|
Re: Probablility Problem
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.
Last edited by Anthony : 15-07-2005 at 12:44.
|