Fixes assignment 6

1 files changed, 8 insertions, 6 deletions

diff --git a/assignment6.tex b/assignment6.tex
index b12af05..bbb3cb2 100644
--- a/assignment6.tex
+++ b/assignment6.tex
@@ -41,7 +41,7 @@
             \item She can choose one combination out of twenty, and then another but not the same out of nineteen. However, we count every combination twice (``$a-b$'' and ``$b-a$''), so we have to take $20\cdot19/2=190$.
         \end{enumerate}
 
-    \item We first pick the five persons, with ${8\choose5}$ possibilities. Then for every combination we have $5!$ possible permutation. This gives ${8\choose5}\cdot5!=806400$. %todo
+    \item We first pick the five persons, with ${8\choose5}$ possibilities. Then for every combination we have $5!$ possible permutation. This gives ${8\choose5}\cdot5!=6720$.
 
     \item \begin{enumerate}
             \item For the diamonds we have $13$ possibilities out of $52$, for the clubs $13$ out of $51$ and for the hearts $13$ out of $50$. That gives $\frac{13}{52}\cdot\frac{13}{51}\cdot\frac{13}{50}=\frac{13^2}{4\cdot50\cdot51} = \frac{169}{10200}$.
@@ -49,7 +49,7 @@
         \end{enumerate}
 
     \item \begin{enumerate}
-            \item We choose the first choose the last digit, out of four ($2,4,6,8$) possibilities. Then, the first four digits can be choosen from $8,7,6$ and $5$ remaining digits. That gives $\frac{8!}{4!} = 1680$ numbers.
+            \item We choose the first choose the last digit, out of four ($2,4,6,8$) possibilities. Then, the first four digits can be chosen from $8,7,6$ and $5$ remaining digits. That gives $\frac{8!}{4!} = 1680$ numbers.
             \item We choose two odd digits with $\frac{5\cdot4}{2}=10$ possibilities. Then we choose the even digits with $\frac{4\cdot3\cdot2}{3!} = 4$ possibilities. Then we can permute these digits in $5!$ possible ways. This gives $10\cdot4\cdot5! = 4800$ possibilities.
             \item \begin{enumerate}[label=(\alph*)]
                 \item The first four digits aren't restricted. The last digit has four possibilities ($2,4,6,8$). That gives $9^4\cdot4=26244$ numbers.
@@ -57,7 +57,9 @@
                 \end{enumerate}
         \end{enumerate}
 
-    \item \begin{enumerate}
+    \item I'm assuming a lexicographical ordering.
+        
+        \begin{enumerate}
             \item $26^4\cdot10^2 = 45697600$ possible license plates.
             \item In the range DAA-00-A till FZZ-99-Z would be $3\cdot26^2\cdot10^2\cdot26 = 5272800$ license plates. In the range DAA-00-A till DAZ-99-Z would be $26\cdot10^2\cdot26 = 67600$ license plates. In the range DBA-00-A till DBA-99-Z would be $10^2\cdot26=2600$ license plates. In the range DBB-00-A till DBB-00-Z would be $26$ license plates and in the range DBB-01-A till DBB-01-B are $2$. In total that gives us $5272800 - 67600 - 2600 - 26 - 2 = 5202572$ license plates.
             \item DFA-78-V is within the range, so it belongs to a moped. DAF-78-A is not in the range, so it belongs to a car. DCF-78-V is in the range, so it belongs to a moped.
@@ -85,11 +87,11 @@
                     \item $\{TH,HT,HH\}$.
                     \item $\{HT,HH\}$.
                 \end{enumerate}
-            \item $P_1 : \pazocal P(S) \to [0,1]$ with $P_1(A) = |A|\cdot\frac14$.
+            \item $P_1 : \pazocal P(S) \to [0,1]$ with $P_1(A) = |A|\cdot\frac14$, since every singleton event occurs with the same probability.
         \end{enumerate}
 
     \item \begin{proof}
-            We know $P(S) = 1$. Furthermore, $S$ and $\emptyset$ are mutually exclusive events since $S\cap\emptyset=\emptyset$. Therefore, $P(S\cup\emptyset) = P(S) + P(\emptyset)$. But we know $P(S\cup\emptyset)=P(S)=1$. This then reduces to $1 = 1 + P(\emptyset)$, which means $P(\emptyset)=0$.
+            We know $P(S) = 1$. Furthermore, $S$ and $\emptyset$ are mutually exclusive events since $S\cap\emptyset=\emptyset$. Therefore, $P(S\cup\emptyset) = P(S) + P(\emptyset)$. But we know $P(S\cup\emptyset)=P(S)=1$. Then the equality $P(S\cup\emptyset) = P(S) + P(\emptyset)$ reduces to $1 = 1 + P(\emptyset)$, which means $P(\emptyset)=0$.
         \end{proof}
 
     \setcounter{enumi}{13}
@@ -97,7 +99,7 @@
     \item \begin{enumerate}
             \item The probability that we have an Ace on first draw but not on second draw.
             \item The probability that we have an Ace on first draw or on second draw (or both).
-            \item The porbability that we don't draw an Ace.
+            \item The probability that we don't draw an Ace at all.
             \item The probability that we draw an Ace on first or third draw (or both), but not on second draw.
         \end{enumerate}