1.1 Probability Space

Probability Space

A probability space is a triple $(\Omega ,\mathcal{F},\mathbb{P})$ used to model random experiments and the associated outcomes. It consists of

1. A sample space $\text{Upomega}$, which represents the set of all possible single outcomes (sample points) $\omega \in \Omega$ of a random experiment. An event $A\subseteq \Omega$ is a subset of the sample space, and consists of one or more mutually exclusive sample points.
2. A
3. A probability measure $\mathbb{P}:\Omega \to \mathbb{R}$ for each sample point (outcome) $\text{upomega}\in \text{Upomega}$, which follows the following axioms:

• (Non-Negativity) $0\le \mathbb{P}[\text{upomega}]\le 1$ for all $\text{upomega}\in \text{Upomega}$.

• (Sum to One) ${\sum }_{\text{upomega}\in \text{Upomega}}^{}\mathbb{P}[\text{upomega}]=1$.

Completeness (Sum to One) $\omega$ of the experiment, it will always be an element of the sample space. In other words, $P[\Omega ]=1$.

The sum to one condition also means that a probability space must be collectively exhaustive so that no matter the outcome

Properties

(Subsets) If $A\subset B$, then $\mathbb{P}[A]<\mathbb{P}[B]$.

Examples

Birthday Paradox

We want to calculate the probability of at least two people sharing the same birthday in a group of n people. We start by saying that $S=\{1,...,365\}$ since there are $365$ days to sample $n$ from. Then, $|\text{Upomega}|=365^n$ (combinatorial sequence), since for each additional sample point, there are 365 available choices.

We let A be the event that at least two people share the same birthday. Then we need to calculate $\mathbb{P}[A]$. However, it is easier to do so using the complement of A, $\mathbb{P}[\bar{A}]$, since it only takes into account instances where everyone has a different birthday. Since this is the case, $|\bar{A}|=\frac{365!}{(365-n)!}$ (FRC w/o replacement) since each person in the group has to have a different birthday. Therefore,
$\mathbb{P}[A]=1-\mathbb{P}[\bar{A}]=1-\frac{|\bar{A}|}{|\text{Upomega}|}=1-\frac{365!}{(365-n)!\times 365^n}$
This means that in a group of 60 people, there is a 99% chance of two or more people sharing the same birthday.