11
1.2. Basic Probability
Sample Space: the set of all possible outcomes, denoted by  .
Event: an event is a collection (set) of sample points (possible outcomes).
Basic requirement for assigning probabilities:
A1 , A2 ,, An , are events. Then,
(i) P Ai   0;
(ii) P   0; P  1;
  
(iii) If A1 , A2 ,, An , are disjoint, then P  Ai    P Ai  .
 i 1  i 1
Formula of the conditional probability:
P( A | B) 
P( A  B)
P( B)
P( B | A) 
and
P( A  B)
P( A)
.
Mutually Exclusive Event: two events having no sample points in common is
called mutually exclusive events. That is, if A and B are mutually
exclusive events, then A  B    empty event
Independent Events:
P ( A | B )  P ( A) or
P ( B | A)  P ( B )
Dependent Events:
P ( A | B )  P ( A)
or
P( B | A)  P( B)
Results:
1.
P( Ac )  1  P( A)
2.
If A and B are mutually exclusive events, then
P( A  B)  0 and P( A  B)  P( A)  P( B) .
3.
(addition law) For any two events A and B,
4.
If A and B are independent events, then
P( A  B)  P( A)  P( B)  P( A  B)
P( A  B)  P APB .
5. Bayes’s Theorem
Let A1 , A2 , , An be mutually exclusive events and
A1  A2   An   ,
11
12
then
P( Ai | B) 
P( Ai  B)
P( B)
P( Ai ) P( B | Ai )
.............. 
P( A1 ) P( B | A1 )  P( A2 ) P( B | A2 )    P( An ) P( B | An )
i  1,2,  , n .
,
[Derivation of Bayes’s theorem (general)]:
A1
A2
B∩A1
B∩A2
B∩A1
B∩A2
B∩A1
B∩A2
………………..
B
An
B∩An
B∩An
……………
B∩An
Since P( B  Ai )  P( Ai ) P( B | Ai ) and
P( B)  P( B  A1 )  P( B  A2 )   P( B  An )
.......  P( A1 ) P( B | A1 )  P( A2 ) P( B | A2 )    P( An ) P( B | An )
,
thus,
P( Ai | B) 
P( B  Ai )
P( Ai ) P( B | Ai )

P( B)
P( A1 ) P( B | A1 )  P( A2 ) P( B | A2 )    P( An ) P( B | An )
Example 1:
You are given the following information on Events A, B, C, and D.
P(A)=0.1,P(C)=0.4, P(B)=0.3, P(C|B)=0.6, P(A∩C)=0.04, P(B|A)=0.9
(i) Compute P Ac  C .
(ii) Compute P C c  B .




(iii) Are A and B mutually exclusive? Explain.
(iv) Are A and C independent? Explain.
[solution:]
12
13


(i) P Ac  C  PC   P A  C   0.4  0.04  0.36 .
(ii)
PC c  B   1  P B C  C   1  PC   PB  C   1  0.4  0.18
 1  0.22  0.78
since PB  C   PBPC | B  0.3  0.6  0.18
(iii) Since P A  B  P APB | A  0.1 0.9  0.09  0 , A and B are not mutually
exclusive.
(iv) Since P A  C   0.04  P APC   0.1 0.4 , A and B are independent.
Example 2:
In a survey of MBA students, the following data were obtained on “students’ first
reason to the school in which they matriculated”.
Quality
Cost
Other
Total
Full Time
400
400
50
850
Part Time
400
600
150
1150
800
1000
200
2000
(i) Develop a joint probability table for these data.
(ii) Given a full time student, what is the probability that school quality is the
first reason for choosing a school?
(iii)Let A denote the event that a student is full time and let B denote the event
that the student lists school cost as the first reason for applying. Are events A
and B independent? Justify your answer.
[solution:]
(i)
Quality
Cost
Other
Total
Full Time
400/2000=0.2
400/2000=0.2
50/2000=0.025
850/2000=0.425
Part Time
400/2000=0.2
600/2000=0.3
150/2000=0.075 1150/2000=0.575
800/2000=0.4
1000/2000=0.5
200/2000=0.1
2000/2000=1
(ii) A: full time; B: school quality. Then,
P A  B 
0.2
P  B | A 

 0.4705 .
P  A
0.425
(iii)
Since
P A  B  0.2  0.2125  0.5  0.425  P APB ,
they
are
not
independent.
Example 3:
In a random sample of Tung Hai University students 50% indicated they are
business majors, 40% engineering majors, and 10% other majors. Of the
business majors, 60% were female; whereas, 30% of engineering majors were
females. Finally, 20% of the other majors were female. Given that a person is
female, what is the probability that she is an engineering major?
13
14
[solution:]
Let
A1: the students are engineering majors
 A1  A2  A3  
A2: the students are business majors
A3: the students are other majors.
B: the students are female.
Originally, we know
P( A1)  0.4, P( A2)  0.5, P( A3)  0.1, P( B | A1)  0.3, P( B | A2)  0.6, P( B | A3)  0.2 .
Then, by Bayes’ theorem,
P( A1) P( B | A1)
P( A1) P( B | A1)  P( A2) P( B | A2)  P( A3) P( B | A3)
0.4  0.3

 0.2727.
0.4  0.3  0.5  0.6  0.1  0.2
P( A1 | B) 
14