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*
FinalFor 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 3Probability
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
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
g) No, since P(A B)≠0
Contents1. Conditional Probability2. The Multiplicative Rule and Independent
Events3. Bayes’s Rule
3.5
Conditional Probability
Conditional Probability1. 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)
S
BlackAce
Conditional Probability Using Venn Diagram
Black ‘Happens’: Eliminates All Other Outcomes
Event (Ace Black)
(S)Black
Conditional Probability Using Two–Way Table
Experiment: Draw 1 Card. Note Kind & Color.
Revised Sample Space
ColorType Red Black Total
Ace 2 2 4Non-Ace 24 24 48
Total 26 26 52
Event
Event
Event Event
P(Ace Black) 2 / 52 2P(Ace | Black) = P(Black) 26 / 52 26
Using the table then the formula, what’s the probability?
Thinking Challenge
1. P(A|D) =
2. P(C|B) =
EventEvent C D TotalA 4 2 6B 1 3 4
Total 5 5 10
Solution*
Using the formula, the probabilities are:
P A D P A B P D
25
510
25
P C B P C B P B
110
410
14
P(D)=P(AD)+P(BD)=2/10+3/10
P(B)=P(BD)+P(BC)=3/10+1/10
3.6
The Multiplicative Ruleand 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 ExampleExperiment: Draw 1 Card. Note Kind & Color. Color
Type Red Black TotalAce 2 2 4Non-Ace 24 24 48Total 26 26 52
4 2 252 4 52
P(Ace Black) = P(Ace)∙P(Black | Ace)
• For two events A and B, we have following probabilities:
• P(BA)=0.3, P(Ac)=0.6, P(Bc)= 0.8• Are events A and B mutually exclusive?• Find P(AB).
Thinking Challenge
• P(BA)=0.3, P(Ac)=0.6, P(Bc)= 0.8• Are events A and B mutually exclusive?• No, since we have P(BA) which is not zero.• P(AB)=P(A)+P(B)-P(AB)• P(A)=1- P(Ac)=1-0.6=0.4• P(B)=1- P(Bc)=1-0.8=0.2• P(BA)= P(AB) / P(A) =0.3 P(AB)=P(BA)*P(A)=0.3*0.4=0.12• P(AB)=P(A)+P(B)-P(AB)=0.4+0.2-0.12=0.48
Solution*
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
1. P(C B) =
2. P(B D) =
3. P(A B) =
EventEvent C D TotalA 4 2 6B 1 3 4
Total 5 5 10
Using the multiplicative rule, what’s the probability?
Solution*
Using the multiplicative rule, the probabilities are:
P C B P C P B C 510
15
1
10
P B D P B P D B 410
35
625
P A B P A P B A 0
Tree DiagramExperiment: Select 2 pens from 20 pens: 14 blue & 6 red. Don’t replace.
Dependent!
BB
RR
BBRR
BB
RR6/20
5/19
14/19
14/206/19
13/19
P(R R)=P(R_1)P(R_2R_1)
=(6/20)(5/19) =3/38
P(R B)= P(R_1)P(B_2R_1)
=(6/20)(14/19) =21/95
P(B R)= P(B_1)P(R_2B_1)
=(14/20)(6/19) =21/95
P(B B)= P(B_1)P(B_2B_1)
=(14/20)(13/19) =91/190
a) A and C, B and CSince AC is empty spaceSince BC is empty space._________________________________b) If P(AB)=P(A)P(B) then they are
independent.P(AB)=P(3)=0.3P(A)P(B)=[P(1)+P(2)+P(3)][P(4)+P(3)] =0.55*0.4=0.22P(AB)≠ P(A)P(B)A and B are not
independentIf we check the other pairs, we find that
they are not independent, either._________________________________c) P(AB)=P(1)+P(2)+P(3)+P(4)=0.65 using additive rule;P(AB)=P(A)+P(B)-
P(AB)=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(AI+)=0.9, P(BI+)=0.95P(AI-)=0.2, P(BI-)=0.1
b) P(ABI+)= P(AI+)P(BI+) = 0.9*0.95=0.855
c) P(ABI-) = P(AI-)P(BI-) = 0.2*0.1=0.02
d) P(ABI+)= P(AI+)+P(BI+)-P(ABI+)
= 0.9+0.95-0.855=0.995
3.7
Bayes’s Rule
Bayes’s RuleGiven k mutually exclusive and exhaustive events B1, B1, . . . Bk , such thatP(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 ExampleA 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
Factory Factory IIII
Factory Factory II0 .6
0.02
0.98
0 .4 0.01
0.99
DefectiveDefective
DefectiveDefective
GoodGood
GoodGood
P(I | D) P(I )P(D | I )
P(I )P(D | I ) P(II )P(D | II )
0.60.020.60.02 0.40.01
0.75
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)