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 for
all possible events. However, some prior knowledge about the
sample space is available. (While in Statistics, the prior knowledge
is unavailable and we want to find it.)
Determining Probabilities
In Probability, our main focus is to determine the probabilities for
all possible events. However, some prior knowledge about the
sample space is available. (While in Statistics, the prior knowledge
is 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), let
E1 , E2 , . . . denote all the simple events. If we know P(Ei ) for each
i, then for any event A,
X
P(A) =
P(Ei )
all Ei s that are in A
Determining Probabilities
In Probability, our main focus is to determine the probabilities for
all possible events. However, some prior knowledge about the
sample space is available. (While in Statistics, the prior knowledge
is 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), let
E1 , E2 , . . . denote all the simple events. If we know P(Ei ) for each
i, then for any event A,
X
P(A) =
P(Ei )
all Ei s that are in A
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 ).)
Determining Probabilities
Example 2.15:
Determining Probabilities
Example 2.15:
During off-peak hours a commuter train has five cars. Suppose a
commuter is twice as likely to select the middle car(#3) as to
select either adjacent car(#2 or #4)(1) , and is twice as likely to
select 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 three
middle cars?
Determining Probabilities
Example 2.15:
During off-peak hours a commuter train has five cars. Suppose a
commuter is twice as likely to select the middle car(#3) as to
select either adjacent car(#2 or #4)(1) , and is twice as likely to
select 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 three
middle cars?
First we need to determine the probabilities for all the simple
events:
Determining Probabilities
Example 2.15:
During off-peak hours a commuter train has five cars. Suppose a
commuter is twice as likely to select the middle car(#3) as to
select either adjacent car(#2 or #4)(1) , and is twice as likely to
select 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 three
middle cars?
First we need to determine the probabilities for all the simple
events:
Let pi = P({car i is selected}) = P(Ei ).
Determining Probabilities
Example 2.15:
During off-peak hours a commuter train has five cars. Suppose a
commuter is twice as likely to select the middle car(#3) as to
select either adjacent car(#2 or #4)(1) , and is twice as likely to
select 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 three
middle cars?
First we need to determine the probabilities for all the simple
events:
Let pi = P({car i is selected}) = P(Ei ).
Then condition (1) tells us p3 = 2p2 = 2p4 ,
Determining Probabilities
Example 2.15:
During off-peak hours a commuter train has five cars. Suppose a
commuter is twice as likely to select the middle car(#3) as to
select either adjacent car(#2 or #4)(1) , and is twice as likely to
select 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 three
middle cars?
First we need to determine the probabilities for all the simple
events:
Let pi = P({car i is selected}) = P(Ei ).
Then condition (1) tells us p3 = 2p2 = 2p4 , and condition (2) tells
us p2 = 2p1 = 2p5 = p4 .
Determining Probabilities
Example 2.15:
During off-peak hours a commuter train has five cars. Suppose a
commuter is twice as likely to select the middle car(#3) as to
select either adjacent car(#2 or #4)(1) , and is twice as likely to
select 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 three
middle cars?
First we need to determine the probabilities for all the simple
events:
Let pi = P({car i is selected}) = P(Ei ).
Then condition (1) tells us p3 = 2p2 = 2p4 , and condition (2) tells
us p2 = 2p1 = 2p5 = p4 .
By Axiom 2 and 3,
1=
5
X
i=1
P(Ei ) = p1 + 2p1 + 4p1 + 2p1 + p1 = 10p1
Determining Probabilities
Example 2.15 (continued):
SincePp3 = 2p2 = 2p4 , p2 = 2p1 = 2p5 = p4 and
1 = 5i=1 P(Ei ) = p1 + 2p1 + 4p1 + 2p1 + p1 = 10p1 ,
Determining Probabilities
Example 2.15 (continued):
SincePp3 = 2p2 = 2p4 , p2 = 2p1 = 2p5 = p4 and
1 = 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.
Determining Probabilities
Example 2.15 (continued):
SincePp3 = 2p2 = 2p4 , p2 = 2p1 = 2p5 = p4 and
1 = 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}, then
P(A) = P(E2 ) + P(E3 ) + P(E4 ) = p2 + p3 + p4 = 0.8.
Determining Probabilities
I
Equally Likely Outcomes: experiments whose outcomes
have exactly the same probabilitiy.
Determining Probabilities
I
Equally Likely Outcomes: experiments whose outcomes
have exactly the same probabilitiy.
In that case, the probability p for each simple event Ei is
determined by the size of the sample space N, i.e.
p = P(Ei ) =
1
N
Determining Probabilities
I
Equally Likely Outcomes: experiments whose outcomes
have exactly the same probabilitiy.
In that case, the probability p for each simple event Ei is
determined 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
X
i=1
P(Ei ) =
N
X
i=1
p =N ·p
Determining Probabilities
Examples:
Determining Probabilities
Examples:
tossing a fair coin: N = 2 and P({H}) = P({T }) = 1/2;
Determining Probabilities
Examples:
tossing a fair coin: N = 2 and P({H}) = P({T }) = 1/2;
tossing a fair die: N = 6 and
P({1}) = P({2}) = P({3}) = P({4}) = P({5}) = P({6}) = 1/6;
Determining Probabilities
Examples:
tossing a fair coin: N = 2 and P({H}) = P({T }) = 1/2;
tossing a fair die: N = 6 and
P({1}) = P({2}) = P({3}) = P({4}) = P({5}) = P({6}) = 1/6;
randomly selecting a student from 25 students: N = 25 and
p = 1/25.
Determining Probabilities
Counting Techniques:
Determining Probabilities
Counting Techniques:
If the sample space is finite and all the outcomes are equally likely
to happen, then the formula
X
P(A) =
P(Ei )
all Ei s that are in A
simplifies to
N(A)
N
where Ei is any simple event, N is the number of outcomes of the
sample space and N(A) is the number of outcomes contained in
event A.
P(A) =
Determining Probabilities
Counting Techniques:
If the sample space is finite and all the outcomes are equally likely
to happen, then the formula
X
P(A) =
P(Ei )
all Ei s that are in A
simplifies to
N(A)
N
where Ei is any simple event, N is the number of outcomes of the
sample space and N(A) is the number of outcomes contained in
event A.
Determining the probability of A ⇒ counting N(A).
P(A) =