axiomatic probability - university of utah

Axiomatic Probability

Proposition

For any three events A, B, and C,

P(A ∪ B ∪ C ) =P(A) + P(B) + P(C )

− P(A ∩ B)− P(B ∩ C )− P(C ∩ A)

+ P(A ∩ B ∩ C )

A Venn Diagram interpretation:

Determining Probabilities

In Probability, our main focus is to determine the probabilities forall possible events. However, some prior knowledge about thesample space is available. (While in Statistics, the prior knowledgeis unavailable and we want to find it.)

For a sample space that is either finite or “countably infinite”(meaning the outcomes can be listed in an infinite sequence), letE1, E2, . . . denote all the simple events. If we know P(Ei ) for eachi , then for any event A,

P(A) =∑

all Ei s that are in A

P(Ei )

Here the knowledge for P(Ei ) is given and we want to find P(A).(In statistics, we want to find the knowledge about P(Ei ).)


In Probability, our main focus is to determine the probabilities forall possible events. However, some prior knowledge about thesample space is available. (While in Statistics, the prior knowledgeis unavailable and we want to find it.)For a sample space that is either finite or “countably infinite”(meaning the outcomes can be listed in an infinite sequence), letE1, E2, . . . denote all the simple events. If we know P(Ei ) for eachi , then for any event A,

P(A) =∑


P(Ei )

Here the knowledge for P(Ei ) is given and we want to find P(A).(In statistics, we want to find the knowledge about P(Ei ).)


Example 2.15:

During off-peak hours a commuter train has five cars. Suppose acommuter is twice as likely to select the middle car(#3) as toselect either adjacent car(#2 or #4)(1), and is twice as likely toselect either adjacent car as to select either end car(#1 or #5)(2).What is the probability for the commuter to select one of the threemiddle cars?First we need to determine the probabilities for all the simpleevents:Let pi = P({car i is selected}) = P(Ei ).Then condition (1) tells us p3 = 2p2 = 2p4, and condition (2) tellsus p2 = 2p1 = 2p5 = p4.By Axiom 2 and 3,

1 =5∑

i=1

P(Ei ) = p1 + 2p1 + 4p1 + 2p1 + p1 = 10p1


Example 2.15:During off-peak hours a commuter train has five cars. Suppose acommuter is twice as likely to select the middle car(#3) as toselect either adjacent car(#2 or #4)(1), and is twice as likely toselect either adjacent car as to select either end car(#1 or #5)(2).What is the probability for the commuter to select one of the threemiddle cars?

First we need to determine the probabilities for all the simpleevents:Let pi = P({car i is selected}) = P(Ei ).Then condition (1) tells us p3 = 2p2 = 2p4, and condition (2) tellsus p2 = 2p1 = 2p5 = p4.By Axiom 2 and 3,

1 =5∑

i=1

P(Ei ) = p1 + 2p1 + 4p1 + 2p1 + p1 = 10p1


Example 2.15:During off-peak hours a commuter train has five cars. Suppose acommuter is twice as likely to select the middle car(#3) as toselect either adjacent car(#2 or #4)(1), and is twice as likely toselect either adjacent car as to select either end car(#1 or #5)(2).What is the probability for the commuter to select one of the threemiddle cars?First we need to determine the probabilities for all the simpleevents:

Let pi = P({car i is selected}) = P(Ei ).Then condition (1) tells us p3 = 2p2 = 2p4, and condition (2) tellsus p2 = 2p1 = 2p5 = p4.By Axiom 2 and 3,

1 =5∑

i=1

P(Ei ) = p1 + 2p1 + 4p1 + 2p1 + p1 = 10p1


Example 2.15:During off-peak hours a commuter train has five cars. Suppose acommuter is twice as likely to select the middle car(#3) as toselect either adjacent car(#2 or #4)(1), and is twice as likely toselect either adjacent car as to select either end car(#1 or #5)(2).What is the probability for the commuter to select one of the threemiddle cars?First we need to determine the probabilities for all the simpleevents:Let pi = P({car i is selected}) = P(Ei ).

Then condition (1) tells us p3 = 2p2 = 2p4, and condition (2) tellsus p2 = 2p1 = 2p5 = p4.By Axiom 2 and 3,

1 =5∑

i=1

P(Ei ) = p1 + 2p1 + 4p1 + 2p1 + p1 = 10p1


Example 2.15:During off-peak hours a commuter train has five cars. Suppose acommuter is twice as likely to select the middle car(#3) as toselect either adjacent car(#2 or #4)(1), and is twice as likely toselect either adjacent car as to select either end car(#1 or #5)(2).What is the probability for the commuter to select one of the threemiddle cars?First we need to determine the probabilities for all the simpleevents:Let pi = P({car i is selected}) = P(Ei ).Then condition (1) tells us p3 = 2p2 = 2p4,

and condition (2) tellsus p2 = 2p1 = 2p5 = p4.By Axiom 2 and 3,

1 =5∑

i=1

P(Ei ) = p1 + 2p1 + 4p1 + 2p1 + p1 = 10p1


Example 2.15:During off-peak hours a commuter train has five cars. Suppose acommuter is twice as likely to select the middle car(#3) as toselect either adjacent car(#2 or #4)(1), and is twice as likely toselect either adjacent car as to select either end car(#1 or #5)(2).What is the probability for the commuter to select one of the threemiddle cars?First we need to determine the probabilities for all the simpleevents:Let pi = P({car i is selected}) = P(Ei ).Then condition (1) tells us p3 = 2p2 = 2p4, and condition (2) tellsus p2 = 2p1 = 2p5 = p4.

By Axiom 2 and 3,

1 =5∑

i=1

P(Ei ) = p1 + 2p1 + 4p1 + 2p1 + p1 = 10p1


Example 2.15:During off-peak hours a commuter train has five cars. Suppose acommuter is twice as likely to select the middle car(#3) as toselect either adjacent car(#2 or #4)(1), and is twice as likely toselect either adjacent car as to select either end car(#1 or #5)(2).What is the probability for the commuter to select one of the threemiddle cars?First we need to determine the probabilities for all the simpleevents:Let pi = P({car i is selected}) = P(Ei ).Then condition (1) tells us p3 = 2p2 = 2p4, and condition (2) tellsus p2 = 2p1 = 2p5 = p4.By Axiom 2 and 3,

1 =5∑

i=1

P(Ei ) = p1 + 2p1 + 4p1 + 2p1 + p1 = 10p1


Example 2.15 (continued):Since p3 = 2p2 = 2p4, p2 = 2p1 = 2p5 = p4 and1 =

∑5i=1 P(Ei ) = p1 + 2p1 + 4p1 + 2p1 + p1 = 10p1,

we have p1 = 0.1, p2 = 0.2, p3 = 0.4, p = 0.2 and p5 = 0.1.Furthermore if A = {one of the three middle cars is selected}, thenP(A) = P(E2) + P(E3) + P(E4) = p2 + p3 + p4 = 0.8.



∑5i=1 P(Ei ) = p1 + 2p1 + 4p1 + 2p1 + p1 = 10p1,

we have p1 = 0.1, p2 = 0.2, p3 = 0.4, p = 0.2 and p5 = 0.1.

Furthermore if A = {one of the three middle cars is selected}, thenP(A) = P(E2) + P(E3) + P(E4) = p2 + p3 + p4 = 0.8.



∑5i=1 P(Ei ) = p1 + 2p1 + 4p1 + 2p1 + p1 = 10p1,

we have p1 = 0.1, p2 = 0.2, p3 = 0.4, p = 0.2 and p5 = 0.1.Furthermore if A = {one of the three middle cars is selected}, thenP(A) = P(E2) + P(E3) + P(E4) = p2 + p3 + p4 = 0.8.


I Equally Likely Outcomes: experiments whose outcomeshave exactly the same probabilitiy.

In that case, the probability p for each simple event Ei isdetermined by the size of the sample space N, i.e.

p = P(Ei ) =1

N

This is simply due to the fact

1 = P(S) = P(N⋃

i=1

Ei ) =N∑

i=1

P(Ei ) =N∑

i=1

p = N · p


I Equally Likely Outcomes: experiments whose outcomeshave exactly the same probabilitiy.In that case, the probability p for each simple event Ei isdetermined by the size of the sample space N, i.e.

p = P(Ei ) =1

N

This is simply due to the fact

1 = P(S) = P(N⋃

i=1

Ei ) =N∑

i=1

P(Ei ) =N∑

i=1

p = N · p


Examples:

tossing a fair coin: N = 2 and P({H}) = P({T}) = 1/2;tossing a fair die: N = 6 andP({1}) = P({2}) = P({3}) = P({4}) = P({5}) = P({6}) = 1/6;randomly selecting a student from 25 students: N = 25 andp = 1/25.


Examples:tossing a fair coin: N = 2 and P({H}) = P({T}) = 1/2;

tossing a fair die: N = 6 andP({1}) = P({2}) = P({3}) = P({4}) = P({5}) = P({6}) = 1/6;randomly selecting a student from 25 students: N = 25 andp = 1/25.


Examples:tossing a fair coin: N = 2 and P({H}) = P({T}) = 1/2;tossing a fair die: N = 6 andP({1}) = P({2}) = P({3}) = P({4}) = P({5}) = P({6}) = 1/6;

randomly selecting a student from 25 students: N = 25 andp = 1/25.


Examples:tossing a fair coin: N = 2 and P({H}) = P({T}) = 1/2;tossing a fair die: N = 6 andP({1}) = P({2}) = P({3}) = P({4}) = P({5}) = P({6}) = 1/6;randomly selecting a student from 25 students: N = 25 andp = 1/25.


Counting Techniques:

If the sample space is finite and all the outcomes are equally likelyto happen, then the formula

P(A) =∑


P(Ei )

simplifies to

P(A) =N(A)

N

where Ei is any simple event, N is the number of outcomes of thesample space and N(A) is the number of outcomes contained inevent A.Determining the probability of A⇒ counting N(A).


Counting Techniques:If the sample space is finite and all the outcomes are equally likelyto happen, then the formula

P(A) =∑


P(Ei )

simplifies to

P(A) =N(A)

N

where Ei is any simple event, N is the number of outcomes of thesample space and N(A) is the number of outcomes contained inevent A.

Determining the probability of A⇒ counting N(A).


Counting Techniques:If the sample space is finite and all the outcomes are equally likelyto happen, then the formula

P(A) =∑


P(Ei )

simplifies to

P(A) =N(A)

N

where Ei is any simple event, N is the number of outcomes of thesample space and N(A) is the number of outcomes contained inevent A.Determining the probability of A⇒ counting N(A).

axiomatic probability - university of utah

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