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Example for calculating your
final grade for this course
•
•
•
•
•
•
•
Midterm 1(MT1)= 100 points
Midterm 2(MT2)= 100 points
Homework (HW)=(HW1+…+HW7)/7
Each homework is 100 points
Quiz=(6*16)+4=100
Final=100 points
Grade=0.20*MT1+0.20*MT2+0.20*Quiz+0.15*HW+0.25*
Final
For instance;
• Grade=0.20*80+0.20*70+0.20*88+0.15*95+.25*85=83.1
• In 4 point scale=3.0
Statistics for Business and
Economics
Chapter 3
Probability
a) Ac = {E3, E6, E8}
P(Ac) = P(E3)+P(E6)+P(E8)
=0.2+0.3+0.03=0.53
b) Bc = {E1, E7, E8}
P(Bc) = P(E1)+P(E7)+P(E8)
= 0.1+0.06+0.03=0.19
g) No, since P(A B)≠0
c) Ac B = {E3, E6}
P(Ac B) = P(E3)+P(E6)
= 0.2+0.3=0.50
d) A B = {E1, E2, E3, E4, E5, E6, E7}
P(A B) =1-P(E8)= 1-0.03=0.97
e) A B = {E2, E4, E5}
P(A B) =0.05+0.20+0.06=0.31
f) Ac  Bc = (A B)c = {E8}
P(Ac  Bc) = P((A B)c )= 1- P(A B)
=0.03
Contents
1. Conditional Probability
2. The Multiplicative Rule and Independent
Events
3. Bayes’s Rule
3.5
Conditional Probability
Conditional Probability
1. Event probability given that another event
occurred
2. Revise original sample space to account for
new information
• Eliminates certain outcomes
3. P(A | B) = P(A and B) = P(A  B)
P(B)
P(B)
Conditional Probability Using
Venn Diagram
Ace
Black
S
Event (Ace  Black)
Black ‘Happens’:
Eliminates All
Other Outcomes
Black
(S)
Conditional Probability Using
Two–Way Table
Experiment: Draw 1 Card. Note Kind & Color.
Event
Type
Event
Event
Ace
Non-Ace
Total
Event
Color
Red
Black
2
2
Total
4
24
24
48
26
26
52
Revised
Sample
Space
P(Ace  Black) 2 / 52
2
P(Ace | Black) =


P(Black)
26 / 52 26
Thinking Challenge
Using the table then the formula, what’s the
probability?
1. P(A|D) =
2. P(C|B) =
Event
A
Event
C
D
4
2
Total
6
B
1
3
4
Total
5
5
10
Solution*
Using the formula, the probabilities are:
P A  B ) 2 5 2
P A D ) 


5
P D )
5
10
P(D)=P(AD)+P(BD)=2/10+3/10
P C  B ) 110 1
P C B ) 


4
P B )
4
10
P(B)=P(BD)+P(BC)=3/10+1/10
3.6
The Multiplicative Rule
and Independent Events
Multiplicative Rule
1. Used to get compound probabilities for
intersection of events
2. P(A and B) = P(A  B)
= P(A)  P(B|A)
= P(B)  P(A|B)
3. The key words both and and in the statement
imply and intersection of two events, which in
turn we should multiply probabilities to obtain the
probability of interest.
Multiplicative Rule Example
Experiment: Draw 1 Card. Note Kind & Color.
Color
Type
Ace
Red
Black
2
2
Total
4
Non-Ace
24
24
48
Total
26
26
52
P(Ace  Black) = P(Ace)∙P(Black | Ace)
 4  2  2
    
 52  4  52
Thinking Challenge
• For two events A and B, we have following
probabilities:
• P(BA)=0.3, P(Ac)=0.6, P(Bc)= 0.8
• Are events A and B mutually exclusive?
• Find P(AB).
Solution*
•
•
•
•
•
•
•
P(BA)=0.3, P(Ac)=0.6, P(Bc)= 0.8
Are events A and B mutually exclusive?
No, since we have P(BA) which is not zero.
P(AB)=P(A)+P(B)-P(AB)
P(A)=1- P(Ac)=1-0.6=0.4
P(B)=1- P(Bc)=1-0.8=0.2
P(BA)= P(AB) / P(A)
=0.3 P(AB)=P(BA)*P(A)=0.3*0.4=0.12
• P(AB)=P(A)+P(B)-P(AB)=0.4+0.2-0.12=0.48
Statistical Independence
1. Event occurrence does not affect probability of
another event
• Toss 1 coin twice
2. Causality not implied
3. Tests for independence
• P(A | B) = P(A)
• P(B | A) = P(B)
• P(A  B) = P(A)  P(B)
Thinking Challenge
Using the multiplicative rule, what’s the
probability?
1. P(C  B) =
Event
C
D
4
2
2. P(B  D) =
Event
A
Total
6
3. P(A  B) =
B
1
3
4
Total
5
5
10
Solution*
Using the multiplicative rule, the probabilities
are:
5 1 1
P C  B )  P C ) P B C )   
10 5 10
4 3 6
P B  D )  P B ) P D B )   
10 5 25
P A  B )  P A ) P B A )  0
Tree Diagram
Experiment: Select 2 pens from 20 pens: 14
blue & 6 red. Don’t replace.
Dependent!
5/19
6/20
P(R  R)=P(R_1)P(R_2R_1)
=(6/20)(5/19) =3/38
R
14/19
14/20
R
6/19
B
R
B
P(R  B)= P(R_1)P(B_2R_1)
=(6/20)(14/19) =21/95
P(B  R)= P(B_1)P(R_2B_1)
=(14/20)(6/19) =21/95
13/19
B
P(B  B)= P(B_1)P(B_2B_1)
=(14/20)(13/19) =91/190
a) A and C, B and C
Since AC is empty space
Since BC is empty space.
_________________________________
b) If P(AB)=P(A)P(B) then they are
independent.
P(AB)=P(3)=0.3
P(A)P(B)=[P(1)+P(2)+P(3)][P(4)+P(3)]
=0.55*0.4=0.22
P(AB)≠ P(A)P(B)A and B are not
independent
If we check the other pairs, we find that
they are not independent, either.
_________________________________
c) P(AB)=P(1)+P(2)+P(3)+P(4)=0.65
using additive rule;
P(AB)=P(A)+P(B)P(AB)=0.55+0.4-0.3=0.65
Let events be
•A=System A sounds an alarm
•B=System B sounds an alarm
•I+=There is an intruder
•I-=There is no intruder
We are given;
P(AI+)=0.9, P(BI+)=0.95
P(AI-)=0.2, P(BI-)=0.1
b) P(ABI+)= P(AI+)P(BI+)
= 0.9*0.95=0.855
c) P(ABI-) = P(AI-)P(BI-)
= 0.2*0.1=0.02
d) P(ABI+)= P(AI+)+P(BI+)-P(ABI+)
= 0.9+0.95-0.855=0.995
3.7
Bayes’s Rule
Bayes’s Rule
Given k mutually exclusive and exhaustive events B1,
B1, . . . Bk , such that
P(B1) + P(B2) + … + P(Bk) = 1,
and an observed event A, then
P(Bi  A)
P(Bi | A) 
P( A)
P(Bi )P( A | Bi )

P(B1 )P( A | B1 )  P(B2 )P( A | B2 )  ...  P(Bk )P( A | Bk )
•Bayes’s rule is useful for finding one conditional probability
when other conditional probabilities are already known.
Bayes’s Rule Example
A company manufactures MP3 players at two
factories. Factory I produces 60% of the MP3
players and Factory II produces 40%. Two
percent of the MP3 players produced at Factory
I are defective, while 1% of Factory II’s are
defective. An MP3 player is selected at random
and found to be defective. What is the
probability it came from Factory I?
Bayes’s Rule Example
0 .6
0 .4
Factory
I
Factory
II
0.02
Defective
0.98
Good
0.01
Defective
0.99
Good
P( I | D) 
P(I )P( D | I )
0.6  0.02

 0.75
P(I )P( D | I )  P(II )P(D | II ) 0.6  0.02  0.4  0.01
Let events be
•U+=Athlete uses testosterone
•U- = Athlete do not use testosterone
•T+=Test is positive
•T- = Test is negative
We are given;
•P(U+)=100/1000=0.1
•P(T+ U+)=50/100=0.5
•P(T+ U-)=9/900=0.01
a) P(T+ U+)=0.5 sensitivity of
the drug test
b) P(T- U-)=1-P(T+ U-)
=1-0.01=0.99
specificity of th e drug test
Ex. 3.84, cont. (sol.)
c)
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