Urn problem involving removement of half of the balls An urn contains $n$ heliotrope and $n$ tangerine balls. Half the balls are removed and placed in a box. One of those remaining in the urn is
removed. What is the probability that it is tangerine?
It is not a homework problem. I am just curious about the result. Such problem comes from the book "Elementary probability" from David Stirzaker. Any help or hint is greatly appreciated.
 A: Note the events like


*

*$B=\{\text{tangerine ball is extracted with the 2nd extraction}\}$ 

*$A_k=\{k - \text{tangerine balls exactly are extracted with the 1st extraction}\}$
And apply total probability
$$P(B)=\sum\limits_{k=0}^n P(B \mid A_k)\cdot P(A_k)$$
$P(B \mid A_k)$ are easy to calculate, $n$ balls were extracted 1st time, $k$ were tangerine, so $n-k$ tangerine balls were left, thus $P(B \mid A_k)=\frac{n-k}{n}$.
And $$P(A_k)=\frac{\binom{n}{n-k}\cdot \binom{n}{k}}{\binom{2n}{n}}$$
Finally
$$P(B)=\frac{1}{\binom{2n}{n}}\left(\sum\limits_{k=0}^n\frac{n-k}{n}\binom{n}{n-k}\binom{n}{k}\right)=
\frac{1}{\binom{2n}{n}}\left(\sum\limits_{k=0}^n\color{blue}{\frac{n-k}{n}\cdot\frac{n!}{(n-k)!k!}}\cdot\binom{n}{k}\right)=\\
\frac{1}{\binom{2n}{n}}\left(\sum\limits_{k=0}^n\color{blue}{\frac{(n-1)!}{(n-k-1)!k!}}\cdot\binom{n}{k}\right)=
\frac{1}{\binom{2n}{n}}\left(\sum\limits_{k=0}^n\color{blue}{\binom{n-1}{k}}\cdot\color{red}{\binom{n}{k}}\right)=\\
\frac{1}{\binom{2n}{n}}\left(\sum\limits_{k=0}^n\binom{n-1}{k}\cdot\color{red}{\binom{n}{n-k}}\right)=...$$
applying Vandermonde's identity
$$...=\frac{\binom{2n-1}{n}}{\binom{2n}{n}}=\frac{1}{2}$$
A: The answer is $\frac 12$.  Indeed, each ball is equally likely to be chosen by this process and half the balls have the desired color.  Had there been $a$ of the desired color and $2n-a$ of the undesired color, the answer would have been $\frac a{2n}$.
Note:  it's clear from symmetry that each ball is likely to be selected this way, but it's not difficult to explicitly compute the probability that a given ball is selected.  Indeed, for a given ball to be selected it must first not be deleted (probability $\frac 12$) and second it must be chosen from the $n$ survivors (probability $\frac 1n$).  Thus the probability that a given ball is selected is $\frac 1{2n}$ as desired.
