Lecture - Probability MT22, I
Notes
- An experiment is a set $\Omega$ of possible outcomes, e.g. a specific outcome $\omega \in \Omega$.
- $\Omega$ is the sample space.
- An event is a subset of $\Omega$, and probabilities can be assiged to events. E.g. flipping tails might be ${T}$.
- If $\omega$ is the observed outcome of an experiment, then $A$ ‘occurs’ if $\omega \in A$.
- $A^C$ occurs if A does not occur.
- $A \backslash B$, A occurs and B does not.
- $\mathbb{P}(A) = \frac{ \vert A \vert }{ \vert \Omega \vert }$ for a finite sample space.
- Counting is important to probability for finite sample spaces because you need to work out the number of outcomes relevant.
- Combinatorics is the “theory of counting”.
- For $X _ 1 X _ 1 X _ 1 X _ 2 X _ 2 \ldots X _ K …$ here each $X _ i$ occurs $m _ i$ times, then the number of ways of arranging them is given by $\frac{n!}{m _ 1!m _ 2!m _ 3!\ldots m _ k!}$ where $n$ is the total number of elements.
- The way to think about this is by considering what it would be like if each $X _ i$ were unique and how they fall into $m _ 1$ groups.
- For the case where $k = 2$, this is just the formula for the binomail coefficient. In general, it extends into the idea of a “multinomal” coefficient.
- ${}^n C _ m$ is the same as the number of ways of choosing a team of size $m$ from a squad of size $n$.
- Consider arranging $m$ ticks and $n-m$ crosses against each person in the squad.
Flashcards
In probability, what does $\Omega$ represent?
The set of all possible outcomes, i.e. the sample space.
What’s the sample space for a single coin flip?
\[\\{H, T\\}\]
What is an event, in terms of the sample space $\Omega$?
A subset of the sample space.
If the observed outcome of an experiment is $\omega$, $\omega \in \Omega$, what does it mean for event A to occur?
\[\omega \in A\]
In the case where $\Omega$ is finite and all outcomes are equally likely, what’s the formula for the probability of $A$?
\[\frac{|A|}{|\Omega|}\]
How many permutations are there of $n$ distinguishable objects?
\[n!\]
What’s the intuitive reason ${}^n C _ k = {}^n C _ {n-k}$?
Because choosing $n$ objects is the same as not choosing $n - k$ objects.