# Probability calculation

## 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$Ω$ referred to as.

example

The event space is created when a die is thrown once

$Ω={1,2,3,4,5,6}.$

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

$Ω={(head,head),(head,number),(number,head),(number,number)},$

whereby $(i,j),i,j=$“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 $≥$ 3 the event

$A.={3,4,5,6}.$

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

$A.={(head,number),(head,head)}$

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

$B.={(head,head),(number,head)}.$

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

$A.∪B.={(head,head),(head,number),(number,head)}.$

The average of $A.$ and $B.$, $A.∩B.$, is the event that a "head" appears on the first and second throw:

$A.∩B.={(head,head)}.$

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

$0≤p(A.)≤1$

enough. is $A.$ the whole event space $Ω$, then applies

$p(Ω)=1.$

The probability of the impossible event $∅$ amounts to

$p(∅)=0.$

Close two events $A.$ and $B.$ mutually exclusive, i.e. $A.∩B.=∅$ ($A.$ and $B.$ are disjoint), then applies

$p(A.∪B.)=p(A.)+p(B.).$

Generally for $n$ mutually exclusive events in pairs $A.1,A.2,…,A.n$ is applicable

$p(A.1∪A.2∪…∪A.n)=p(A.1)+p(A.2)+…+p(A.n).$
example

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

$A.={(1,4)}∪{(4,1)}∪{(2,3)}∪{(3,2)}$

and the probability accordingly

$p(A.)=p({(1,4)})+p({(4,1)})+p({(2,3)})+p({(3,2)})=136+136+136+136=19.$

An event $A.$ and its complementary event $A.¯$ are by definition mutually exclusive

$A.∩A.¯=∅,A.∪A.¯=Ω$

and thus is

$p(A.)+p(A.¯)=1.$
theorem
Close two events $A.$ and $B.$not mutually exclusive, i.e. $A.∩B.≠∅$, then applies
$p(A.∪B.)=p(A.)+p(B.)-p(A.∩B.).$

One draws $p(A.∩B.)$ so that the natural events of the average are not recorded twice.

proof

$A.∪B.$ can be written as the union of two disjoint sets

$A.∪B.=A.∪(A.¯∩B.)⇒p(A.∪B.)=p(A.)+p(A.¯∩B.).$

For $B.$ applies similarly

$B.=(A.∩B.)∪(A.¯∩B.)⇒p(B.)=p(A.∩B.)+p(A.¯∩B.).$

Combine the two equations to get $p(A.¯∩B.)$ to be eliminated, the result is the addition theorem

$p(A.∪B.)=p(A.)+p(B.)-p(A.∩B.).$
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

$Ω={KK,KZ,ZK,ZZ},$

whereby $KK=(head,head)$ etc. is. The event that the first coin shows "heads" is

$A.={KK,KZ}$

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

$B.={KK,ZK}$

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

$C.=A.∪B..$

events $A.$ and $B.$ are not mutually exclusive. After the addition theorem is

$p(C.)=p(A.)+p(B.)-p(A.∩B.).$

the end $A.∩B.={KK}$ surrendered $p(A.∩B.)=0,25$. So is

$p(C.)=0,5+0,5-0,25=0,75.$