What is the appropriate probability distribution to model this situation? I want to model a random variable which represents the number of failures before success in a repeated Bernoulli trial. I will conduct only utmost N trails and I am guaranteed of one success (only one success possible) if N trials are conducted. What is the appropriate distribution? If N can go to infinity I think geometric  distribution  models the situation, but in my case N is finite and is known in advance. 
EDIT
My situation is such that the random variable has the geometric distribution properties, in the sense that, the first trial is most likely to successful, the success rate decreasing as trials increases. But within N trials, I am sure of success. 
 A: Assuming I've understood your question correctly, the distribution should be uniform over the N trials.
What you want is $P(k-1$ failures before $1^{st}$ success $|$ only 1 success in $N$ trials$)$. This is the same as $P($success at k^{th} location $|$ 1 success and $N-1$ failures in $N$ trials$)$.
P(success at $k^{th}$ location $|$ 1 success and $N-1$ failures) = P(success at $k^{th}$ location AND 1 success and $N-1$ failures)/P(1 success and $N-1$ failures).
the numerator is $p(1-p)^{N-1}$. The denominator is $Np(1-p)^{N-1}$.
This intuitively makes sense because if you guarantee exactly one success in N trials, there are N possible locations where you can put the successful trial and they are all equally likely.
A: Suppose you throw a die repeatedly and the probability is $p$ that you get an ace on each trial.  Then the probability distribution of the number of failures before the first success is a geometric distribution.  Let $X$ be the number of failures before the first success.  Then $\Pr(X = x) = (1-p)^x p$, for $x\in\{0,1,2,3,\ldots\}$.
But now let's look at the conditional probability distribution of $X$ given the event $A$ that exactly one success occurs in the first $N$ trials.  We're looking for
$$
\begin{align}
\Pr(X = x \mid A) & = \frac{\Pr(X=x\  \&\  A)}{\Pr(A)} \\  \\
&  = \frac{\overbrace{(1-p)\cdot (1-p)\cdot (1-p)\cdots (1-p)}^{x-1}\  \  p\  \  \overbrace{(1-p)\cdots(1-p)}^{N-x} }{\binom N 1 p^1(1-p)^{N-1}} \\  \\
&  = \frac{1}{\binom N 1} = \frac 1 N.
\end{align}
$$
The bottom line is $1/N$, no matter what number $x$ is.  It doesn't depend on $x$.  In other words, the conditional distribution is a uniform distribution, regardless of the value of $p$, and notwithstanding the fact that the unconditional distribution is a geometric distribution.
