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, I need to figure out two things: the total number of possible three-point sets, and the number of those sets that are colinear (lie on the same straight line). Then, the probability would be the number of colinear triplets divided by the total number of triplets.

Let me start with the total number of possible three-point sets. Since there are 9 points, the number of ways to choose 3 is given by the combination formula C(n, k) = n! / (k! * (n - k)!). So, plugging in n=9 and k=3:

Total triplets = C(9, 3) = 9! / (3! * 6!) = (9 * 8 * 7) / (3 * 2 * 1) = 84. So there are 84 possible ways to choose three points.

Now, the next part is figuring out how many of these triplets are colinear. This seems a bit trickier. Let me visualize the grid. It's a 3x3 grid, so like a tic-tac-toe board. The points are arranged in rows, columns, and diagonals. But there might be other lines as well, like diagonals with different slopes or maybe other lines that pass through three points.

First, let me count the obvious ones: the rows, columns, and the two main diagonals.

Starting with rows: there are 3 rows, each containing 3 points. So, each row contributes 1 colinear triplet. Therefore, 3 rows * 1 = 3 triplets.

Similarly, columns: there are 3 columns, each with 3 points. So another 3 triplets. So that's 3 + 3 = 6 so far.

Then the main diagonals: there are two main diagonals, each with 3 points. So that's 2 more triplets. So total so far is 6 + 2 = 8.

But wait, are there any other lines that pass through three points? For example, lines with different slopes?

Let me think. In a 3x3 grid, are there other lines that contain exactly three points?

For example, consider lines with slope 1/2 or 2, or maybe other slopes. Let me check.

Let me try to visualize the grid. Let me label the points with coordinates. Let's say the grid has coordinates from (0,0) to (2,2). 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 think about lines that pass through three points. We already considered horizontal, vertical, and main diagonals. Are there other lines?

For example, consider the line that goes through (0,1), (1,2), and (2,1). Wait, does that line pass through three points? Let me check.

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

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

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

What about lines with slope 2? Let's see. For example, starting at (0,0), going up 2, right 1: (0,0), (1,2). But then the next point would be (2,4), which is outside the grid. So only two points. Similarly, starting at (0,1), going up 2, right 1: (0,1), (1,3) which is outside. Not helpful.

How about slope 1/2? Starting at (0,0), going right 2, up 1: (0,0), (2,1). Then next would be (4,2), outside. So only two points. Similarly, starting at (0,1), going right 2, up 1: (0,1), (2,2). Again, only two points. Similarly, starting at (1,0), going right 2, up 1: (1,0), (3,1) which is outside. So no.

What about slope -2? Starting at (0,2), going right 1, down 2: (0,2), (1,0). Then next would be (2, -2), outside. So only two points. Similarly, slope -1/2? Starting at (0,2), going right 2, down 1: (0,2), (2,1). Then next would be (4,0), outside. So again, only two points.

So maybe there are no lines with slopes other than 0, infinity, 1, -1 that pass through three points?

Wait, but maybe there are other lines? For example, consider the line that goes through (0,0), (1,1), (2,2) – that's already counted. How about (0,1), (1,0), (2,-1)? But (2,-1) is outside the grid. Similarly, (0,1), (1,2), (2,3) – again outside. So no.

Wait, but maybe there are lines that aren't aligned with the grid or main diagonals but still pass through three points? Let me think. For example, is there a line that goes through (0,0), (1,2), and another point?

Let me check. The line through (0,0) and (1,2) has slope 2. The equation is y = 2x. Let's see if any other grid points lie on this line. For x=2, y=4, which is outside. For x=0.5, y=1, but that's not a grid point. So only two points.

How about the line through (0,1) and (1,0)? Slope is -1. Equation is y - 1 = -1(x - 0) => y = -x + 1. Let's check for other points. When x=2, y = -2 + 1 = -1, which is outside. When x= -1, y=2, which is outside. So only two points.

How about the line through (0,0) and (2,1)? Slope is 1/2. Equation y = (1/2)x. For x=1, y=0.5, not a grid point. So only two points.

Similarly, line through (0,2) and (1,1): slope is -1. Equation y - 2 = -1(x - 0) => y = -x + 2. For x=2, y= -2 + 2 = 0. So (2,0) is on this line. So the points (0,2), (1,1), (2,0). Wait, that's the other main diagonal! Which we already counted. So that's already included in the two main diagonals.

Wait, so maybe there are no other lines with three points? Let me check another possibility. How about the line through (1,0), (1,1), (1,2)? That's a vertical line, already counted in columns. Similarly for horizontal lines.

Wait, so maybe the only lines that have three points are the rows, columns, and two main diagonals? So that would be 3 + 3 + 2 = 8 lines, each containing exactly one triplet. Therefore, there are 8 colinear triplets?

But wait, hold on. Wait, let me think again. Are there other lines? For example, in some grids, there are more lines. Wait, for example, in a 3x3 grid, there are also lines that are not aligned with the grid or the main diagonals but still pass through three points. Wait, but I can't think of any. Let me check again.

Wait, suppose we take points (0,0), (1,1), (2,2) – that's the main diagonal. Similarly, (0,2), (1,1), (2,0). What about points like (0,1), (1,1), (2,1)? That's a horizontal line, already counted. How about (1,0), (1,1), (1,2)? Vertical line, already counted.

Wait, maybe there are lines with different slopes? For example, let me check if there's a line that goes through (0,0), (1,2), and another point. Wait, we saw that (0,0), (1,2) is slope 2, and the next point would be (2,4), which is outside. So no. How about (0,1), (1,2), (2,3)? Again, (2,3) is outside. Similarly, (0,0), (2,1), (something)? The line through (0,0) and (2,1) is y = (1/2)x. The next point would be (4, 2), which is outside. So no.

Alternatively, maybe lines that are not straight? No, we are talking about straight lines.

Wait, maybe there are lines that pass through three points in a different way? For example, in some cases, maybe three points that are not in a straight line? No, we are looking for colinear.

Wait, maybe I need to check all possible lines. Let me think of all possible lines that can be formed by three points in the grid.

Alternatively, maybe there are more than 8 lines. Wait, let me think again. For example, in addition to the rows, columns, and diagonals, are there other lines?

Wait, here's an idea. What about the lines that go through the center point and two other points? For example, the center is (1,1). If I take (0,0), (1,1), (2,2) – that's the main diagonal. Similarly, (0,2), (1,1), (2,0). But what about other lines through the center? For example, (0,1), (1,1), (2,1) – that's a horizontal line. Similarly, (1,0), (1,1), (1,2) – vertical. So those are already counted.

What about lines that go through the center with other slopes? For example, slope 2? Let's see. The line through (1,1) with slope 2 would be y - 1 = 2(x - 1). So y = 2x - 1. Let's check if any other grid points lie on this line. For x=0, y = -1, which is outside. For x=1, y=1. For x=2, y=3, which is outside. So only (1,1) and maybe another? Wait, if x=0.5, y=0, but that's not a grid point. So no. Similarly, slope 1/2? y - 1 = (1/2)(x - 1). For x=0, y = 1 - 0.5 = 0.5, not a grid point. For x=1, y=1. For x=2, y=1 + 0.5 = 1.5, not a grid point. So again, only the center point.

How about slope -2? y - 1 = -2(x - 1). So y = -2x + 3. For x=0, y=3, outside. For x=1, y=1. For x=2, y= -1, outside. So again, only the center.

Slope -1/2? y - 1 = -1/2(x - 1). For x=0, y = 1 + 0.5 = 1.5, not a grid point. For x=1, y=1. For x=2, y = 1 - 0.5 = 0.5, not a grid point. So again, only the center.

So no other lines through the center with three points. So maybe there are no other lines?

Wait, but let me think again. What about lines that are not passing through the center? For example, take points (0,0), (1,1), (2,2) – already counted. How about (0,0), (0,1), (0,2) – vertical line, already counted. How about (0,0), (1,0), (2,0) – horizontal line, already counted. How about (0,0), (1,2), (2,1)? Wait, earlier I thought that wasn't colinear, but let me check again.

Wait, (0,0), (1,2), (2,1). Let me check if they are colinear. 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. Since the slopes are different, they are not colinear. So that's not a line.

How about (0,1), (1,0), (2, -1)? Again, (2, -1) is outside. Not valid.

How about (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, already counted.

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

Is there any other line? Let me think of all possible combinations. Maybe there are lines that are not aligned with the grid or diagonals but still pass through three points. Wait, for example, in a 3x3 grid, is there a line that goes through three points with slope 2/1? Wait, we saw that slope 2 only gives two points. Similarly for slope 1/2. How about slope 3? That would be even steeper. For example, from (0,0) to (1,3), but (1,3) is outside. Similarly, slope 1/3? From (0,0) to (3,1), which is outside. So no.

Alternatively, maybe there are lines that are not straight? No, we are talking about straight lines.

Wait, maybe I need to check all possible lines. Let me think of all possible lines that can be formed by three points. Since there are only 9 points, maybe there's a systematic way to check.

First, the rows: 3 lines.

Columns: 3 lines.

Diagonals: 2 lines.

Now, are there any other lines? Let me think of lines that have exactly three points. For example, in some cases, maybe a line that is not horizontal, vertical, or diagonal but still passes through three points. Wait, but in a 3x3 grid, is that possible?

Wait, here's an idea. Let me check if there are any other lines. For example, take points (0,0), (1,1), (2,2) – that's the main diagonal. What about points (0,1), (1,2), (2,3)? But (2,3) is outside. Similarly, (0,1), (1,0), (2,-1) – outside. How about (1,0), (1,1), (1,2) – vertical, already counted.

Wait, how about points (0,0), (1,2), (2,1)? Wait, we checked that before, they are not colinear. How about (0,0), (2,1), (something)? The line through (0,0) and (2,1) is y = (1/2)x. The only grid points on this line are (0,0) and (2,1). Similarly, the line through (0,1) and (2,0) is y = (-1/2)x + 1. The grid points on this line are (0,1), (1, 0.5) which is not a grid point, and (2, 0). So only two points.

Wait, so maybe there are no other lines with three points? Then the total number of colinear triplets is 8?

But wait, I remember that in some problems, there are more lines. Wait, let me check again. Maybe I missed some diagonals?

Wait, for example, in a 3x3 grid, there are also diagonals of different lengths. But we are only considering lines that have exactly three points. So, for example, the main diagonals have three points. But are there other diagonals? For example, in a 3x3 grid, if you consider diagonals of length 2, but they don't have three points. So no.

Wait, but maybe there are other lines with three points? Let me think of coordinates. Let me list all possible lines that pass through at least two points and check if they pass through a third.

Let me take all pairs of points and see if they form a line with a third point.

But that might take a while, but maybe manageable.

First, there are C(9, 2) = 36 pairs of points. For each pair, check if there's a third point on the same line.

But maybe there's a smarter way. Let me think.

Alternatively, think of all possible lines that can be formed in the grid. For a 3x3 grid, the number of lines is more than 8? Wait, according to some references, in a 3x3 grid, there are 8 lines that contain exactly three points. But I need to confirm.

Wait, actually, I think there are more. Wait, for example, in addition to the rows, columns, and two main diagonals, there are also four more lines that have three points. Wait, how?

Wait, let me think of the lines that go through the center and two other points. Wait, but we already considered those. For example, the horizontal, vertical, and main diagonals. But maybe there are other lines through the center?

Wait, for example, if you take the center (1,1) and the points (0,1) and (2,1), that's the horizontal line. Similarly, (1,0), (1,1), (1,2) is vertical. The main diagonals are (0,0), (1,1), (2,2) and (0,2), (1,1), (2,0). Are there other lines through the center?

For example, if I take (0,1), (1,1), (2,1) – horizontal. (1,0), (1,1), (1,2) – vertical. What about (0,0), (1,1), (2,2) – main diagonal. (0,2), (1,1), (2,0) – other main diagonal. What about other lines?

Wait, suppose I take (0,0), (1,1), (2,2) – already counted. How about (0,1), (1,2), (2,3)? But (2,3) is outside. How about (0,1), (1,0), (2,-1)? Outside. How about (1,0), (1,1), (1,2) – vertical. How about (0,0), (1,0), (2,0) – horizontal. How about (0,2), (1,2), (2,2) – horizontal. How about (0,0), (0,1), (0,2) – vertical. How about (2,0), (2,1), (2,2) – vertical. So all the rows and columns are covered.

Wait, maybe there are other lines with three points? For example, if you take points (0,0), (1,2), and (2,1). Wait, we checked that before, they are not colinear. How about (0,0), (2,1), and (something)? The line through (0,0) and (2,1) is y = (1/2)x. The only grid points on this line are (0,0) and (2,1). Similarly, the line through (0,1) and (2,0) is y = (-1/2)x + 1. The grid points on this line are (0,1) and (2,0). So no third point.

How about the line through (0,0) and (1,1)? That's the main diagonal, already counted. Similarly for others.

Wait, maybe there are lines that are not aligned with the grid or diagonals but still pass through three points? For example, in a 3x3 grid, is there a line that goes through (0,0), (1,1), (2,2)? Yes, that's the main diagonal. What about (0,1), (1,2), (2,3)? No, (2,3) is outside. How about (1,0), (1,1), (1,2)? Vertical line, already counted.

Wait, maybe I need to think of all possible lines. Let me think of all possible lines that can be formed by three points. Since there are only 9 points, maybe there are only 8 such lines. But I have a feeling that there are more. Wait, let me check online... Wait, no, I can't do that. I need to figure it out myself.

Wait, here's another approach. Let me count all possible lines with three points. For that, I can think of all possible directions.

First, horizontal lines: 3 lines, each with 3 points.

Vertical lines: 3 lines, each with 3 points.

Diagonals with slope 1: 1 line (from (0,0) to (2,2)).

Diagonals with slope -1: 1 line (from (0,2) to (2,0)).

Are there any other lines with three points?

Wait, what about lines with slope 2? As discussed earlier, they only pass through two points.

Slope 1/2? Also only two points.

Slope -2? Two points.

Slope -1/2? Two points.

Slope 3? Outside.

Slope 1/3? Outside.

Slope -3? Outside.

Slope -1/3? Outside.

So no other slopes with three points.

Wait, but what about lines that are not straight? No, we are talking about straight lines.

Wait, but maybe there are other diagonals? For example, in a 3x3 grid, there are also diagonals of different lengths. But they don't have three points. For example, the diagonal from (0,1) to (1,2) is only two points. Similarly, from (1,0) to (2,1), etc. So those don't have three points.

Wait, so maybe there are only 8 lines with three points? Then the number of colinear triplets is 8? But that seems low. Wait, let me check again.

Wait, if there are 8 lines, each containing exactly one triplet, then there are 8 favorable cases. But I have a feeling that there are more. Wait, let me think again. For example, in the 3x3 grid, there are also lines that are not aligned with the grid or the main diagonals but still pass through three points. Wait, but how?

Wait, here's an idea. Maybe there are lines that pass through three points with different slopes. Wait, for example, take the points (0,0), (1,1), (2,2) – slope 1. Then (0,1), (1,2), (2,3) – but (2,3) is outside. How about (0,0), (1,2), (2, something)? As before, only two points.

Wait, maybe there are lines that pass through three points in a different way. Wait, for example, if you take the points (0,0), (1,1), (2,2) – that's one line. What about (0,0), (1,0), (2,0) – horizontal. (0,0), (0,1), (0,2) – vertical. Similarly for others. So maybe there are only 8 lines?

But I remember that in some problems, like the number of lines in a 3x3 grid, it's more than 8. Wait, let me think. For example, in the 3x3 grid, how many lines are there? Each line is determined by two points. But some lines are counted multiple times. Wait, but if I think of all possible lines, there are more than 8. For example, there are lines that have only two points, but we are only interested in those that have exactly three points.

Wait, so maybe there are only 8 lines with exactly three points? Let me check again. For example, rows: 3, columns: 3, diagonals: 2. Total 8. So that would mean 8 colinear triplets. Therefore, the probability would be 8 / 84 = 2 / 21 ≈ 0.095. But I have a feeling that the answer is different. Wait, maybe I missed some lines?

Wait, let me think of another way. Let me count all possible lines with three points. For that, maybe think of all possible lines that can be formed by three points. For example, in addition to the rows, columns, and main diagonals, are there any other lines?

Wait, here's a thought. What about the lines that go through the center and two other points not aligned with the grid? For example, the center (1,1). If I take (0,0) and (2,2), that's the main diagonal. If I take (0,2) and (2,0), that's the other main diagonal. What about (0,1) and (2,1)? That's horizontal, already counted. Similarly, (1,0) and (1,2) is vertical. What about (0,0) and (1,2)? That's a line with slope 2, but only two points. Similarly, (0,0) and (2,1) – slope 1/2, only two points. So no.

Wait, maybe there are lines that pass through three points if you consider different directions? Wait, for example, in a 3x3 grid, there are also lines that are not straight? No, we are talking about straight lines.

Wait, maybe I need to think of all possible triplets and check if they are colinear. But that would take time, but maybe manageable.

Total triplets: 84. Let me think of how many colinear triplets there are.

First, rows: 3 rows, each with C(3,3)=1 triplet. So 3.

Columns: same, 3.

Diagonals: 2, each with 1 triplet. So 2.

Total so far: 3 + 3 + 2 = 8.

But wait, are there any other lines? For example, suppose there is a line that isn't horizontal, vertical, or diagonal, but still has three points. Let me try to find such a line.

Take points (0,0), (1,1), (2,2) – already counted.

Take points (0,1), (1,2), (2,3) – (2,3) is outside.

Take points (1,0), (1,1), (1,2) – vertical, counted.

Take points (0,0), (1,0), (2,0) – horizontal, counted.

Take points (0,0), (0,1), (0,2) – vertical, counted.

Take points (0,2), (1,1), (2,0) – diagonal, counted.

Take points (0,1), (1,1), (2,1) – horizontal, counted.

Take points (1,0), (1,1), (1,2) – vertical, counted.

Take points (2,0), (2,1), (2,2) – horizontal, counted.

Take points (0,0), (1,2), (2,1) – not colinear.

Take points (0,1), (1,0), (2, -1) – outside.

Take points (1,0), (2,1), (3,2) – outside.

Take points (0,2), (1,1), (2,0) – diagonal, counted.

Take points (0,0), (2,1), (something)? Only two points.

Take points (0,1), (1,0), (2,-1) – outside.

Take points (1,1), (0,0), (2,2) – diagonal, counted.

Take points (1,1), (0,2), (2,0) – diagonal, counted.

Take points (1,1), (0,1), (2,1) – horizontal, counted.

Take points (1,1), (1,0), (1,2) – vertical, counted.

So after checking all these, I can't find any other lines with three points. Therefore, there are only 8 colinear triplets.

But wait, I remember that in some problems, like the one where you have to count the number of lines in a grid, there are more lines. Wait, for example, in a 3x3 grid, there are 12 lines: 3 horizontal, 3 vertical, 2 main diagonals, and 4 more diagonals? Wait, how?

Wait, maybe I was wrong before. Let me think again. For example, in addition to the main diagonals, there are also diagonals of different lengths. Wait, but they don't have three points. For example, the diagonal from (0,1) to (1,2) is only two points. Similarly, from (1,0) to (2,1). So those don't count. So maybe there are only 8 lines with three points.

But then why do I have a feeling that there are more? Maybe I'm confusing with another problem. Let me think of another approach.

Suppose I consider all possible lines that can be formed by three points. Each line is uniquely determined by its slope and intercept. Let me try to find all such lines.

First, horizontal lines: y = 0, y = 1, y = 2. Each has three points. So 3 lines.

Vertical lines: x = 0, x = 1, x = 2. Each has three points. So 3 lines.

Diagonals with slope 1: y = x, y = x - 1, y = x - 2. Wait, let's check:

For slope 1:

- Line y = x passes through (0,0), (1,1), (2,2). That's one line.

- Line y = x - 1 passes through (1,0), (2,1). Only two points.

- Line y = x - 2 passes through (2,0). Only one point.

Similarly, for slope 1, only one line with three points.

Similarly, for slope -1:

- Line y = -x + 2 passes through (0,2), (1,1), (2,0). That's one line.

- Line y = -x + 1 passes through (0,1), (1,0). Two points.

- Line y = -x + 0 passes through (0,0), (1,-1). Outside.

So only one line with slope -1.

So that's 1 + 1 = 2 diagonals.

Now, what about other slopes?

Slope 2:

- Line y = 2x passes through (0,0), (1,2). Next would be (2,4), outside. So only two points.

- Line y = 2x - 1 passes through (1,1), (2,3). (2,3) is outside. So only (1,1).

- Line y = 2x - 2 passes through (1,0), (2,2). So (1,0) and (2,2). Two points.

Slope 1/2:

- Line y = (1/2)x passes through (0,0), (2,1). Only two points.

- Line y = (1/2)x + 1 passes through (0,1), (2,2). Two points.

- Line y = (1/2)x - 1 passes through (2,0). One point.

Slope -2:

- Line y = -2x passes through (0,0), (1,-2). Outside.

- Line y = -2x + 2 passes through (0,2), (1,0). Two points.

- Line y = -2x + 1 passes through (0,1), (1,-1). Outside.

Slope -1/2:

- Line y = (-1/2)x passes through (0,0), (2, -1). Outside.

- Line y = (-1/2)x + 1 passes through (0,1), (2,0). Two points.

- Line y = (-1/2)x + 2 passes through (0,2), (2,1). Two points.

So for slopes other than 0, infinity, 1, -1, there are no lines with three points.

Therefore, total lines with three points are 3 horizontal + 3 vertical + 2 diagonals = 8 lines. Therefore, 8 colinear triplets.

Therefore, the number of favorable outcomes is 8, total outcomes is 84. Therefore, probability is 8/84 = 2/21 ≈ 0.095.

But wait, I have a doubt here because I remember that in some similar problems, the answer is different. Wait, let me check again. Maybe I missed some lines?

Wait, here's another thought. What about the lines that are not aligned with the grid but still pass through three points? For example, in a 3x3 grid, is there a line that goes through (0,0), (1,1), (2,2) – already counted. How about (0,0), (1,0), (2,0) – horizontal. How about (0,0), (0,1), (0,2) – vertical. How about (0,1), (1,1), (2,1) – horizontal. How about (1,0), (1,1), (1,2) – vertical. How about (0,2), (1,1), (2,0) – diagonal. How about (0,0), (1,2), (2,1)? Not colinear. How about (0,1), (1,0), (2,-1)? Outside. How about (1,0), (2,1), (3,2)? Outside. How about (1,2), (2,1), (3,0)? Outside. So no.

Wait, maybe there are lines that pass through three points if you consider different directions? For example, in a 3x3 grid, there are also lines that go through the center and two other points not aligned with the grid. Wait, but we saw that those lines only pass through two points.

Wait, unless there are other lines. Wait, for example, take points (0,0), (1,1), (2,2) – already counted. How about (0,0), (1,2), (2, something)? As before, only two points. How about (0,0), (2,1), (something)? Only two points. How about (0,1), (1,2), (2, something)? Only two points. How about (1,0), (2,1), (something)? Only two points. How about (1,2), (2,1), (something)? Only two points. So no.

Therefore, I think my original conclusion is correct: there are 8 colinear triplets. Therefore, the probability is 8/84, which simplifies to 2/21.

But wait, let me check with another approach. Let me count all possible lines with three points again.

Rows: 3 lines.

Columns: 3 lines.

Diagonals: 2 lines.

Total: 8 lines. Each line has exactly one triplet. Therefore, 8 triplets.

But wait, here's a different thought. What if there are lines that have more than three points? But in a 3x3 grid, all lines have at most three points. So each line with three points contributes exactly one triplet. Therefore, 8 triplets.

But I have a feeling that maybe there are more lines. Wait, let me think of the following: in addition to the rows, columns, and diagonals, there are also lines that are not aligned with the grid but still pass through three points. Wait, but how?

Wait, here's an example. Take the points (0,0), (1,1), (2,2) – that's a diagonal. What about the points (0,1), (1,2), (2,3)? But (2,3) is outside. How about (0,1), (1,0), (2,-1)? Outside. How about (1,0), (1,1), (1,2) – vertical. How about (0,0), (1,0), (2,0) – horizontal. How about (0,0), (0,1), (0,2) – vertical. How about (0,2), (1,1), (2,0) – diagonal. How about (0,2), (1,2), (2,2) – horizontal. How about (2,0), (2,1), (2,2) – horizontal. How about (1,0), (1,1), (1,2) – vertical. How about (0,0), (1,1), (2,2) – diagonal. How about (0,2), (1,1), (2,0) – diagonal. So that's all.

Wait, so maybe there are only 8 lines. Therefore, 8 triplets. Therefore, probability is 8/84 = 2/21. But I need to check if that's correct.

Wait, but I found a resource once that said in a 3x3 grid, there are 8 lines with three points. So maybe that's correct. Therefore, the answer is 8/84 = 2/21.

But let me check with another method. Let me calculate the total number of colinear triplets.

First, rows: 3 rows, each with 3 points. So C(3,3) = 1 per row. Total 3.

Columns: same, 3.

Diagonals: 2 diagonals, each with 3 points. So 2.

Total: 3 + 3 + 2 = 8. So that's 8.

But wait, here's a different thought. What about other diagonals? For example, in a 3x3 grid, there are also diagonals of length 2, but they don't have three points. So no.

Alternatively, maybe there are more diagonals if we consider different slopes. Wait, but we checked slopes 0, infinity, 1, -1, and others don't work.

Alternatively, maybe there are lines that pass through three points if you consider different directions. Wait, for example, in a 3x3 grid, there are also lines that go through three points with slope 2/1, but as we saw, they only pass through two points.

Wait, unless there's a line that passes through three points with a different slope. Wait, for example, slope 3/1? From (0,0) to (1,3), which is outside. No.

Alternatively, slope 1/1, which is already considered.

So, I think my conclusion is correct. Therefore, the probability is 8/84, which simplifies to 2/21.

But let me check with another approach. Let me think of all possible triplets and see how many are colinear.

Total triplets: 84.

Number of colinear triplets: ?

Let me think of the number of lines with three points: 8. Each line has exactly one triplet. Therefore, 8 triplets. Therefore, probability is 8/84 = 2/21 ≈ 0.095.

But I have a doubt because I remember that in some similar problems, the answer is different. Wait, for example, in the problem where you have a 3x3 grid and you choose three points, the probability that they are colinear is 8/84 = 2/21. But maybe I'm confusing with another problem.

Alternatively, maybe I missed some lines. Wait, let me think again. Suppose there are lines that pass through three points not aligned with the grid. For example, take points (0,0), (1,2), and (2, something). Wait, (0,0), (1,2), and (2, something). The line through (0,0) and (1,2) has slope 2. The equation is y = 2x. For x=2, y=4, which is outside. So no. How about points (0,1), (1,2), (2,3)? (2,3) is outside. How about (0,1), (1,0), (2,-1)? Outside. How about (1,0), (2,1), (3,2)? Outside. How about (1,1), (2,2), (3,3)? Outside. So no.

Wait, maybe there are lines that pass through three points if you consider different slopes? For example, slope 3/2? Let me check. Take points (0,0) and (2,3). But (2,3) is outside. How about (0,1) and (2,2)? Slope is (2-1)/(2-0) = 1/2. The line is y = (1/2)x + 1. For x=0, y=1; x=1, y=1.5; x=2, y=2. So only (0,1) and (2,2) are grid points. So only two points.

How about slope 2/1? From (0,0) to (1,2). As before, only two points.

Slope 1/1: already counted.

Slope -1/1: already counted.

Slope -2/1: from (0,2) to (1,0). Equation y = -2x + 2. For x=0, y=2; x=1, y=0; x=2, y=-2. So only two points.

Slope -1/2: from (0,2) to (2,1). Equation y = (-1/2)x + 2. For x=0, y=2; x=2, y=1; x=1, y=1.5. So only two points.

So no other lines with three points.

Therefore, I think my original conclusion is correct. There are 8 colinear triplets. Therefore, the probability is 8/84, which simplifies to 2/21.

But let me check with another method. Let me calculate the total number of colinear triplets by considering all possible lines.

First, horizontal lines: 3 lines, each with 3 points. Each line contributes 1 triplet. So 3.

Vertical lines: same, 3.

Diagonals with slope 1: 1 line, 1 triplet.

Diagonals with slope -1: 1 line, 1 triplet.

Total: 3 + 3 + 1 + 1 = 8. So same as before.

Therefore, the probability is 8/84 = 2/21.

But wait, I just found a resource (in my mind) that says that in a 3x3 grid, there are 8 lines with three points, so the answer is 8/84. But I want to make sure.

Alternatively, maybe I made a mistake in the total number of triplets? Let me check again.

Total number of ways to choose 3 points from 9: C(9,3) = 84. Correct.

Number of colinear triplets: 8. So probability is 8/84 = 2/21 ≈ 0.095.

But I have a feeling that maybe there are more lines. Wait, here's a different idea. What about the lines that pass through three points in the following way: for example, the line that goes through (0,0), (1,1), (2,2) – that's one. What about the line that goes through (0,1), (1,1), (2,1) – horizontal. What about (1,0), (1,1), (1,2) – vertical. What about (0,0), (0,1), (0,2) – vertical. What about (0,2), (1,1), (2,0) – diagonal. What about (0,0), (1,0), (2,0) – horizontal. What about (0,0), (0,1), (0,2) – vertical. What about (0,2), (0,1), (0,0) – vertical. Wait, these are all the same lines.

Wait, maybe there are other lines if you consider different directions. Wait, for example, in a 3x3 grid, there are also lines that go through three points if you consider the center and two other points. But we already considered those. For example, the center is part of the horizontal, vertical, and two diagonals.

Wait, unless there are other lines through the center. For example, if you take the center (1,1) and two other points not aligned with the grid. But as we saw, those lines only pass through two points.

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

But to be thorough, let me think of another way. Suppose I list all possible colinear triplets.

Rows:

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

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

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

Columns:

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)

That's 8 triplets. So yes, exactly 8. Therefore, the probability is 8/84 = 2/21.

Therefore, the answer should be 2/21.

But wait, I just found a contradiction. Wait, in some references, the number of lines in a 3x3 grid is 12, but that includes lines with only two points. But we are only interested in lines with three points. So if there are 12 lines in total, but only 8 of them have three points, then the answer is 8/84.

Alternatively, maybe there are more lines with three points. Wait, let me think again. Suppose there are lines with three points that are not horizontal, vertical, or diagonal. For example, in a 3x3 grid, is there a line that goes through three points with slope 2/1? No, as discussed. How about slope 1/1? Already counted. How about slope 3/1? No. How about slope 1/3? No. So no.

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

But to be 100% sure, let me check with an example. Suppose I have the grid:

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

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

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

Now, list all possible colinear triplets:

Rows:

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

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

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

Columns:

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

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

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

Diagonals:

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

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

That's 8 triplets. Any others?

For example, is there a line that goes through (0,0), (1,1), (2,2)? Yes, already counted.

Is there a line that goes through (0,1), (1,1), (2,1)? Yes, column 1, already counted.

Is there a line that goes through (1,0), (1,1), (1,2)? Yes, column 0, already counted.

Is there a line that goes through (0,0), (0,1), (0,2)? Yes, row 0, already counted.

Is there a line that goes through (0,2), (1,2), (2,2)? Yes, row 2, already counted.

Is there a line that goes through (2,0), (2,1), (2,2)? Yes, row 2, already counted.

Is there a line that goes through (0,0), (1,0), (2,0)? Yes, column 0, already counted.

Is there a line that goes through (0,1), (1,0), (2,-1)? No, because (2,-1) is outside.

Is there a line that goes through (0,1), (1,2), (2,3)? No, (2,3) is outside.

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

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

Is there a line that goes through (0,0), (1,2), (2, something)? No, only two points.

Therefore, there are no other colinear triplets. Therefore, the total is indeed 8.

Therefore, the probability is 8/84 = 2/21.

But wait, I just thought of something. What about the line that goes through (0,0), (1,1), (2,2)? That's one triplet. What about the line that goes through (0,0), (1,0), (2,0)? That's another triplet. Similarly for others. So yes, 8 triplets.

Therefore, I think the answer is 2/21.

But let me check with another approach. Suppose I calculate the number of lines with three points, which is 8, and each line has exactly one triplet. Therefore, 8 triplets. Therefore, probability is 8/84 = 2/21.

Yes, that seems correct.

Therefore, the probability is 2/21.

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

To determine the probability that three randomly chosen points from a 3×3 grid lie on the same straight line, we proceed in two main steps: **counting the total number of possible three-point sets**, and **counting the number of those sets that are colinear**.

---

### **Step 1: Total Number of Three-Point Sets**

The grid contains 9 points. The number of ways to choose any 3 points from these 9 is given by the combination formula:

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

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

---

### **Step 2: Counting Colinear Triplets**

We now count how many of these 84 sets lie on the same straight line. A line in the grid can be horizontal, vertical, or diagonal. We examine each type:

#### **Horizontal Lines**
There are 3 rows, each containing 3 points. Each row forms a horizontal line with exactly 3 points. So, we have:

$$
3 \text{ horizontal lines} \Rightarrow 3 \text{ colinear triplets}
$$

#### **Vertical Lines**
Similarly, there are 3 columns, each with 3 points. Each column forms a vertical line with exactly 3 points. So:

$$
3 \text{ vertical lines} \Rightarrow 3 \text{ colinear triplets}
$$

#### **Diagonal Lines**
There are two main diagonals:
- From (0,0) to (1,1) to (2,2)
- From (0,2) to (1,1) to (2,0)

Each of these diagonals contains exactly 3 points, so:

$$
2 \text{ diagonal lines} \Rightarrow 2 \text{ colinear triplets}
$$

#### **Other Lines**
We now check if there are any other lines that pass through **exactly three** points in the grid. These would be lines with slopes other than 0, ∞, 1, or -1. For example, lines with slopes like 2, 1/2, -2, or -1/2.

However, upon checking all such lines, we find that **no other line** in the 3×3 grid contains **exactly three** grid points. For instance:
- A line with slope 2 through (0,0) and (1,2) would pass through (2,4), which is outside the grid.
- A line with slope 1/2 through (0,0) and (2,1) would pass through (4,2), which is also outside the grid.

Thus, **no additional lines** with three points exist.

---

### **Step 3: Total Colinear Triplets**

Adding up the colinear triplets from all valid lines:

$$
3 \text{ (horizontal)} + 3 \text{ (vertical)} + 2 \text{ (diagonal)} = 8 \text{ colinear triplets}
$$

---

### **Step 4: Calculating the Probability**

The probability that a randomly chosen set of three points lies on the same straight line is the ratio of favorable outcomes to total outcomes:

$$
\text{Probability} = \frac{\text{Number of colinear triplets}}{\text{Total number of triplets}} = \frac{8}{84} = \frac{2}{21}
$$

---

### **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, I need to figure out two things: the total number of possible three-point sets, and the number of those sets that are colinear (lie on the same straight line). Then, the probability would be the number of colinear triplets divided by the total number of triplets.

Let me start with the total number of possible three-point sets. Since there are 9 points, the number of ways to choose 3 is given by the combination formula C(n, k) = n! / (k! * (n - k)!). So, plugging in n=9 and k=3:

Total triplets = C(9, 3) = 9! / (3! * 6!) = (9 * 8 * 7) / (3 * 2 * 1) = 84. So there are 84 possible ways to choose three points.

Now, the next part is figuring out how many of these triplets are colinear. This seems a bit trickier. Let me visualize the grid. It's a 3x3 grid, so like a tic-tac-toe board. The points are arranged in rows, columns, and diagonals. But there might be other lines as well, like diagonals with different slopes or maybe other lines that pass through three points.

First, let me count the obvious ones: the rows, columns, and the two main diagonals.

Starting with rows: there are 3 rows, each containing 3 points. So, each row contributes 1 colinear triplet. Therefore, 3 rows * 1 = 3 triplets.

Similarly, columns: there are 3 columns, each with 3 points. So another 3 triplets. So that's 3 + 3 = 6 so far.

Then the main diagonals: there are two main diagonals, each with 3 points. So that's 2 more triplets. So total so far is 6 + 2 = 8.

But wait, are there any other lines that pass through three points? For example, lines with different slopes?

Let me think. In a 3x3 grid, are there other lines that contain exactly three points?

For example, consider lines with slope 1/2 or 2, or maybe other slopes. Let me check.

Let me try to visualize the grid. Let me label the points with coordinates. Let's say the grid has coordinates from (0,0) to (2,2). 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 think about lines that pass through three points. We already considered horizontal, vertical, and main diagonals. Are there other lines?

For example, consider the line that goes through (0,1), (1,2), and (2,1). Wait, does that line pass through three points? Let me check.

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

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

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

What about lines with slope 2? Let's see. For example, starting at (0,0), going up 2, right 1: (0,0), (1,2). But then the next point would be (2,4), which is outside the grid. So only two points. Similarly, starting at (0,1), going up 2, right 1: (0,1), (1,3) which is outside. Not helpful.

How about slope 1/2? Starting at (0,0), going right 2, up 1: (0,0), (2,1). Then next would be (4,2), outside. So only two points. Similarly, starting at (0,1), going right 2, up 1: (0,1), (2,2). Again, only two points. Similarly, starting at (1,0), going right 2, up 1: (1,0), (3,1) which is outside. So no.

What about slope -2? Starting at (0,2), going right 1, down 2: (0,2), (1,0). Then next would be (2, -2), outside. So only two points. Similarly, slope -1/2? Starting at (0,2), going right 2, down 1: (0,2), (2,1). Then next would be (4,0), outside. So again, only two points.

So maybe there are no lines with slopes other than 0, infinity, 1, -1 that pass through three points?

Wait, but maybe there are other lines? For example, consider the line that goes through (0,0), (1,1), (2,2) – that's already counted. How about (0,1), (1,0), (2,-1)? But (2,-1) is outside the grid. Similarly, (0,1), (1,2), (2,3) – again outside. So no.

Wait, but maybe there are lines that aren't aligned with the grid or main diagonals but still pass through three points? Let me think. For example, is there a line that goes through (0,0), (1,2), and another point?

Let me check. The line through (0,0) and (1,2) has slope 2. The equation is y = 2x. Let's see if any other grid points lie on this line. For x=2, y=4, which is outside. For x=0.5, y=1, but that's not a grid point. So only two points.

How about the line through (0,1) and (1,0)? Slope is -1. Equation is y - 1 = -1(x - 0) => y = -x + 1. Let's check for other points. When x=2, y = -2 + 1 = -1, which is outside. When x= -1, y=2, which is outside. So only two points.

How about the line through (0,0) and (2,1)? Slope is 1/2. Equation y = (1/2)x. For x=1, y=0.5, not a grid point. So only two points.

Similarly, line through (0,2) and (1,1): slope is -1. Equation y - 2 = -1(x - 0) => y = -x + 2. For x=2, y= -2 + 2 = 0. So (2,0) is on this line. So the points (0,2), (1,1), (2,0). Wait, that's the other main diagonal! Which we already counted. So that's already included in the two main diagonals.

Wait, so maybe there are no other lines with three points? Let me check another possibility. How about the line through (1,0), (1,1), (1,2)? That's a vertical line, already counted in columns. Similarly for horizontal lines.

Wait, so maybe the only lines that have three points are the rows, columns, and two main diagonals? So that would be 3 + 3 + 2 = 8 lines, each containing exactly one triplet. Therefore, there are 8 colinear triplets?

But wait, hold on. Wait, let me think again. Are there other lines? For example, in some grids, there are more lines. Wait, for example, in a 3x3 grid, there are also lines that are not aligned with the grid or the main diagonals but still pass through three points. Wait, but I can't think of any. Let me check again.

Wait, suppose we take points (0,0), (1,1), (2,2) – that's the main diagonal. Similarly, (0,2), (1,1), (2,0). What about points like (0,1), (1,1), (2,1)? That's a horizontal line, already counted. How about (1,0), (1,1), (1,2)? Vertical line, already counted.

Wait, maybe there are lines with different slopes? For example, let me check if there's a line that goes through (0,0), (1,2), and another point. Wait, we saw that (0,0), (1,2) is slope 2, and the next point would be (2,4), which is outside. So no. How about (0,1), (1,2), (2,3)? Again, (2,3) is outside. Similarly, (0,0), (2,1), (something)? The line through (0,0) and (2,1) is y = (1/2)x. The next point would be (4, 2), which is outside. So no.

Alternatively, maybe lines that are not straight? No, we are talking about straight lines.

Wait, maybe there are lines that pass through three points in a different way? For example, in some cases, maybe three points that are not in a straight line? No, we are looking for colinear.

Wait, maybe I need to check all possible lines. Let me think of all possible lines that can be formed by three points in the grid.

Alternatively, maybe there are more than 8 lines. Wait, let me think again. For example, in addition to the rows, columns, and diagonals, are there other lines?

Wait, here's an idea. What about the lines that go through the center point and two other points? For example, the center is (1,1). If I take (0,0), (1,1), (2,2) – that's the main diagonal. Similarly, (0,2), (1,1), (2,0). But what about other lines through the center? For example, (0,1), (1,1), (2,1) – that's a horizontal line. Similarly, (1,0), (1,1), (1,2) – vertical. So those are already counted.

What about lines that go through the center with other slopes? For example, slope 2? Let's see. The line through (1,1) with slope 2 would be y - 1 = 2(x - 1). So y = 2x - 1. Let's check if any other grid points lie on this line. For x=0, y = -1, which is outside. For x=1, y=1. For x=2, y=3, which is outside. So only (1,1) and maybe another? Wait, if x=0.5, y=0, but that's not a grid point. So no. Similarly, slope 1/2? y - 1 = (1/2)(x - 1). For x=0, y = 1 - 0.5 = 0.5, not a grid point. For x=1, y=1. For x=2, y=1 + 0.5 = 1.5, not a grid point. So again, only the center point.

How about slope -2? y - 1 = -2(x - 1). So y = -2x + 3. For x=0, y=3, outside. For x=1, y=1. For x=2, y= -1, outside. So again, only the center.

Slope -1/2? y - 1 = -1/2(x - 1). For x=0, y = 1 + 0.5 = 1.5, not a grid point. For x=1, y=1. For x=2, y = 1 - 0.5 = 0.5, not a grid point. So again, only the center.

So no other lines through the center with three points. So maybe there are no other lines?

Wait, but let me think again. What about lines that are not passing through the center? For example, take points (0,0), (1,1), (2,2) – already counted. How about (0,0), (0,1), (0,2) – vertical line, already counted. How about (0,0), (1,0), (2,0) – horizontal line, already counted. How about (0,0), (1,2), (2,1)? Wait, earlier I thought that wasn't colinear, but let me check again.

Wait, (0,0), (1,2), (2,1). Let me check if they are colinear. 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. Since the slopes are different, they are not colinear. So that's not a line.

How about (0,1), (1,0), (2, -1)? Again, (2, -1) is outside. Not valid.

How about (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, already counted.

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

Is there any other line? Let me think of all possible combinations. Maybe there are lines that are not aligned with the grid or diagonals but still pass through three points. Wait, for example, in a 3x3 grid, is there a line that goes through three points with slope 2/1? Wait, we saw that slope 2 only gives two points. Similarly for slope 1/2. How about slope 3? That would be even steeper. For example, from (0,0) to (1,3), but (1,3) is outside. Similarly, slope 1/3? From (0,0) to (3,1), which is outside. So no.

Alternatively, maybe there are lines that are not straight? No, we are talking about straight lines.

Wait, maybe I need to check all possible lines. Let me think of all possible lines that can be formed by three points. Since there are only 9 points, maybe there's a systematic way to check.

First, the rows: 3 lines.

Columns: 3 lines.

Diagonals: 2 lines.

Now, are there any other lines? Let me think of lines that have exactly three points. For example, in some cases, maybe a line that is not horizontal, vertical, or diagonal but still passes through three points. Wait, but in a 3x3 grid, is that possible?

Wait, here's an idea. Let me check if there are any other lines. For example, take points (0,0), (1,1), (2,2) – that's the main diagonal. What about points (0,1), (1,2), (2,3)? But (2,3) is outside. Similarly, (0,1), (1,0), (2,-1) – outside. How about (1,0), (1,1), (1,2) – vertical, already counted.

Wait, how about points (0,0), (1,2), (2,1)? Wait, we checked that before, they are not colinear. How about (0,0), (2,1), (something)? The line through (0,0) and (2,1) is y = (1/2)x. The only grid points on this line are (0,0) and (2,1). Similarly, the line through (0,1) and (2,0) is y = (-1/2)x + 1. The grid points on this line are (0,1), (1, 0.5) which is not a grid point, and (2, 0). So only two points.

Wait, so maybe there are no other lines with three points? Then the total number of colinear triplets is 8?

But wait, I remember that in some problems, there are more lines. Wait, let me check again. Maybe I missed some diagonals?

Wait, for example, in a 3x3 grid, there are also diagonals of different lengths. But we are only considering lines that have exactly three points. So, for example, the main diagonals have three points. But are there other diagonals? For example, in a 3x3 grid, if you consider diagonals of length 2, but they don't have three points. So no.

Wait, but maybe there are other lines with three points? Let me think of coordinates. Let me list all possible lines that pass through at least two points and check if they pass through a third.

Let me take all pairs of points and see if they form a line with a third point.

But that might take a while, but maybe manageable.

First, there are C(9, 2) = 36 pairs of points. For each pair, check if there's a third point on the same line.

But maybe there's a smarter way. Let me think.

Alternatively, think of all possible lines that can be formed in the grid. For a 3x3 grid, the number of lines is more than 8? Wait, according to some references, in a 3x3 grid, there are 8 lines that contain exactly three points. But I need to confirm.

Wait, actually, I think there are more. Wait, for example, in addition to the rows, columns, and two main diagonals, there are also four more lines that have three points. Wait, how?

Wait, let me think of the lines that go through the center and two other points. Wait, but we already considered those. For example, the horizontal, vertical, and main diagonals. But maybe there are other lines through the center?

Wait, for example, if you take the center (1,1) and the points (0,1) and (2,1), that's the horizontal line. Similarly, (1,0), (1,1), (1,2) is vertical. The main diagonals are (0,0), (1,1), (2,2) and (0,2), (1,1), (2,0). Are there other lines through the center?

For example, if I take (0,1), (1,1), (2,1) – horizontal. (1,0), (1,1), (1,2) – vertical. What about (0,0), (1,1), (2,2) – main diagonal. (0,2), (1,1), (2,0) – other main diagonal. What about other lines?

Wait, suppose I take (0,0), (1,1), (2,2) – already counted. How about (0,1), (1,2), (2,3)? But (2,3) is outside. How about (0,1), (1,0), (2,-1)? Outside. How about (1,0), (1,1), (1,2) – vertical. How about (0,0), (1,0), (2,0) – horizontal. How about (0,2), (1,2), (2,2) – horizontal. How about (0,0), (0,1), (0,2) – vertical. How about (2,0), (2,1), (2,2) – vertical. So all the rows and columns are covered.

Wait, maybe there are other lines with three points? For example, if you take points (0,0), (1,2), and (2,1). Wait, we checked that before, they are not colinear. How about (0,0), (2,1), and (something)? The line through (0,0) and (2,1) is y = (1/2)x. The only grid points on this line are (0,0) and (2,1). Similarly, the line through (0,1) and (2,0) is y = (-1/2)x + 1. The grid points on this line are (0,1) and (2,0). So no third point.

How about the line through (0,0) and (1,1)? That's the main diagonal, already counted. Similarly for others.

Wait, maybe there are lines that are not aligned with the grid or diagonals but still pass through three points? For example, in a 3x3 grid, is there a line that goes through (0,0), (1,1), (2,2)? Yes, that's the main diagonal. What about (0,1), (1,2), (2,3)? No, (2,3) is outside. How about (1,0), (1,1), (1,2)? Vertical line, already counted.

Wait, maybe I need to think of all possible lines. Let me think of all possible lines that can be formed by three points. Since there are only 9 points, maybe there are only 8 such lines. But I have a feeling that there are more. Wait, let me check online... Wait, no, I can't do that. I need to figure it out myself.

Wait, here's another approach. Let me count all possible lines with three points. For that, I can think of all possible directions.

First, horizontal lines: 3 lines, each with 3 points.

Vertical lines: 3 lines, each with 3 points.

Diagonals with slope 1: 1 line (from (0,0) to (2,2)).

Diagonals with slope -1: 1 line (from (0,2) to (2,0)).

Are there any other lines with three points?

Wait, what about lines with slope 2? As discussed earlier, they only pass through two points.

Slope 1/2? Also only two points.

Slope -2? Two points.

Slope -1/2? Two points.

Slope 3? Outside.

Slope 1/3? Outside.

Slope -3? Outside.

Slope -1/3? Outside.

So no other slopes with three points.

Wait, but what about lines that are not straight? No, we are talking about straight lines.

Wait, but maybe there are other diagonals? For example, in a 3x3 grid, there are also diagonals of different lengths. But they don't have three points. For example, the diagonal from (0,1) to (1,2) is only two points. Similarly, from (1,0) to (2,1), etc. So those don't have three points.

Wait, so maybe there are only 8 lines with three points? Then the number of colinear triplets is 8? But that seems low. Wait, let me check again.

Wait, if there are 8 lines, each containing exactly one triplet, then there are 8 favorable cases. But I have a feeling that there are more. Wait, let me think again. For example, in the 3x3 grid, there are also lines that are not aligned with the grid or the main diagonals but still pass through three points. Wait, but how?

Wait, here's an idea. Maybe there are lines that pass through three points with different slopes. Wait, for example, take the points (0,0), (1,1), (2,2) – slope 1. Then (0,1), (1,2), (2,3) – but (2,3) is outside. How about (0,0), (1,2), (2, something)? As before, only two points.

Wait, maybe there are lines that pass through three points in a different way. Wait, for example, if you take the points (0,0), (1,1), (2,2) – that's one line. What about (0,0), (1,0), (2,0) – horizontal. (0,0), (0,1), (0,2) – vertical. Similarly for others. So maybe there are only 8 lines?

But I remember that in some problems, like the number of lines in a 3x3 grid, it's more than 8. Wait, let me think. For example, in the 3x3 grid, how many lines are there? Each line is determined by two points. But some lines are counted multiple times. Wait, but if I think of all possible lines, there are more than 8. For example, there are lines that have only two points, but we are only interested in those that have exactly three points.

Wait, so maybe there are only 8 lines with exactly three points? Let me check again. For example, rows: 3, columns: 3, diagonals: 2. Total 8. So that would mean 8 colinear triplets. Therefore, the probability would be 8 / 84 = 2 / 21 ≈ 0.095. But I have a feeling that the answer is different. Wait, maybe I missed some lines?

Wait, let me think of another way. Let me count all possible lines with three points. For that, maybe think of all possible lines that can be formed by three points. For example, in addition to the rows, columns, and main diagonals, are there any other lines?

Wait, here's a thought. What about the lines that go through the center and two other points not aligned with the grid? For example, the center (1,1). If I take (0,0) and (2,2), that's the main diagonal. If I take (0,2) and (2,0), that's the other main diagonal. What about (0,1) and (2,1)? That's horizontal, already counted. Similarly, (1,0) and (1,2) is vertical. What about (0,0) and (1,2)? That's a line with slope 2, but only two points. Similarly, (0,0) and (2,1) – slope 1/2, only two points. So no.

Wait, maybe there are lines that pass through three points if you consider different directions? Wait, for example, in a 3x3 grid, there are also lines that are not straight? No, we are talking about straight lines.

Wait, maybe I need to think of all possible triplets and check if they are colinear. But that would take time, but maybe manageable.

Total triplets: 84. Let me think of how many colinear triplets there are.

First, rows: 3 rows, each with C(3,3)=1 triplet. So 3.

Columns: same, 3.

Diagonals: 2, each with 1 triplet. So 2.

Total so far: 3 + 3 + 2 = 8.

But wait, are there any other lines? For example, suppose there is a line that isn't horizontal, vertical, or diagonal, but still has three points. Let me try to find such a line.

Take points (0,0), (1,1), (2,2) – already counted.

Take points (0,1), (1,2), (2,3) – (2,3) is outside.

Take points (1,0), (1,1), (1,2) – vertical, counted.

Take points (0,0), (1,0), (2,0) – horizontal, counted.

Take points (0,0), (0,1), (0,2) – vertical, counted.

Take points (0,2), (1,1), (2,0) – diagonal, counted.

Take points (0,1), (1,1), (2,1) – horizontal, counted.

Take points (1,0), (1,1), (1,2) – vertical, counted.

Take points (2,0), (2,1), (2,2) – horizontal, counted.

Take points (0,0), (1,2), (2,1) – not colinear.

Take points (0,1), (1,0), (2, -1) – outside.

Take points (1,0), (2,1), (3,2) – outside.

Take points (0,2), (1,1), (2,0) – diagonal, counted.

Take points (0,0), (2,1), (something)? Only two points.

Take points (0,1), (1,0), (2,-1) – outside.

Take points (1,1), (0,0), (2,2) – diagonal, counted.

Take points (1,1), (0,2), (2,0) – diagonal, counted.

Take points (1,1), (0,1), (2,1) – horizontal, counted.

Take points (1,1), (1,0), (1,2) – vertical, counted.

So after checking all these, I can't find any other lines with three points. Therefore, there are only 8 colinear triplets.

But wait, I remember that in some problems, like the one where you have to count the number of lines in a grid, there are more lines. Wait, for example, in a 3x3 grid, there are 12 lines: 3 horizontal, 3 vertical, 2 main diagonals, and 4 more diagonals? Wait, how?

Wait, maybe I was wrong before. Let me think again. For example, in addition to the main diagonals, there are also diagonals of different lengths. Wait, but they don't have three points. For example, the diagonal from (0,1) to (1,2) is only two points. Similarly, from (1,0) to (2,1). So those don't count. So maybe there are only 8 lines with three points.

But then why do I have a feeling that there are more? Maybe I'm confusing with another problem. Let me think of another approach.

Suppose I consider all possible lines that can be formed by three points. Each line is uniquely determined by its slope and intercept. Let me try to find all such lines.

First, horizontal lines: y = 0, y = 1, y = 2. Each has three points. So 3 lines.

Vertical lines: x = 0, x = 1, x = 2. Each has three points. So 3 lines.

Diagonals with slope 1: y = x, y = x - 1, y = x - 2. Wait, let's check:

For slope 1:

- Line y = x passes through (0,0), (1,1), (2,2). That's one line.

- Line y = x - 1 passes through (1,0), (2,1). Only two points.

- Line y = x - 2 passes through (2,0). Only one point.

Similarly, for slope 1, only one line with three points.

Similarly, for slope -1:

- Line y = -x + 2 passes through (0,2), (1,1), (2,0). That's one line.

- Line y = -x + 1 passes through (0,1), (1,0). Two points.

- Line y = -x + 0 passes through (0,0), (1,-1). Outside.

So only one line with slope -1.

So that's 1 + 1 = 2 diagonals.

Now, what about other slopes?

Slope 2:

- Line y = 2x passes through (0,0), (1,2). Next would be (2,4), outside. So only two points.

- Line y = 2x - 1 passes through (1,1), (2,3). (2,3) is outside. So only (1,1).

- Line y = 2x - 2 passes through (1,0), (2,2). So (1,0) and (2,2). Two points.

Slope 1/2:

- Line y = (1/2)x passes through (0,0), (2,1). Only two points.

- Line y = (1/2)x + 1 passes through (0,1), (2,2). Two points.

- Line y = (1/2)x - 1 passes through (2,0). One point.

Slope -2:

- Line y = -2x passes through (0,0), (1,-2). Outside.

- Line y = -2x + 2 passes through (0,2), (1,0). Two points.

- Line y = -2x + 1 passes through (0,1), (1,-1). Outside.

Slope -1/2:

- Line y = (-1/2)x passes through (0,0), (2, -1). Outside.

- Line y = (-1/2)x + 1 passes through (0,1), (2,0). Two points.

- Line y = (-1/2)x + 2 passes through (0,2), (2,1). Two points.

So for slopes other than 0, infinity, 1, -1, there are no lines with three points.

Therefore, total lines with three points are 3 horizontal + 3 vertical + 2 diagonals = 8 lines. Therefore, 8 colinear triplets.

Therefore, the number of favorable outcomes is 8, total outcomes is 84. Therefore, probability is 8/84 = 2/21 ≈ 0.095.

But wait, I have a doubt here because I remember that in some similar problems, the answer is different. Wait, let me check again. Maybe I missed some lines?

Wait, here's another thought. What about the lines that are not aligned with the grid but still pass through three points? For example, in a 3x3 grid, is there a line that goes through (0,0), (1,1), (2,2) – already counted. How about (0,0), (1,0), (2,0) – horizontal. How about (0,0), (0,1), (0,2) – vertical. How about (0,1), (1,1), (2,1) – horizontal. How about (1,0), (1,1), (1,2) – vertical. How about (0,2), (1,1), (2,0) – diagonal. How about (0,0), (1,2), (2,1)? Not colinear. How about (0,1), (1,0), (2,-1)? Outside. How about (1,0), (2,1), (3,2)? Outside. How about (1,2), (2,1), (3,0)? Outside. So no.

Wait, maybe there are lines that pass through three points if you consider different directions? For example, in a 3x3 grid, there are also lines that go through the center and two other points not aligned with the grid. Wait, but we saw that those lines only pass through two points.

Wait, unless there are other lines. Wait, for example, take points (0,0), (1,1), (2,2) – already counted. How about (0,0), (1,2), (2, something)? As before, only two points. How about (0,0), (2,1), (something)? Only two points. How about (0,1), (1,2), (2, something)? Only two points. How about (1,0), (2,1), (something)? Only two points. How about (1,2), (2,1), (something)? Only two points. So no.

Therefore, I think my original conclusion is correct: there are 8 colinear triplets. Therefore, the probability is 8/84, which simplifies to 2/21.

But wait, let me check with another approach. Let me count all possible lines with three points again.

Rows: 3 lines.

Columns: 3 lines.

Diagonals: 2 lines.

Total: 8 lines. Each line has exactly one triplet. Therefore, 8 triplets.

But wait, here's a different thought. What if there are lines that have more than three points? But in a 3x3 grid, all lines have at most three points. So each line with three points contributes exactly one triplet. Therefore, 8 triplets.

But I have a feeling that maybe there are more lines. Wait, let me think of the following: in addition to the rows, columns, and diagonals, there are also lines that are not aligned with the grid but still pass through three points. Wait, but how?

Wait, here's an example. Take the points (0,0), (1,1), (2,2) – that's a diagonal. What about the points (0,1), (1,2), (2,3)? But (2,3) is outside. How about (0,1), (1,0), (2,-1)? Outside. How about (1,0), (1,1), (1,2) – vertical. How about (0,0), (1,0), (2,0) – horizontal. How about (0,0), (0,1), (0,2) – vertical. How about (0,2), (1,1), (2,0) – diagonal. How about (0,2), (1,2), (2,2) – horizontal. How about (2,0), (2,1), (2,2) – horizontal. How about (1,0), (1,1), (1,2) – vertical. How about (0,0), (1,1), (2,2) – diagonal. How about (0,2), (1,1), (2,0) – diagonal. So that's all.

Wait, so maybe there are only 8 lines. Therefore, 8 triplets. Therefore, probability is 8/84 = 2/21. But I need to check if that's correct.

Wait, but I found a resource once that said in a 3x3 grid, there are 8 lines with three points. So maybe that's correct. Therefore, the answer is 8/84 = 2/21.

But let me check with another method. Let me calculate the total number of colinear triplets.

First, rows: 3 rows, each with 3 points. So C(3,3) = 1 per row. Total 3.

Columns: same, 3.

Diagonals: 2 diagonals, each with 3 points. So 2.

Total: 3 + 3 + 2 = 8. So that's 8.

But wait, here's a different thought. What about other diagonals? For example, in a 3x3 grid, there are also diagonals of length 2, but they don't have three points. So no.

Alternatively, maybe there are more diagonals if we consider different slopes. Wait, but we checked slopes 0, infinity, 1, -1, and others don't work.

Alternatively, maybe there are lines that pass through three points if you consider different directions. Wait, for example, in a 3x3 grid, there are also lines that go through three points with slope 2/1, but as we saw, they only pass through two points.

Wait, unless there's a line that passes through three points with a different slope. Wait, for example, slope 3/1? From (0,0) to (1,3), which is outside. No.

Alternatively, slope 1/1, which is already considered.

So, I think my conclusion is correct. Therefore, the probability is 8/84, which simplifies to 2/21.

But let me check with another approach. Let me think of all possible triplets and see how many are colinear.

Total triplets: 84.

Number of colinear triplets: ?

Let me think of the number of lines with three points: 8. Each line has exactly one triplet. Therefore, 8 triplets. Therefore, probability is 8/84 = 2/21 ≈ 0.095.

But I have a doubt because I remember that in some similar problems, the answer is different. Wait, for example, in the problem where you have a 3x3 grid and you choose three points, the probability that they are colinear is 8/84 = 2/21. But maybe I'm confusing with another problem.

Alternatively, maybe I missed some lines. Wait, let me think again. Suppose there are lines that pass through three points not aligned with the grid. For example, take points (0,0), (1,2), and (2, something). Wait, (0,0), (1,2), and (2, something). The line through (0,0) and (1,2) has slope 2. The equation is y = 2x. For x=2, y=4, which is outside. So no. How about points (0,1), (1,2), (2,3)? (2,3) is outside. How about (0,1), (1,0), (2,-1)? Outside. How about (1,0), (2,1), (3,2)? Outside. How about (1,1), (2,2), (3,3)? Outside. So no.

Wait, maybe there are lines that pass through three points if you consider different slopes? For example, slope 3/2? Let me check. Take points (0,0) and (2,3). But (2,3) is outside. How about (0,1) and (2,2)? Slope is (2-1)/(2-0) = 1/2. The line is y = (1/2)x + 1. For x=0, y=1; x=1, y=1.5; x=2, y=2. So only (0,1) and (2,2) are grid points. So only two points.

How about slope 2/1? From (0,0) to (1,2). As before, only two points.

Slope 1/1: already counted.

Slope -1/1: already counted.

Slope -2/1: from (0,2) to (1,0). Equation y = -2x + 2. For x=0, y=2; x=1, y=0; x=2, y=-2. So only two points.

Slope -1/2: from (0,2) to (2,1). Equation y = (-1/2)x + 2. For x=0, y=2; x=2, y=1; x=1, y=1.5. So only two points.

So no other lines with three points.

Therefore, I think my original conclusion is correct. There are 8 colinear triplets. Therefore, the probability is 8/84, which simplifies to 2/21.

But let me check with another method. Let me calculate the total number of colinear triplets by considering all possible lines.

First, horizontal lines: 3 lines, each with 3 points. Each line contributes 1 triplet. So 3.

Vertical lines: same, 3.

Diagonals with slope 1: 1 line, 1 triplet.

Diagonals with slope -1: 1 line, 1 triplet.

Total: 3 + 3 + 1 + 1 = 8. So same as before.

Therefore, the probability is 8/84 = 2/21.

But wait, I just found a resource (in my mind) that says that in a 3x3 grid, there are 8 lines with three points, so the answer is 8/84. But I want to make sure.

Alternatively, maybe I made a mistake in the total number of triplets? Let me check again.

Total number of ways to choose 3 points from 9: C(9,3) = 84. Correct.

Number of colinear triplets: 8. So probability is 8/84 = 2/21 ≈ 0.095.

But I have a feeling that maybe there are more lines. Wait, here's a different idea. What about the lines that pass through three points in the following way: for example, the line that goes through (0,0), (1,1), (2,2) – that's one. What about the line that goes through (0,1), (1,1), (2,1) – horizontal. What about (1,0), (1,1), (1,2) – vertical. What about (0,0), (0,1), (0,2) – vertical. What about (0,2), (1,1), (2,0) – diagonal. What about (0,0), (1,0), (2,0) – horizontal. What about (0,0), (0,1), (0,2) – vertical. What about (0,2), (0,1), (0,0) – vertical. Wait, these are all the same lines.

Wait, maybe there are other lines if you consider different directions. Wait, for example, in a 3x3 grid, there are also lines that go through three points if you consider the center and two other points. But we already considered those. For example, the center is part of the horizontal, vertical, and two diagonals.

Wait, unless there are other lines through the center. For example, if you take the center (1,1) and two other points not aligned with the grid. But as we saw, those lines only pass through two points.

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

But to be thorough, let me think of another way. Suppose I list all possible colinear triplets.

Rows:

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

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

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

Columns:

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)

That's 8 triplets. So yes, exactly 8. Therefore, the probability is 8/84 = 2/21.

Therefore, the answer should be 2/21.

But wait, I just found a contradiction. Wait, in some references, the number of lines in a 3x3 grid is 12, but that includes lines with only two points. But we are only interested in lines with three points. So if there are 12 lines in total, but only 8 of them have three points, then the answer is 8/84.

Alternatively, maybe there are more lines with three points. Wait, let me think again. Suppose there are lines with three points that are not horizontal, vertical, or diagonal. For example, in a 3x3 grid, is there a line that goes through three points with slope 2/1? No, as discussed. How about slope 1/1? Already counted. How about slope 3/1? No. How about slope 1/3? No. So no.

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

But to be 100% sure, let me check with an example. Suppose I have the grid:

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

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

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

Now, list all possible colinear triplets:

Rows:

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

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

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

Columns:

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

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

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

Diagonals:

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

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

That's 8 triplets. Any others?

For example, is there a line that goes through (0,0), (1,1), (2,2)? Yes, already counted.

Is there a line that goes through (0,1), (1,1), (2,1)? Yes, column 1, already counted.

Is there a line that goes through (1,0), (1,1), (1,2)? Yes, column 0, already counted.

Is there a line that goes through (0,0), (0,1), (0,2)? Yes, row 0, already counted.

Is there a line that goes through (0,2), (1,2), (2,2)? Yes, row 2, already counted.

Is there a line that goes through (2,0), (2,1), (2,2)? Yes, row 2, already counted.

Is there a line that goes through (0,0), (1,0), (2,0)? Yes, column 0, already counted.

Is there a line that goes through (0,1), (1,0), (2,-1)? No, because (2,-1) is outside.

Is there a line that goes through (0,1), (1,2), (2,3)? No, (2,3) is outside.

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

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

Is there a line that goes through (0,0), (1,2), (2, something)? No, only two points.

Therefore, there are no other colinear triplets. Therefore, the total is indeed 8.

Therefore, the probability is 8/84 = 2/21.

But wait, I just thought of something. What about the line that goes through (0,0), (1,1), (2,2)? That's one triplet. What about the line that goes through (0,0), (1,0), (2,0)? That's another triplet. Similarly for others. So yes, 8 triplets.

Therefore, I think the answer is 2/21.

But let me check with another approach. Suppose I calculate the number of lines with three points, which is 8, and each line has exactly one triplet. Therefore, 8 triplets. Therefore, probability is 8/84 = 2/21.

Yes, that seems correct.

Therefore, the probability is 2/21.

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

To determine the probability that three randomly chosen points from a 3×3 grid lie on the same straight line, we proceed in two main steps: **counting the total number of possible three-point sets**, and **counting the number of those sets that are colinear**.

---

### **Step 1: Total Number of Three-Point Sets**

The grid contains 9 points. The number of ways to choose any 3 points from these 9 is given by the combination formula:

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

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

---

### **Step 2: Counting Colinear Triplets**

We now count how many of these 84 sets lie on the same straight line. A line in the grid can be horizontal, vertical, or diagonal. We examine each type:

#### **Horizontal Lines**
There are 3 rows, each containing 3 points. Each row forms a horizontal line with exactly 3 points. So, we have:

$$
3 \text{ horizontal lines} \Rightarrow 3 \text{ colinear triplets}
$$

#### **Vertical Lines**
Similarly, there are 3 columns, each with 3 points. Each column forms a vertical line with exactly 3 points. So:

$$
3 \text{ vertical lines} \Rightarrow 3 \text{ colinear triplets}
$$

#### **Diagonal Lines**
There are two main diagonals:
- From (0,0) to (1,1) to (2,2)
- From (0,2) to (1,1) to (2,0)

Each of these diagonals contains exactly 3 points, so:

$$
2 \text{ diagonal lines} \Rightarrow 2 \text{ colinear triplets}
$$

#### **Other Lines**
We now check if there are any other lines that pass through **exactly three** points in the grid. These would be lines with slopes other than 0, ∞, 1, or -1. For example, lines with slopes like 2, 1/2, -2, or -1/2.

However, upon checking all such lines, we find that **no other line** in the 3×3 grid contains **exactly three** grid points. For instance:
- A line with slope 2 through (0,0) and (1,2) would pass through (2,4), which is outside the grid.
- A line with slope 1/2 through (0,0) and (2,1) would pass through (4,2), which is also outside the grid.

Thus, **no additional lines** with three points exist.

---

### **Step 3: Total Colinear Triplets**

Adding up the colinear triplets from all valid lines:

$$
3 \text{ (horizontal)} + 3 \text{ (vertical)} + 2 \text{ (diagonal)} = 8 \text{ colinear triplets}
$$

---

### **Step 4: Calculating the Probability**

The probability that a randomly chosen set of three points lies on the same straight line is the ratio of favorable outcomes to total outcomes:

$$
\text{Probability} = \frac{\text{Number of colinear triplets}}{\text{Total number of triplets}} = \frac{8}{84} = \frac{2}{21}
$$

---

### **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 $8/84 = \boxed{\frac{2}{21}}$.