What is the probability that the special ball is chosen? I know this question is already posted here, but I doubting my own solution which is I know is wrong. It is quite basic, but I want to learn probability from scratch, so I am posting my question.
Question:

An urn contains $n$ balls, one of which is special. If $k$ of these balls are withdrawn one at a time, with each selection being equally likely to be any of the balls that remain at the time, what is the probability that the special ball is chosen?

My approach:
The number of ways in which we can select $k$ balls from a set of $n$ balls equals:
$$\binom{n}{k}$$
Out of these $ \binom{n}{k}$, we have a single special ball.

As such, the required probability equals: $$\frac {1} {\binom{n}{k}}$$

What am I doing wrong?

You have to count the number of selections of $k$ balls that contains the special ball : there are $\binom{n-1}{k-1}$, so probability is 
$$\frac{\binom{n-1}{k-1}}{\binom nk} = \frac kn$$
which can be found some simpler way.
