## Basic concepts

The starting point of the probability calculation is the concept of the random experiment, e.g. The process of tossing a coin or dice once or several times. Such a process is carried out according to certain rules, can be repeated as often as desired and its result depends on chance. The totality of all possible elementary results of a random experiment (elementary events) forms a set, which is used as an event space or sample space$\Omega $ referred to as.

- example

The event space is created when a die is thrown once

- $$\Omega =\{1,2,3,4,5,6\}.$$

If you toss a coin twice, you get a lot of ordered pairs

- $$\Omega =\{(\text{head},\text{head}),(\text{head},\text{number}),(\text{number},\text{head}),(\text{number},\text{number})\},$$

whereby $(\mathit{i},\mathit{j}),\mathit{i},\mathit{j}=$“Head”, “tails” represents an elementary event.

The term event is understood to be a subset of the event space. An example is throwing a number $\ge $ 3 the event

- $$\mathit{A.}=\{3,4,5,6\}.$$

When tossing a coin twice, the event is that a "head" occurs on the first toss:

- $$\mathit{A.}=\{(\text{head},\text{number}),(\text{head},\text{head})\}$$

and that a "head" occurs on the second throw:

- $$\mathit{B.}=\{(\text{head},\text{head}),(\text{number},\text{head})\}.$$

The union of events $\mathit{A.}$ and $\mathit{B.}$, $\mathit{A.}\cup \mathit{B.}$, is the event that a "head" occurs on the first or second throw:

- $$\mathit{A.}\cup \mathit{B.}=\{(\text{head},\text{head}),(\text{head},\text{number}),(\text{number},\text{head})\}.$$

The average of $\mathit{A.}$ and $\mathit{B.}$, $\mathit{A.}\cap \mathit{B.}$, is the event that a "head" appears on the first and second throw:

- $$\mathit{A.}\cap \mathit{B.}=\{(\text{head},\text{head})\}.$$

We denote the probability for the event $\mathit{A.}$ of a random experiment with the real number $\mathit{p}(\mathit{A.})$who have the condition

- $$0\le \mathit{p}(\mathit{A.})\le 1$$

enough. is $\mathit{A.}$ the whole event space $\Omega $, then applies

- $$\mathit{p}(\Omega )=1.$$

The probability of the impossible event $\varnothing $ amounts to

- $$\mathit{p}(\varnothing )=0.$$

Close two events $\mathit{A.}$ and $\mathit{B.}$ mutually exclusive, i.e. $\mathit{A.}\cap \mathit{B.}=\varnothing $ ($\mathit{A.}$ and $\mathit{B.}$ are disjoint), then applies

- $$\mathit{p}(\mathit{A.}\cup \mathit{B.})=\mathit{p}(\mathit{A.})+\mathit{p}(\mathit{B.}).$$

Generally for $\mathit{n}$ mutually exclusive events in pairs ${\mathit{A.}}_{1},{\mathit{A.}}_{2},\dots ,{\mathit{A.}}_{\mathit{n}}$ is applicable

- $$\mathit{p}({\mathit{A.}}_{1}\phantom{\rule{.4em}{0ex}}\cup \phantom{\rule{.4em}{0ex}}{\mathit{A.}}_{2}\phantom{\rule{.4em}{0ex}}\cup \phantom{\rule{.4em}{0ex}}\dots \phantom{\rule{.4em}{0ex}}\cup \phantom{\rule{.4em}{0ex}}{\mathit{A.}}_{\mathit{n}})=\mathit{p}({\mathit{A.}}_{1})+\mathit{p}({\mathit{A.}}_{2})+\dots +\mathit{p}({\mathit{A.}}_{\mathit{n}}).$$

- example

What is the probability that if you roll a dice twice, the sum of the two dice will be 5? The event $\mathit{A.}$ for it is

- $$\mathit{A.}=\{(1,4)\}\phantom{\rule{.4em}{0ex}}\cup \phantom{\rule{.4em}{0ex}}\{(4,1)\}\phantom{\rule{.4em}{0ex}}\cup \phantom{\rule{.4em}{0ex}}\{(2,3)\}\phantom{\rule{.4em}{0ex}}\cup \phantom{\rule{.4em}{0ex}}\{(3,2)\}$$

and the probability accordingly

- $$\begin{array}{rcl}\mathit{p}(\mathit{A.})& =& \mathit{p}(\{(1,4)\})+\mathit{p}(\{(4,1)\})+\mathit{p}(\{(2,3)\})+\mathit{p}(\{(3,2)\})\\ & =& \frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}\\ & =& \frac{1}{9}.\end{array}$$

An event $\mathit{A.}$ and its complementary event $\overline{\mathit{A.}}$ are by definition mutually exclusive

- $$\mathit{A.}\cap \overline{\mathit{A.}}=\varnothing ,\phantom{\rule{1em}{0ex}}\mathit{A.}\cup \overline{\mathit{A.}}=\Omega $$

and thus is

- $$\mathit{p}(\mathit{A.})+\mathit{p}(\overline{\mathit{A.}})=1.$$

- theorem
- Close two events $\mathit{A.}$ and $\mathit{B.}$not mutually exclusive, i.e. $\mathit{A.}\cap \mathit{B.}\ne \varnothing $, then applies

- $$\mathit{p}(\mathit{A.}\cup \mathit{B.})=\mathit{p}(\mathit{A.})+\mathit{p}(\mathit{B.})-\mathit{p}(\mathit{A.}\cap \mathit{B.}).$$

One draws $\mathit{p}(\mathit{A.}\cap \mathit{B.})$ so that the natural events of the average are not recorded twice.

- proof

$\mathit{A.}\cup \mathit{B.}$ can be written as the union of two disjoint sets

- $$\mathit{A.}\cup \mathit{B.}=\mathit{A.}\cup (\overline{\mathit{A.}}\cap \mathit{B.})\phantom{\rule{1em}{0ex}}\Rightarrow \phantom{\rule{1em}{0ex}}\mathit{p}(\mathit{A.}\cup \mathit{B.})=\mathit{p}(\mathit{A.})+\mathit{p}(\overline{\mathit{A.}}\cap \mathit{B.}).$$

For $\mathit{B.}$ applies similarly

- $$\mathit{B.}=(\mathit{A.}\cap \mathit{B.})\cup (\overline{\mathit{A.}}\cap \mathit{B.})\phantom{\rule{1em}{0ex}}\Rightarrow \phantom{\rule{1em}{0ex}}\mathit{p}(\mathit{B.})=\mathit{p}(\mathit{A.}\cap \mathit{B.})+\mathit{p}(\overline{\mathit{A.}}\cap \mathit{B.}).$$

Combine the two equations to get $\mathit{p}(\overline{\mathit{A.}}\cap \mathit{B.})$ to be eliminated, the result is the addition theorem

- $$\mathit{p}(\mathit{A.}\cup \mathit{B.})=\mathit{p}(\mathit{A.})+\mathit{p}(\mathit{B.})-\mathit{p}(\mathit{A.}\cap \mathit{B.}).$$

- example

What is the probability that at least one coin will show "heads" when two coins are tossed? The event space of the random experiment is

- $$\Omega =\{\mathit{K}\mathit{K},\mathit{K}\mathit{Z},\mathit{Z}\mathit{K},\mathit{Z}\mathit{Z}\},$$

whereby $\mathit{K}\mathit{K}=(\text{head},\text{head})$ etc. is. The event that the first coin shows "heads" is

- $$\mathit{A.}=\{\mathit{K}\mathit{K},\mathit{K}\mathit{Z}\}$$

and is the probability $\mathit{p}(\mathit{A.})=\mathrm{0,5}$. The event that the second coin shows "heads" is

- $$\mathit{B.}=\{\mathit{K}\mathit{K},\mathit{Z}\mathit{K}\}$$

and is the probability $\mathit{p}(\mathit{B.})=\mathrm{0,5}$. The searched event that at least one coin shows "heads" is

- $$\mathit{C.}=\mathit{A.}\cup \mathit{B.}.$$

events $\mathit{A.}$ and $\mathit{B.}$ are not mutually exclusive. After the addition theorem is

- $$\mathit{p}(\mathit{C.})=\mathit{p}(\mathit{A.})+\mathit{p}(\mathit{B.})-\mathit{p}(\mathit{A.}\cap \mathit{B.}).$$

the end $\mathit{A.}\cap \mathit{B.}=\{\mathit{K}\mathit{K}\}$ surrendered $\mathit{p}(\mathit{A.}\cap \mathit{B.})=\mathrm{0,25}$. So is

- $$\mathit{p}(\mathit{C.})=\mathrm{0,5}+\mathrm{0,5}-\mathrm{0,25}=\mathrm{0,75}.$$