# 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* * * *95+.25*85=83.1 In 4 point scale=3.0

Statistics for Business and Economics
Chapter 3 Probability

P(Ac) = P(E3)+P(E6)+P(E8) =0.2+0.3+0.03=0.53
a) Ac = {E3, E6, E8} P(Ac) = P(E3)+P(E6)+P(E8) = =0.53 b) Bc = {E1, E7, E8} P(Bc) = P(E1)+P(E7)+P(E8) = =0.19 g) No, since P(A B)≠0 c) Ac B = {E3, E6} P(Ac B) = P(E3)+P(E6) = =0.50 d) A B = {E1, E2, E3, E4, E5, E6, E7} P(A B) =1-P(E8)= =0.97 e) A B = {E2, E4, E5} P(A B) = =0.31 f) Ac  Bc = (A B)c = {E8} P(Ac  Bc) = P((A B)c )= 1- P(A B) =0.03

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

Conditional Probability
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
Black ‘Happens’: Eliminates All Other Outcomes Ace Black Black S (S) Event (Ace  Black)

Conditional Probability Using Two–Way Table
Experiment: Draw 1 Card. Note Kind & Color. Color Type Red Black Total Ace 2 4 Non-Ace 24 48 26 52 Event Revised Sample Space

Thinking Challenge Using the table then the formula, what’s the probability? P(A|D) = P(C|B) = Event C D Total A 4 2 6 B 1 3 5 10

Solution* Using the formula, the probabilities are:
P(D)=P(AD)+P(BD)=2/10+3/10 P(B)=P(BD)+P(BC)=3/10+1/10

The Multiplicative Rule and Independent Events
3.6 The Multiplicative Rule and Independent Events

Multiplicative Rule 1. Used to get compound probabilities for intersection of events P(A and B) = P(A  B) = 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 Red Black Total Ace 2 2 4 Non-Ace 24 24 48 Total 26 26 52 P(Ace  Black) = P(Ace)∙P(Black | Ace)

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)= =0.48

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? Event C D Total A 4 2 6 B 1 3 5 10 P(C  B) = P(B  D) = P(A  B) =

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(R  R)=P(R_1)P(R_2R_1) =(6/20)(5/19) =3/38 5/19 R 6/20 P(R  B)= P(R_1)P(B_2R_1) =(6/20)(14/19) =21/95 14/19 B R 6/19 P(B  R)= P(B_1)P(R_2B_1) =(14/20)(6/19) =21/95 14/20 B 13/19 B P(B  B)= P(B_1)P(B_2B_1) =(14/20)(13/19) =91/190

A and C, B and C Since AC is empty space Since BC is empty space. _________________________________ 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)= =0.65

A=System A sounds an alarm B=System B sounds an alarm
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+) = =0.995

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 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 0 .4 Factory II 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

Ex. 3.84, cont. (sol.) c)