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 If $k$ of these balls are withdrawn one at a time, with each selection being equally likely to be any of the balls that remain at the time, what is the probability that the special ball is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

Can we say the following?

$\# \Omega={n\choose k}$

$A\equiv special\ ball\ was\ chosen $

$\# A^C={n-1 \choose k}$

$P(A)=1-\frac{{n-1 \choose k}}{{n\choose k}}$

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    $\begingroup$ The expression looks OK. This expression simplifies a lot. There are simpler direct approaches. You should not say that the probability the special ball was not chosen is $\binom{n-1}{k}$. This is the number of ways to not choose the special ball, for the probability we divide by $\binom{n}{k}$, which you later did. $\endgroup$ May 12, 2015 at 17:24
  • $\begingroup$ To emphasise what @André Nicolas said, you should simplify this. You could start with $1-\dfrac{{n-1 \choose k}}{{n\choose k}} = 1- \dfrac{\frac{(n-1)!}{k!(n-k-1)!}}{\frac{n!}{k!(n-k)!}}= 1- \dfrac{(n-1)!(n-k)}{n!}$ and so on until you get something simple (and I suspect it will be obvious once you have seen it). $\endgroup$
    – Henry
    May 12, 2015 at 23:21

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Yes, that's one way to do it.   Another way to do it is calculate the probability of selecting $k-1$ of the $n-1$ not special balls (and one of the one special balls) out of all the ways to choose $k$ of $n$ balls.   Then expand and cancel terms.

$$\begin{align}\mathsf P(A) & = \dfrac{\dbinom{1}{1}\dbinom{n-1}{k-1}}{\dbinom{n}{k}} \\[1ex] & = \dfrac{(n-1)!/(k-1)!(n-k)!}{n!/k!(n-k)!} \\[1ex] & = \dfrac k n \end{align}$$

And in hindsight you should see that this is the probability that the ball will be among the first $k$ of $n$ places if you line the balls up such that the special ball has an equal probability of being in each place.

Keep this in mind for the future.   Sometimes it is easier not to count the ways to select balls into places, but to count ways to select the places to for the balls to be put.

If things look complicated, take a step back and see if you can look at the problem from a different angle.

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