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Example for calculating your final grade for this course

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- 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(AB)≠0

c) Ac B = {E3, E6}

P(Ac B) = P(E3)+P(E6)

= 0.2+0.3=0.50

d) AB = {E1, E2, E3, E4, E5, E6, E7}

P(AB) =1-P(E8)= 1-0.03=0.97

e) AB = {E2, E4, E5}

P(AB) =0.05+0.20+0.06=0.31

f) Ac Bc = (AB)c = {E8}

P(Ac Bc) = P((AB)c )= 1- P(AB) =0.03

- Conditional Probability
- The Multiplicative Rule and Independent Events
- Bayes’s Rule

Conditional Probability

1. Event probability given that another event occurred

2. Revise original sample space to account fornew information

- Eliminates certain outcomes
3.P(A|B) =P(A and B)=P(A B)P(B) P(B)

Black ‘Happens’: Eliminates All Other Outcomes

Ace

Black

Black

S

(S)

Event (Ace Black)

Event

Event

Color

Type

Total

Red

Black

2

2

4

Ace

Event

24

48

Non-Ace

24

Event

26

52

26

Total

Experiment: Draw 1 Card. Note Kind & Color.

Revised Sample Space

Event

Event

C

D

Total

4

2

6

A

1

3

4

B

5

5

10

Total

Using the table then the formula, what’s the probability?

- P(A|D) =
- P(C|B) =

Using the formula, the probabilities are:

P(D)=P(AD)+P(BD)=2/10+3/10

P(B)=P(BD)+P(BC)=3/10+1/10

The Multiplicative Rule

and Independent Events

1.Used to get compound probabilities for intersection of events

- P(A and B) = P(AB)= P(A) P(B|A) = P(B) P(A|B)
- 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.

Experiment: Draw 1 Card. Note Kind & Color.

Color

Type

Total

Red

Black

2

2

4

Ace

24

24

48

Non-Ace

26

26

52

Total

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).

- 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

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(AB) = P(A) P(B)

Event

Event

C

D

Total

4

2

6

A

1

3

4

B

5

5

10

Total

- P(CB) =
- P(BD) =
- P(AB) =

Using the multiplicative rule, what’s the probability?

Using the multiplicative rule, the probabilities are:

Experiment: Select 2 pens from 20 pens: 14 blue & 6 red. Don’t replace.

Dependent!

R

P(RR)=P(R_1)P(R_2R_1)

=(6/20)(5/19) =3/38

5/19

R

6/20

P(RB)= P(R_1)P(B_2R_1)

=(6/20)(14/19) =21/95

B

14/19

R

P(BR)= P(B_1)P(R_2B_1)

=(14/20)(6/19) =21/95

6/19

14/20

B

13/19

B

P(BB)= P(B_1)P(B_2B_1)

=(14/20)(13/19) =91/190

- A and C, B and C
Since AC is emptyspace

Since BC is emptyspace.

_________________________________

- If P(AB)=P(A)P(B) thentheyareindependent.
P(AB)=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(AB)≠ P(A)P(B)A and B are not independent

Ifwechecktheotherpairs, wefindthattheyare not independent, either.

_________________________________

c) P(AB)=P(1)+P(2)+P(3)+P(4)=0.65

usingadditiverule;

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+)=0.95

P(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

Bayes’s Rule

Given k mutually exclusive and exhaustive events B1, B1, . . . Bk , such thatP(B1) + P(B2) + … + P(Bk) = 1,and an observed event A, then

- Bayes’s rule is useful for finding one conditional probability when other conditional probabilities are already known.

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?

Defective

0.02

Factory I

0 .6

0.98

Good

Defective

0.01

Factory II

0 .4

0.99

Good

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

c)