Answer
The set of chips in the urn is:
\[\begin{align}
& \bigcup\limits_{k=1}^{\infty }{({{A}_{k}}-\{k\})}=\bigcup\limits_{k=1}^{\infty }{{{A}_{k}}-\bigcup\limits_{k=1}^{\infty }{\{k\}}} \\
& =\varnothing
\end{align}\]
Work Step by Step
Let ${{A}_{1}}=\{11,12,....,20\}$ be the set of chips placed in the urn at $\frac{1}{{{2}^{2}}}$minute until midnight and chip number 2 is quickly removed.
Let ${{A}_{2}}=\{21,22,....,30\}$ be the set of chips placed in the urn at $\frac{1}{{{2}^{3}}}$minute until midnight and chip number 3 is quickly removed and so on.
In general, ${{A}_{k}}$ is the set of chips placed in the urn at $\frac{1}{{{2}^{k}}}$ minutes until midnight and chip number k is quickly removed.
Therefore, the set of chips in the urn is:
\[\begin{align}
& \bigcup\limits_{k=1}^{\infty }{({{A}_{k}}-\{k\})}=\bigcup\limits_{k=1}^{\infty }{{{A}_{k}}-\bigcup\limits_{k=1}^{\infty }{\{k\}}} \\
& =\varnothing
\end{align}\]
Since \[\bigcup\limits_{k=1}^{\infty }{{{A}_{k}}}\]is a subset of \[\bigcup\limits_{k=1}^{\infty }{\{k\}}\].
We can calculate the number of chips in the urn at midnight using the set of chips in the urn as:
\[\begin{align}
& =[1+[2(20-1)+{{2}^{2}}]+[3(30-1)+{{3}^{2}}] \\
& =1+42+114 \\
& =157 \\
\end{align}\]
