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Example for calculating your final grade for this coursePowerPoint Presentation

Example for calculating your final grade for this course

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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

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

Contents

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

3.5

Conditional Probability

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)

Conditional Probability Using Venn Diagram

Black ‘Happens’: Eliminates All Other Outcomes

Ace

Black

Black

S

(S)

Event (Ace Black)

Event

Color

Type

Total

Red

Black

2

2

4

Ace

Event

24

48

Non-Ace

24

Event

26

52

26

Total

Conditional Probability Using Two–Way TableExperiment: Draw 1 Card. Note Kind & Color.

Revised Sample Space

Event

C

D

Total

4

2

6

A

1

3

4

B

5

5

10

Total

Thinking ChallengeUsing the table then the formula, what’s the probability?

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

Solution*

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

Multiplicative Rule

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.

Multiplicative Rule Example

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)

Thinking Challenge

- 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).

Solution*

- 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

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

Event

C

D

Total

4

2

6

A

1

3

4

B

5

5

10

Total

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

Using the multiplicative rule, what’s the probability?

Solution*

Using the multiplicative rule, the probabilities are:

Tree Diagram

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

- 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

3.7

Bayes’s Rule

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.

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

Defective

0.02

Factory I

0 .6

0.98

Good

Defective

0.01

Factory II

0 .4

0.99

Good

- 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

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