Ball withdrawn from a box. White ball by kth draw. A box contains m white and n black balls. Balls are drawn at random one at a time without replacement. Find the probability of encountering a white ball by the kth draw.

This is our tutorial sheet's question and I don't know the answer. Thanks in advance.

Edit : I corrected the title and I misunderstood the question. I read 'at' instead of 'by'. With 'by', the question is very easy, still thank you for your answers. If you have solution for 'at', please answer. Again thank you for your efforts.
The probability of encountering a white ball by the $k^{th}$ is equal to $1 - P$ where $P$ is the probability that the first $k$ balls were all black. The latter probability is easier to compute. For example, the probability of the first ball being black is $\frac{n}{m+n}$. The probability that the second ball is also black is $\frac{n-1}{m+n-1}$. So, the probability that the first two balls are both black is $\frac{n(n-1)}{(m+n)(m+n-1)}$. The probability that the first $k$ balls are all black is similarly given by $\frac{n(n-1)...(n-k+1)}{(m+n)(m+n-1)...(m+n-k+1)}$. 

Note that if $k &gt; n$, then the probability becomes 1 because clearly you will have to draw a white ball by the $k^{th}$ draw (a version of Pigeon Hole Principle).
The problem is to find the probability of getting a white ball in $k$-th draw. May be for the first time or may be not... so we may have already gotten a white ball in first $k-1$ times. Therefore we have to consider them. If we consider '$i$' as no. of times a black ball can come in those $k-1$ times(as $k$-th is assumed to be a white ball) $i=0,1,2,.....k-1$.

Total probability is : $(m/m+n) + (n/m+n)(m/m+n-1) + (n/m+n)(n-1/m+n-1)(m/m+n-2) + 
                      (n/m+n)(n-1/m+n-1)(n-2/m+n-2)(m/m+n-3) + ...+ 
                      (n/m+n)(n-1/m+n-1)(n-2/m+n-2)...(n-k+2/m+n-k+2)(m/m+n-k+1)$