Question

a box contains fewer than twenty marbles. if you reach into the box and randomly pull out two marbles without replacing them, you have 50 percent chance of getting two marbles. how many blue marbles are in the box?

Answer 1

If there are x blue marbles out of a total of n marbles, 
then the probability of a blue on first draw is  x/n, and the 
probability of a blue on the second draw is (x-1)/(n-1).  
Then we require:
   x/n (x-1)/(n-1) = 1/2
   2x(x-1) = n(n-1).
If n = 18, say, then  
   2x(x-1) = (18)(17)
   x(x-1) = 153
   x^2 - x - 153 = 0
   x = [1 +or- sqrt(1+612)]/2 = 12.879.  
This is not an integer, so n cannot equal 18.
If we take the general case with n marbles, then
   2x^2 - 2x - n(n-1) = 0
   x = [2 + sqrt(4 + 8n(n-1))]/4.
We require 
   4 + 8n(n-1) 
to be a perfect square, or  
   1 + 2n(n-1) 
to be a perfect square, thus
   2n^2 - 2n + 1 
to be a perfect square.
n = 4 gives a value of 25 to this expression.  Then 
   x = [2+2sqrt(25)]/4
     = [2 + 10]/4
     = 12/4
     = 3.
So if we have 4 marbles, 3 of which are blue, we should have a 
probability equal to 1/2 that if we draw out 2 then both will be blue.

Check: (3/4)(2/3) = 2/4 = 1/2.