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!
There being 1 equation simply means that it has an infinite number of real solutions. But we don't want any real solution. The solutions need to be positive integers, and the problem asks for the smallest one, so you need to experiment.

Simplify the notation a bit by making $M$ the total number of balls. Your equation is $\frac{n}{M}\frac{n-1}{M-1} = \frac{1}{3}$, or $3n(n-1) = M(M - 1)$. Since everything is in integers in this last equation, you must have 3 divides either $M$ or $M-1$. What is the smallest value of $M$ such that 3 divides $M$ or $M-1$? And does that value of $M$ allow you to find an integer $n$ satisfying $3n(n-1) = M(M - 1)$?
there is another solution to this problem:

Let a and b denote the number of white and black balls in the box. We can thus define  the probability of the event as.

$$P(W_2|W_1)P(W_1) = ((a-1)/(a+b-1))\cdot (a/(a+b))$$
As $a/(a+b) &gt; (a-1)/(a+b-1)$
Since $\min(x,y)\le GM(x,y)\le\max(x,y)$
we can say that  $((a-1)/(a+b-1))^2 &lt; 1/3 &lt; (a/(a+b))^2$

Solving these inequalities we can say that 

$$b / (\sqrt3-1) &lt; a &lt;  (b-1+\sqrt3) / (\sqrt3-1)$$
if we compute $b = 1$ we find an interval of $1.366 &lt; a &lt; 2.366$ so we can say that $a = 2$ and the minimum number of balls is $3$.

You can compute as many $b$ as you wish...