I ran across this on another forum and we’re running in circles trying to figure it out:
Scenario: I have two lines that intersect, both are 8 inches long. (Assume “8 points on the line” ignore the zero->one) The lines intersect at an inch mark.
There are 4 dots distributed along line A, randomly, but always fall on an inch mark (not in between). So at any given inch mark, there is a 50% chance of finding a dot.
There are 2 dot distributed along line B, randomly, but always fall on an inch mark. So at any given inch mark on line B, there is a 25% chance of finding a dot.
How do I calculate the odds of finding a dot where the lines interesect?
**Example **of the lines:
Exhaustively, you can count the number of yeses and divide by the number of nos. In this case, counting nos is easier. Think of your lines there as a grid, with the dots extending as colored lines. You’ve got 64 total possibilities (8 x 8), and you’ve got an area there that doesn’t have any colored lines which is where you get a no cause there isn’t a colored dot there. This area happens to be 24 dots large (6 x 4) ((8-2) x (8-4)).
So. 24/64 odds of it being a no. 40/64 odds of it being a yes. 5/8 = 62.5%
The math with just the percentages goes:
50% of it being blue
25% of it being red
50%*25% of it being red and blue
50%*75% of it not being blue or red
You will have to forgive me, I havent done real math in about two months (yay burning myself out on Diff Eqs), but arent there a lot more than 64 possibilities? There are 64 ways the lines can intersect (X1,Y1 ; X1,Y2 ; … ; X8,Y8). Then, there is the fact that the dots have variable locations.
Here is the part I amy need correcting on: On the X line, there are 36 different arrangements for the four dots. On the Y line, there are 28 different arrangements for the two dots. (The 36 is the number I question).
Therefore, for each of the 64 possible intersections, there are 36 x 28 different dot arrangements. In stating that, there is a 25% chance that the point of intersection on the Y line will have a dot, and a 50% chance that the point of intersection on the X line will have a dot.
Therefore, there is a 37.5% chance that the point of intersection will have both dots.
Upon rereading your post, I may have misunderstood, do both lines have to have dots or just one line. If it is just one line having a dot at the point of intersection, then the odds are 62.5%. 50% of the time blue will not have one, 75% of the time red will not have one. Thus 62.5% of the time one if them will have one.
Can two dots on the same line end up on the same point? That is very important…
I think that we’re making this problem harder than it realy is. If we assume that the answer to Daniel’s question is “no”, the dots on either line cannot occupy the same point, then the answer is:
for line 1, there is a 50% probability that the intersection falls on a dot
for line 2, there is a 25% probability that the intersection falls on a dot
for both to occur, the probablility is 50% x 25% (or mathematically)
0.5 x 0.25 = .125, so there is a 12.5% probability (1 in 8) that both lines will be on dots at the intersection.
Or am I wrong?
(don’t think so)
OK, I think I was wrong. I too, misread the problem. As I see it now, you are looking for the probability of ANY dot being at the intersection, not both lines having a dot there. This changes everything.
It is easy to see that there is a 50% chance of finding a dot on the first line and a 25% chance for the second, but the probability of either or is additive (with a twist), not multiplicative.
To start, there is a 50% probability that the first line will intersect on a dot and a 25% chance for the second so, if we add these together, we get a 75% chance of hitting a dot … BUT WAIT …, 50% of the time, when the second line is on a dot, the first line will ALREADY BE ON A DOT so we have to subtract these from the probability. So the final equation becomes:
50% + 25% - (50% x 25%) – or mathematically
0.50 + 0.25 - (0.50 x 0.25) = .625 == or 62.5% probability of EITHER line intersecting on a dot.
Geeez, I hate making mistakes!