sample space
A sample space $$\Omega$$ is a non-empty set
event
Given a sample space $$\Omega$$, we say that any subset $$E \subseteq \Omega$$ is an event
probability measure
Given a sample space $$\Omega$$ and a function $$P : \Omega \to \left [ 0 , 1 \right ]$$ , then we say that $$P$$ is a probability measure if the following holds
• $$P \left ( \Omega \right )$$ $$= 1$$
• $$\forall E_{1} , E_{2} , \ldots \subseteq \Omega$$ such that $$E_{1} , E_{2} , \ldots$$ are all pairwise disjoint
$$P \left ( \bigcup_{i = 1}^{\infty} E_{i} \right ) = \sum_{i = 1}^{\infty} P \left ( E_{i} \right )$$
finite pairwise disjoint probability
Suppose that $$A_{1} , A_{2} , \ldots , A_{n}$$ are pairwise disjoint, then $$P \left ( A_{1} \cup \ldots \cup A_{n} \right ) = \sum_{i = 1}^{n} P \left ( A_{i} \right )$$
In the definition of a probability measure, for $$i \in \left \lbrace 1 , \ldots , n \right \rbrace$$ we set $$E_{i} = A_{i}$$ and then for any $$j > n$$ we set $$E_{j} = \emptyset$$, then we can see that $$E_{1} , E_{2} , \ldots$$ is pairwise disjoint, and thus
 $$P \left ( A_{1} \cup A_{2} \cup \ldots \cup A_{n} \right )$$ $$=$$ $$P \left ( A_{1} \cup A_{2} \cup \ldots \cup A_{n} \cup \emptyset \cup \emptyset \cup \ldots \right )$$ $$=$$ $$P \left ( \bigcup_{i = 1}^{\infty} E_{i} \right )$$ $$=$$ $$\sum_{i = 1}^{\infty} P \left ( E_{i} \right )$$ $$=$$ $$\sum_{i = 1}^{n} P \left ( E_{i} \right ) + P \left ( \emptyset \right ) + P \left ( \emptyset \right ) + P \left ( \emptyset \right ) + \ldots$$ $$=$$ $$\sum_{i = 1}^{n} P \left ( E_{i} \right ) + 0 + 0 + 0 + \ldots$$ $$=$$ $$\sum_{i = 1}^{n} P \left ( A_{i} \right )$$
probability of a single point with equally likely outcomes is zero
Suppose that our sample space is $$\left [ 0 , 1 \right ]$$ and there is some $$c \in \left [ 0 , 1 \right ]$$ such that for any $$x \in \left [ 0 , 1 \right ]$$, $$P \left ( \left \lbrace x \right \rbrace \right ) = c$$, then $$c = 0$$
Suppose that $$c \ne 0$$, and note that $$P \left ( \Omega \right ) = P \left ( \left [ \left . 0 , 1 \right . \right ] \right ) = \sum_{x \in \left [ 0 , 1 \right ]} P \left ( x \right ) = \sum_{x \in \left [ 0 , 1 \right ]} c$$, but note that no matter how small the value of $$c$$, since we are summing it uncountably many times with itself the sum always goes to infinity, but at the same time, $$P \left ( \Omega \right ) = 1$$ and thus we have a contradiction, so $$c = 0$$.
random variable
Suppose that $$\Omega$$ is a sample space. A random variable is a function $$X : \Omega \to \mathbb{R}$$
inverse image notation
Suppose that $$X : \Omega \to \mathbb{R}$$ is a random variable and that $$y \in \mathbb{R}$$, then we define the string of characters $$X = y$$ to be the set $$X^{- 1} \left ( \left \lbrace y \right \rbrace \right )$$
conditional probability
Suppose that $$A , B \subseteq \Omega$$, with $$P \left ( B \right ) \gt 0$$, then we define the conditional probability of $$A$$ given $$B$$ as
$$P \left ( A \mid B \right ) = \frac{P \left ( A \cap B \right )}{P \left ( B \right )}$$