example for calculating your final grade for this course

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)