Fix incorrect solution to Chapter 1 Mixed Problem 57 by adding the combinatorial part for Principle of Inclusion-Exclusion

Author: skntrl

src/chapters/1/sections/mixed_practice/problems/57.tex

@@ -3,15 +3,14 @@
 with Caesar. We can compute the desired probability using inclusion exclusion.
 
 Since every molecule in the universe is equally likely to be shared with Caesar, 
-and we assume our breath samples molecules with replacement, 
-$P(\bigcap\limits_{i=1}^{n}A_{i}) = (\frac{1}{10^{22}})^{n}$. 
+and we assume our breath samples molecules with replacement, for any $k$ specific molecules, 
+$P(\bigcap\limits_{i=1}^{k}A_{i_{k}}) = (\frac{10^{22}}{10^{44}})^{k} = (\frac{1}{10^{22}})^{k}$, where $A_{i_{k}}$ is index variable to represent a specific, randomly chosen subset of $k$ events out of the total ${10^{22}}$.
 
 Thus,
 
 \begin{flalign}
-P(\bigcup\limits_{i=1}^{10^{22}}A_{i}) & = 
-\sum_{i=1}^{10^{22}}(-1)^{i+1}\left(\frac{1}{10^{22}}\right)^{i} \nonumber && \\
-& = \left(1 - \frac{1}{10^{22}}\right)^{10^{22}} \nonumber && \\
-& \approx e^{-1} \nonumber 
+P(\bigcup\limits_{i=1}^{10^{22}}A_{i}) & = \sum_{k=1}^{10^{22}}(-1)^{k+1}\binom{10^{22}}{k}\left(\frac{1}{10^{22}}\right)^{k} \nonumber && \\
+& = 1 - \sum_{k=0}^{10^{22}}\binom{10^{22}}{k}\left(-\frac{1}{10^{22}}\right)^{k} \nonumber && \\
+& = 1 - \left(1 - \frac{1}{10^{22}}\right)^{10^{22}} \nonumber && \\
+& \approx 1 - e^{-1} \nonumber 
 \end{flalign}
-