, the region x + y ≤ 4 within the rectangle is actually a triangle with vertices at (0, 0), (0, 4), and (4, 0). Because beyond x=4, even if y is 0, x=4 is the maximum for x + y ≤ 4. Wait, let me check that.

Let me take some example points. For example, (0,0): 0 + 0 = 0 ≤ 4, which is true. (0,4): 0 + 4 = 4 ≤ 4, which is true. (4, 0): 4 + 0 = 4 ≤ 4, which is true. Now, if I take a point like (5, 0): 5 + 0 = 5 > 4, which is not allowed. Similarly, (0, 5) is outside the original region because y can only go up to 4. So the region where x + y ≤ 4 is indeed a triangle with vertices at (0, 0), (0, 4), and (4, 0). Therefore, the area of this triangle is (base * height)/2.

Wait, base and height? Let me think. If we consider the triangle with vertices at (0,0), (0,4), and (4,0), the base can be along the y-axis from (0,0) to (0,4), which is 4 units. The height would be along the x-axis from (0,0) to (4,0), which is 4 units. But since it's a right triangle, the area is (base * height)/2 = (4 * 4)/2 = 8. Is that correct? Let me confirm with another approach.

Alternatively, the area under the line x + y = 4 from x=0 to x=4. For each x between 0 and 4, y can go from 0 up to 4 - x. Therefore, integrating with respect to x, the area would be the integral from 0 to 4 of (4 - x) dx. Let me compute that integral:

Integral of 4 dx from 0 to 4 is 4x evaluated from 0 to 4, which is 4*4 - 4*0 = 16.

Integral of x dx from 0 to 4 is (1/2)x² evaluated from 0 to 4, which is (1/2)(16) - 0 = 8.

Therefore, the integral of (4 - x) dx from 0 to 4 is 16 - 8 = 8. Yep, that matches the area of the triangle. So that area is 8.

Therefore, the probability is area of favorable region divided by total area, so 8 / 32. Simplifying that gives 1/4. Wait, is that correct? Wait, 8 divided by 32 is indeed 1/4. But wait, let me double-check everything because sometimes there might be a mistake in the area of the region.

Wait, the total area is 8*4=32, correct. The region where x + y ≤ 4 is a triangle with area 8, so the probability is 8/32=1/4. But let me make sure that there isn't any other area beyond that triangle where x + y ≤ 4. Wait, since x can go up to 8, but if x is greater than 4, say x=5, then y would have to be negative to satisfy x + y ≤4, but since y is non-negative, there are no such points. So indeed, the entire region where x + y ≤4 is that triangle. Therefore, area is 8, so probability is 1/4. Therefore, is the answer 1/4? Wait, but I need to make sure.

Alternatively, maybe I can think of the coordinates. Let me visualize the rectangle with x from 0 to 8 and y from 0 to 4. The line x + y = 4 cuts through the rectangle. At x=0, y=4; at y=0, x=4. So that line is only within the rectangle from (0,4) to (4,0). So the area below that line is a triangle, as previously thought. So yes, area is 8. Hence, probability is 8/32=1/4.

Wait, but just to be thorough, let me consider if there are other regions. For example, if the line x + y = 4 extended beyond the rectangle? Well, no. Since in the rectangle, the maximum x is 8 and maximum y is 4. The line x + y =4 would not extend beyond (4,0) in the x-direction because if you go beyond x=4, say x=5, then y would have to be -1, which is outside the rectangle. Similarly, if y=5, which is outside the rectangle, x would have to be -1. Therefore, there's no other regions in the rectangle that satisfy x + y ≤4 except that triangle.

Therefore, probability is 1/4. Wait, but just to be 100% sure, maybe I can compute the area another way. Suppose I think about the region x ≥0, y ≥0, x ≤8, y ≤4, and x + y ≤4. So this is the intersection of all these regions. Since x and y are positive and their sum is less than or equal to 4, so x can go from 0 to 4, and for each x, y goes from 0 to 4 -x. Therefore, the area is the integral from x=0 to 4 of (4 -x) dx, which we already did and got 8. Therefore, same answer. So the area is 8. Then the total area is 32. So 8/32=1/4.

But wait, let me check if the rectangle is from 0 to 8 in x and 0 to 4 in y. So the area is 8*4=32. So the probability is 1/4. So answer is 1/4? Wait, but maybe I made a mistake here. Maybe I need to think again.

Alternatively, let me think about the coordinates. For example, if I have a square (but it's actually a rectangle) with sides 8 and 4. The line x + y =4 divides the rectangle. The area where x + y ≤4 is a triangle with area 8 as calculated before. But let me think again: if I consider the rectangle from (0,0) to (8,4). The line x + y =4 intersects the rectangle at (0,4) and (4,0). So the area below the line is the triangle with vertices at (0,0), (0,4), (4,0). So yes, as before. Therefore, area is 8. Hence, probability is 8/32=1/4.

But let me think if there's a different interpretation. For instance, maybe the problem is considering continuous probability, so it's uniform distribution over the rectangle. Therefore, probability is the area ratio. So I think that is correct. Therefore, maybe the answer is 1/4. But wait, let me check with another method.

Alternatively, maybe I can use geometry. The region x + y ≤4 in the first quadrant is a triangle with legs of length 4. Therefore, area is (4*4)/2=8. Then divide by the total area of the rectangle, which is 8*4=32, so 8/32=1/4. So seems consistent. Therefore, the probability is 1/4.

Wait, but I remember sometimes problems like this can have different answers if the line intersects the rectangle in a different way. Let me check if the line x + y =4 actually intersects the rectangle. At x=0, y=4 which is on the top edge of the rectangle. At y=0, x=4 which is on the right edge of the rectangle (since x goes up to 8). So yes, the line goes from the top left corner (0,4) to the middle of the right edge (4,0). Therefore, the area under the line is the triangle. Therefore, yes, area is 8.

Alternatively, maybe I can use coordinates to calculate the area. Let's think of coordinates. The triangle has coordinates (0,0), (0,4), (4,0). Using the formula for area of a triangle with coordinates, which is |(x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2))/2|. Let me plug in the coordinates:

x1 = 0, y1 = 0; x2 = 0, y2 = 4; x3 = 4, y3 = 0.

So area is |0*(4 - 0) + 0*(0 - 0) + 4*(0 - 4)| / 2 = |0 + 0 + 4*(-4)| / 2 = | -16 | / 2 = 16 / 2 = 8. So same result. Therefore, area is 8. So probability is 8/32 = 1/4. Therefore, the probability is 1/4.

But now, just to be absolutely thorough, let me consider if there's a possibility that I have misread the problem. The problem says "a point (x, y) is randomly selected such that 0 ≤ x ≤ 8 and 0 ≤ y ≤ 4". So it's the rectangle as I considered. The question is the probability that x + y ≤ 4. So, yes, the area where x + y ≤4 is 8, total area 32. Therefore, answer is 1/4. So maybe the answer is 1/4? Wait, but wait, let me check with some other approach.

Alternatively, think of it as a uniform distribution over the rectangle. So the probability density function is 1/(area) = 1/32. The probability that x + y ≤4 is the integral over x from 0 to 4, and for each x, y from 0 to 4 - x. So integrating 1/32 over that area. Which is the same as area times 1/32, so 8*(1/32) = 1/4. Yep, same answer.

Alternatively, maybe think of it as a ratio. Since the region is a triangle with area 8, and total area is 32, so ratio is 1/4. Seems straightforward.

But now I'm thinking, maybe the problem is from a math competition or something, and maybe the answer is different? Wait, maybe I need to check with specific values. For example, if I take x from 0 to 4, and y from 0 to 4 - x. Let me compute the area again.

For x between 0 and 4, y goes from 0 to 4 - x. Therefore, the area is ∫₀⁴ (4 - x) dx = [4x - (x²)/2]₀⁴ = (16 - 8) - 0 = 8. Yep, same answer. So that seems right.

Alternatively, if I consider that for each x between 0 and 4, the length in y is decreasing from 4 to 0. So average length is (4 + 0)/2 = 2. Therefore, area is average length multiplied by the range in x, which is 4. Therefore, 2 * 4 = 8. Therefore, same answer. So that's another way to see it.

Therefore, I think this is solid. The probability is 1/4. Therefore, the answer should be \boxed{\dfrac{1}{4}}.

Wait, but just to be 100% sure, let me check with a different approach. Imagine if the rectangle was a square, say 4x4. Then the area where x + y ≤4 would be 8, and total area 16, so probability 1/2. But in our case, the rectangle is 8x4, so it's twice as wide. However, the line x + y =4 is still only intersecting at x=4. So the area under the line is same as in the 4x4 square? Wait, no. Wait, if the rectangle was 4x4, then the area where x + y ≤4 would be the same triangle, area 8. But total area is 16, so probability 1/2. But in our problem, the rectangle is 8x4, so the same triangle area is 8, total area 32, so probability 1/4. So that seems to make sense. It's like the rectangle is twice as wide, so the probability is half of the previous case. So that seems logical.

Alternatively, if I think of scaling. If you have a rectangle that's twice as wide but same height, then the area where x + y ≤4 would not scale in the same way. But since the line x + y =4 is fixed, the area under it would be same as in the smaller rectangle. So in our case, the area is 8, total area is 32. So 1/4. So yes.

Alternatively, if I have a rectangle that's 8x4, and I want to find the area where x + y ≤4. Since x can be up to 8, but the condition x + y ≤4 restricts x to be at most 4. So the x ranges from 0 to 4, and y from 0 to 4 -x. So that's the same as if we had a smaller rectangle 4x4. So the area is same as 4x4 square's area where x + y ≤4, which is 8, but since the total area is 32 instead of 16, the probability is 8/32=1/4. So that makes sense.

Therefore, I think I have verified this multiple ways, so I can be confident that the answer is 1/4.

**Final Answer**
The probability is \boxed{\dfrac{1}{4}}.

To solve this problem, we are asked to find the probability that a randomly selected point $(x, y)$ in the region defined by $0 \le x \le 8$ and $0 \le y \le 4$ satisfies the condition:

$$
x + y \le 4
$$

---

### **Step 1: Understand the Total Region**

We are selecting points from a rectangular region in the $xy$-plane with:

- $x$ ranging from 0 to 8
- $y$ ranging from 0 to 4

This defines a **rectangle** with:

$$
\text{Length along } x\text{-axis} = 8, \quad \text{Length along } y\text{-axis} = 4
$$

So the **total area** of this rectangle is:

$$
\text{Total Area} = 8 \times 4 = 32
$$

---

### **Step 2: Understand the Favorable Region**

We now want to determine the **area of the region** where the condition $x + y \le 4$ is satisfied, given the same domain $0 \le x \le 8$, $0 \le y \le 4$.

This inequality describes the area **below** the line $x + y = 4$ in the first quadrant.

Let's graphically consider the bounds of this region:

- The line $x + y = 4$ intersects the $x$-axis at $(4, 0)$ and the $y$-axis at $(0, 4)$
- The region $x + y \le 4$ in the positive quadrant is the **triangle** with vertices at:

  $$
  (0, 0), \quad (0, 4), \quad (4, 0)
  $$

So, this is a **right triangle** with base = 4 and height = 4 (along the axes), and its area is:

$$
\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 4 = 8
$$

---

### **Step 3: Calculate the Probability**

Now that we have both the **favorable area** (where $x + y \le 4$) and the **total area** (of the rectangle), the desired probability is:

$$
\text{Probability} = \frac{\text{Favorable Area}}{\text{Total Area}} = \frac{8}{32} = \frac{1}{4}
$$

---

### **Final Answer**

$$
\boxed{\dfrac{1}{4}}
$$

Solution:
Rewrite $x+y\leq 4$ as $y\leq 4-x$.  This inequality is satisfied by the points on and under the line $y=4-x$.  Sketching this line along with the $4\times 8$ rectangle determined by the inequalities $0\leq x\leq 8$ and $0\leq y\leq 4$, we find that the points satisfying $x+y\leq 4$ are those in the shaded triangle (see figure).  The  area of the triangle is $\frac{1}{2}(4)(4)=8$ square units, and the area of the rectangle is $(4)(8)=32$ square units, so the probability that a randomly selected point would fall in the shaded triangle is $\boxed{\frac{1}{4}}$.

[asy] import graph; size(200); defaultpen(linewidth(0.7)+fontsize(10));

dotfactor=4;

real f(real x) { return 4-x; }

pair A=(0,4), B=(8,4), C=(8,0), D=(0,0); pair[] dots={A,B,C,D};

fill(A--(4,0)--D--cycle,gray(0.7)); draw(A--B--C);

xaxis(xmin=-3,xmax=9,Ticks(" ",1.0, begin=false, end=false, NoZero, Size=3), Arrows(4), above=true);

yaxis(ymin=-1,ymax=5,Ticks(" ",1.0,begin=false, end=false, NoZero, Size=3), Arrows(4), above=true);

draw(graph(f,-0.8,4.5),Arrows(4)); label("$x+y=4$",(-2.2,5.2));[/asy]