1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
|
\title{Eksamnens Noter}
The universal set or sample space is the set everything, and is denoted $S$.
Therefore the probability of hitting $S$ is $P(S) = 1$.
This is the first of 3 axioms repeated below.
\begin{enumerate}
\item For any event $A$, $P(A) \geq 0$.
\item The probability of hitting sample space is always 1, $P(S) = 1$.
\item If events $A_1, A_2, ...$ are \textbf{disjoint} event, then
\begin{equation}
P(A_1 \cup A_2 ...) = P(A_1) + P(A_2)\,.
\end{equation}
\end{enumerate}
The last axiom requires that the events $A_n$ are disjoint.
If they aren't one should subtract the part they have in common.
This is called the \emph{Inclusion-Exclusion Principle}.
\begin{principle}
The \emph{Inclusion-Exclusion Principle} is defined as
\begin{equation}
P(A \cup B) = P(A) + P(B) - P(A \cap B)\,.
\end{equation}
Definition with 3 events can be found in the in the book.
\end{principle}
\section{Counting}
The probability of a event $A$ can be found by
\begin{equation}
P(A) = \frac {|A|} {|S|}\,.
\end{equation}
It is therefore required to count how many elements are in $S$ and $A$.
The most simple method is the \emph{multiplication principle}.
\begin{principle}[Multiplication principle]
Let there be $r$ random experiments, where the $k$'th experiment has $n_k$ outcomes.
Then there are
\begin{equation}
n_1 \cdot n_2 \cdot ... \cdot n_r
\end{equation}
possible outcomes over all $r$ experiments.
\end{principle}
Possible outcomes when choosing $k$ objects from a basket with $n$ objects.
\begin{itemize}
\item Sampling with Replacement and with Ordering
\[
n^k
\]
\item Sampling without Replacement and with Ordering
\[
n(n - 1) \dots (n - k - 1)
\]
\item Sampling without Replacement and without Ordering
\[
\frac{n(n-1) \dots (n - k - 1)}{k!} = \frac {n!} {(n- k)! k!} = \binom n k
\]
\item Sampling with Replacement and without Ordering
\[
\binom {n - 1 - k} k
\]
\end{itemize}
\section{Independence of Events}
Describes whether the outcome of event $B$, changes the changes of event $A$ occuring.
If this is false, it can be said that $P(A|B) = P(A)$.
\begin{definition}[Independence]
Two events are independent if
\[
P(A \cap B) = P(A) P(B)
\]
\end{definition}
\section{Mixed Definitions}
\begin{definition}[Conditional Probability]
The probability that $A$ occurs given that $B$ occurs:
\[
P(A|B) = \frac {P(A \cap B)} {P(B)}\,,\:P(B) > 0
\]
\end{definition}
\begin{definition}[Baye's Rule]
For any two events $A$ and $B$, where $P(A) \neq 0$, we have that
\[
P(B|A) = \frac {P(A|B) P(B)} {P(A)}\,.
\]
\end{definition}
|
