Chemistry

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 spaceΩ 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

0p(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.1A.2A.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.


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