If you take two balls randomly out of a jar of colored balls, there is a 50% chance that the balls will both be red.
What is the total percentage of red balls in the jar?
I assume that you do not replace the first red ball before drawing the second, because if you do, the chances are 1/(√2) =(√2)/2 Which means that for every red ball, there are .414... other balls, but the number of balls must be an integer
Assume that there are r red balls and o balls that are not red.
The chance of drawing a red ball is r/(r + o)
The chance of then drawing a second red ball is (r - 1)/(r + o - 1)
The chance of drawing two red balls thus is [r/(r + o)][(r - 1)/(r + o - 1)] = [r(r - 1)]/[(r + o)(r + o - 1)}. the terms of the problem equates this chance to 1/2
A couple of observations: (r - 1)/(r + o -1) is smaller than r/(r + o), since there are fewer red balls but the same number of other balls.
If the two chances were the same, they would have to be 1/√2 = (√)/2
This means that (r - 1)/(r + o -1) < (√2)/2 < r/(r + o)
So, for r and r+o, we are looking for two integers such that 2r² is slightly more than (r+o)²
First approximation: r = 3, o=1 [2(3²) = 18; 4² = 16]
[r/(r+o)][(r-1)/(r+o-1)] = [3/4][2/3] = 1/2
Could I have been lucky enough to find it on my first approximation?
Second approximation r=5, o=2 [2(5²)=50; 7²=49]
[r/(r+o)][(r-1)/(r+o-1)] = [5/7][4/6] = 20/42 = 10/21 = .476
so i guess I lucky with my first guess: there are four balls in the jar, three of them are red.
BTW there are exactly four balls and exactly three red. If there were twice as many, or some larger multiple, the second probability would no longer equal 2/3 (for 6/8 it would be 5/7; for 9/12 it would be 8/11, etc)
Posted by TomM
on 2002-10-18 00:01:45