An urn contains a number of colored balls, with equal numbers of each color. Adding 20 balls of a new color to the urn would not change the probability of drawing (without replacement) two balls of the same color. How many balls are in the urn? (Before the extra balls are added.)

n: no. of colors
x: no. of balls of each color
Total No. of balls = n*x

1) before adding 20 balls
no. of ways 2 balls of same color can be chosen
= n*C(x,2)
Probability=n*C(x,2)/C(n*x,2)

2) after adding 20 balls
total no. of balls = n*x+20
no. of ways 2 balls of same color can be chosen
=n*C(x,2)+C(20,2)
Probability={n*C(x,2)+C(20,2)}/C(n*x+20,2)

According to given condition:
n*C(x,2)/C(n*x,2)={n*C(x,2)+C(20,2)/C(n*x+20,2)}

upon simplification, we get
(x-1)(n*x+19)(n*x+20)=(n*x*(x-1)+380)(n*x-1)
=>(n²*x²*(x-1)+39*n*x*(x-1)+380*(x-1))=n*x*(x-1)*(n*x-1)
+380*(n*x-1)
=>(n²*x²*(x-1)+39*n*x*(x-1)+380*(x-1))=n²*x²*(x-1)-
n*x*(x-1)+380*(n*x-1)
=>40*n*x*(x-1)=380*(n-1)*x
=>2*n*(x-1)=19*(n-1) -- (#)
From the equation, n divides 19*(n-1), but n is 
co-prime to (n-1). So, n divides 19
So, n = 1 or 19. If n=1,x=1 which is not true because
1st prob will be 0 but 2nd prob will not be 0.
=> n=19 => From eq(#) 
2*(x-1)=(19-1)
=> x-1=18/2 = 9 => x=10
No. of balls in the urn = n*x=19*10=190.

Posted by Praneeth on 2008-11-12 07:02:32