Smallest no of balls in the box? A box contains white and black balls. When two balls are drawn without replacement suppose the probability that both are white is  1 /3. 
a) Find smallest number of balls in the box ? 
b) How small can the total number of balls be if black balls are even in  number  ? 
I considered no of white balls as 'n' and no of black balls as 'm'. 
So for first two balls to be white it would be 
{n/(m+n)}*{(n-1)/(m+n-1)} = 1/3 . But how to solve further ? we have only one equation 
Can anyone please explain in detail. 
Thanks in advance!  

$$\frac{n}{m+n}\frac{n-1}{m+n-1} = \frac{1}{3}$$
$$\frac{n(n-1)}{(m+n)(m+n-1)} = \frac{1}{3}$$
$$\frac{n(n-1)}{m^2 + 2nm + n^2 - m - n} = \frac{1}{3}$$
$$\frac{n(n-1)}{(m+n)^2 - (m+n)} = \frac{1}{3}$$
Let us now set $m+n = B$, the number of total balls
$$3(n^2-n) = B^2 - B$$
$$B^2 - B - 3(n^2-n) = 0$$
$$B = \frac{1 \pm \sqrt{1+12(n^2-n)}}{2}$$
$$m = \frac{1 \pm \sqrt{1+12(n^2-n)}}{2}-n$$
Since $m$ must be an integer, $1+12(n^2-n)$ must be some $k^2$ for some whole number $k$. We throw in some values for $n$ until we get an integer. The first one we get is $n=2$, yielding $k=5$.
$$m = \frac{1 \pm \sqrt{25}}{2}-2$$
$m$ must be positive, throwing out one solution of the quadratic
$$\therefore n=2, \,\,m = 1$$
