diff --git a/is/UB1/ISUB1.pdf b/is/UB1/ISUB1.pdf
index 2a82d91..52fef6a 100644
--- a/is/UB1/ISUB1.pdf
+++ b/is/UB1/ISUB1.pdf
Binary files differ
diff --git a/is/UB1/ISUB1.tex b/is/UB1/ISUB1.tex
index 71831be..b5a7b9e 100644
--- a/is/UB1/ISUB1.tex
+++ b/is/UB1/ISUB1.tex
@@ -69,7 +69,7 @@
 %Kommando für Aufgaben
 %\Aufgabe{AufgTitel}{Punktezahl}
 \newcommand{\Aufgabe}[2]{\stepcounter{n}
-    \indent\textbf{Exercise \arabic{n}: #1} (#2 Points)\\}
+    \textbf{Exercise \arabic{n}: #1} (#2 Points)\\}
 
 
 
@@ -95,13 +95,19 @@
                 one pair:\\
                 ${13\choose 1}{3\choose 2}{12 \choose 3}\cdot 3^3=231660$\\
                 $\mathds{P}(1\ pair|\neg Spades)=\frac{231660}{575757}=\frac{1980}{4921}\approx 0.402$
-                \item $A:=1\ pair,\ B:=\neg Spades$\\
-                $P(A|B)=\frac{P(B|A)P(A)}{P(B)}$\\
-                $P(A)=\frac{352}{833},\ P(B)=\frac{39}{52}\cdot\frac{38}{51}\cdot\frac{37}{50}\cdot\frac{36}{49}\cdot\frac{35}{48}=\frac{2109}{9520}$\\
-                $P(B|A)=1-P(\neg B| A)=1-\left(\frac{1}{4}+\frac{3}{4}\frac{1}{4}+\left(\frac{3}{4}\right)^2\frac{1}{4}+\left(\frac{3}{4}\right)^3\frac{1}{4}+\left(\frac{3}{4}\right)^4\frac{1}{3}\right)=1-\frac{101}{128}=\frac{27}{128}\approx 0.211$\\
-                $P(A|B)=\frac{\frac{27}{128}\cdot \frac{352}{833}}{\frac{2109}{9520}}=\frac{1980}{4921}\approx 0.0402$%TODO
+                \item \begin{enumerate}[(i)]
+                    \item $A:=1\ pair,\ B:=\neg Spades$\\
+                        $P(A|B)=\frac{P(B|A)P(A)}{P(B)}$\\
+                        $P(A)=\frac{352}{833},\ P(B)=\frac{39}{52}\cdot\frac{38}{51}\cdot\frac{37}{50}\cdot\frac{36}{49}\cdot\frac{35}{48}=\frac{2109}{9520}$\\
+                        $P(B|A)=1-P(\neg B| A)=1-\left(\frac{1}{4}+\frac{3}{4}\frac{1}{4}+\left(\frac{3}{4}\right)^2\frac{1}{4}+\left(\frac{3}{4}\right)^3\frac{1}{4}+\left(\frac{3}{4}\right)^4\frac{1}{3}\right)=1-\frac{101}{128}=\frac{27}{128}\approx 0.211$\\
+                        $P(A|B)=\frac{\frac{27}{128}\cdot \frac{352}{833}}{\frac{2109}{9520}}=\frac{1980}{4921}\approx 0.402$
+                    \item $A:=2\ pairs,\ P(A)=\frac{198}{4165}$, B as above\\
+                        $P(B|A)=1-\left(\frac{1}{4}+\frac{3}{4}\cdot\frac{1}{4}+\left(\frac{3}{4}\right)^2\cdot\frac{1}{4}+\left(\frac{3}{4}\right)^3\cdot\frac{1}{3}+\left(\frac{3}{4}\right)^3\frac{2}{3}\cdot\frac{1}{3}\right)=\frac{3}{16}$\\
+                        $P(A|B)=\frac{\frac{3}{16}\cdot \frac{198}{4165}}{\frac{2109}{9520}}=\frac{198}{4921}\approx0.0402$
+                \end{enumerate}
             \end{enumerate}
         \end{enumerate}
+\pagebreak
     \Aufgabe{Random Variables}{2+2+2=6}
         \begin{enumerate}
             \item $F\left(\frac{3}{2}\right)-F\left(\frac{1}{2}\right)=\frac{3}{4}-\frac{1}{4}=\frac{1}{2}$
@@ -119,10 +125,18 @@
                 0 & if\ x<0\\
                 \frac{\sqrt{2}}{2}\sqrt{x} & if\ 0\leq x\leq 2\\
                 1 & if\ x>2
-            \end{cases}$
+            \end{cases}$%TODO: adjust intervals
         \end{enumerate}
     \Aufgabe{Keep Rolling!}{4}
-        $\mathds{E}(X)=\frac{1}{\frac{1}{6}}=6$ (geometric distribution)\\
-        $\mathds{E}(X)=?$%TODO: Explain
+        \begin{align*}
+            \mathds{E}(X)
+            &=\sum\limits_{i=1}^{\infty} i\cdot p\cdot \left(1-p\right)^{i-1}\\
+            &=p\cdot\sum\limits_{i=1}^{\infty} i\left(1-p\right)^{i-1}\\
+            &=p\cdot \frac{d}{dp}\left[ -\sum\limits_{i=0}^{\infty}\left(1-p\right)^i\right] &\text{(geometric series)}\\
+            &=p\cdot \frac{d}{dp}\left(-\frac{1}{p}\right)\\
+            &=p\cdot \frac{1}{p^2}\\
+            &=\frac{1}{p}
+        \end{align*}
+        $\Rightarrow\mathds{E}(X)=\frac{1}{p}=\frac{1}{\frac{1}{6}}=6$
 \end{document}