All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 The odds stay unchanged (Posted on 2004-06-17)
On a certain island each of the inhabitants is a member of one of the two existing clubs.

The membership distribution is such that when two random people meet, the probability of those two belonging to the same club is equal to the probability of them belonging to distinct clubs.

When 100 newcomers arrive on the island and each enrolls in one of the two clubs, the distribution still retains this feature. How many people belong to either club?

 See The Solution Submitted by Ady TZIDON Rating: 4.2727 (11 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 hints - partial solution | Comment 7 of 21 |

If you draw up all the members of Club Alpha in one columns and all the members of Club Beta and draw paths between each person to every other person, then the number of paths between A and B is A•B, and the number of paths that are only in A or only in B are (A²-A)/2 + (B²-B)/2.

What we want to find is when A•B = (A²-A)/2 + (B²-B)/2.

We are told that at some time, C people join Club Alpha and D people join Club Beta, so the new equation becomes:  (A+C)•(B+D) = ((A+C)²-(A+C))/2 + ((B+D)²-(B+D))/2 and we are further told that C+D = 100.

Now we have three equations and four unknowns... ...I think I'll have to write a program to solve this one.

 Posted by Erik O. on 2004-06-17 14:41:36

 Search: Search body:
Forums (0)