Consider an experiment.
The state space \(S\) is the set of all possible outcomes.
Example: We roll a die. The possible outcomes are \(S=\{1,2,3,4,5,6\}\).
Example: We measure wind speed (in m/s). The state space is \([0,\infty)\).
An event is a subset \(A\subseteq S\) of the sample space.
Example: Rolling a die and getting an even number is the event \(A=\{2,4,6\}\).
Example: Measuring a wind speed of at least 5m/s is the event \([5,\infty)\).
Consider two events \(A\) and \(B\).
The union \(A\cup B\) of is the event that either \(A\) or \(B\) occurs.
The intersection \(A\cap B\) of is the event that both \(A\) and \(B\) occurs.
Example: We roll a die and consider the events \(A=\{2,4,6\}\) that we get an even number and \(B=\{4,5,6\}\) that we get at least 4. Then
\(A\cup B = \{2,4,5,6\}\)
\(A\cap B = \{4,6\}\)
\(A^c = \{1,3,5\}\)
Consider two events \(A\) and \(B\).
If \(A\) and \(B\) are mutually exclusive (never occur at the same time, i.e. \(A\cap B=\emptyset\)), then
\[ P(A\cup B) = P(A) + P(B). \] 
\[P(A\cup B) = P(A) + P(B) = \tfrac{1}{6} + \tfrac{1}{6} = \tfrac{1}{3}. \]
\[ P(A\cup B) = P(A) + P(B) - P(A\cap B).\]
\[P(A\cup B) = P(A) + P(B) -P(A\cap B)= \tfrac{1}{3} + \tfrac{1}{3} - \tfrac{1}{6} = \tfrac{1}{2}.\]
Consider events \(A\) and \(B\).
The conditional probability of \(A\) given \(B\) is defined by \[ P(A|B) = \frac{P(A\cap B)}{P(B)}\] if \(P(B)>0\).
Example: We toss a coin two times. The possible outcomes are \(S=\{HH,HT,TH,TT\}\). Each outcome has probability \(\tfrac{1}{4}\). What is the probability of at least one head if we know there was at least one tail?
Let \(X\) be a discrete stochastic variable which can take the values \(x_1,x_2,\ldots\)
The distribution of \(X\) is given by the probability function, which is given by \[f(x_i)=P(X=x_i), \quad i=1,2,\ldots\]
Example: We throw a coin three times and let \(X\) be the number of heads. The possible outcomes are \[S=\{HHH,HHT,HTH,HTT,THH,THT,TTH,TTT\}.\] The probability function is
Let \(X\) be a discrete random variable with probability function \(f\). The distribution function of \(X\) is given by \[F(x)=P(X\leq x) = \sum_{y\leq x} f(y), \quad x\in \mathbb{R}.\]
Example: For the three coin tosses, we have
The distribution of a continuous random variable \(X\) is given by a probability density function \(f\), which is a function satisfying
\(f(x)\) is defined for all \(x\) in \(\mathbb{R}\),
\(f(x)\geq 0\) for all \(x\) in \(\mathbb{R}\),
\(\int_{-\infty}^{\infty} f(x)dx = 1\).
The probability that \(X\) lies between the values \(a\) and \(b\) is given by
\[P(a<X<b) = \int_a^b f(x) dx.\]
Notes:
Condition 3. ensures that \(P(-\infty < X < \infty) = 1\).
The probability of \(X\) assuming a specific value \(a\) is zero, i.e. \(P(X=a)=0\).
The uniform distribution on the interval \((A,B)\) has density \[
f(x)=
\begin{cases}
\frac{1}{B-A} & A \leq x \leq B \\
0 & \text{otherwise}
\end{cases}
\] 
Example: If \(X\) has a uniform distribution on \((0,1)\), find \(P(\tfrac{1}{3}<X\leq \tfrac{2}{3})\).
\[P\left(\tfrac{1}{3}<X\leq \tfrac{2}{3}\right) =P\left(\tfrac{1}{3}<X < \tfrac{2}{3}\right) + P\left(X = \tfrac{2}{3}\right)\\ = \int_{1/3}^{2/3}f(x)dx + 0 =\int_{1/3}^{2/3}1dx = \frac{1}{3}.\]
Let \(X\) be a continuous random variable with probability density \(f\). The distribution function of \(X\) is given by \[F(x)=P(X\leq x) = \int_{-\infty}^{x} f(y) dy, \quad x\in \mathbb{R}.\]
