View Single Post
  #3   Spotlight this post!  
Unread 15-07-2005, 12:36
Anthony's Avatar
Anthony Anthony is offline
FIRST Addict ... Er ... Mentor
FRC #0900 (Team Infinity)
Team Role: Mentor
 
Join Date: Oct 2003
Rookie Year: 2001
Location: Durham, NC
Posts: 59
Anthony has a spectacular aura aboutAnthony has a spectacular aura aboutAnthony has a spectacular aura about
Send a message via AIM to Anthony
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.
Reply With Quote