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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

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Statistics for Business and Economics Chapter 3 Probability

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a) A c = {E3, E6, E8} P(A c ) = P(E3)+P(E6)+P(E8) = =0.53 b) B c = {E1, E7, E8} P(B c ) = P(E1)+P(E7)+P(E8) = =0.19 c) A c B = {E3, E6} P(A c 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) A c B c = (A B) c = {E8} P(A c B c ) = P((A B) c )= 1- P(A B) =0.03 g) No, since P(A B)≠0

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Contents 1.Conditional Probability 2.The Multiplicative Rule and Independent Events 3.Bayes’s Rule

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3.5 Conditional Probability

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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)

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S BlackAce Conditional Probability Using Venn Diagram Black ‘Happens’: Eliminates All Other Outcomes Event (Ace Black) (S)(S) Black

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Conditional Probability Using Two–Way Table Experiment: Draw 1 Card. Note Kind & Color. Revised Sample Space Color Type RedBlack Total Ace 224 Non-Ace Total Event

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Using the table then the formula, what’s the probability? Thinking Challenge 1. P(A|D) = 2. P(C|B) = Event CDTotal A 426 B

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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

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3.6 The Multiplicative Rule and Independent Events

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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.

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Multiplicative Rule Example Experiment: Draw 1 Card. Note Kind & Color. Color Type RedBlack Total Ace 224 Non-Ace Total P(Ace Black) = P(Ace)∙P(Black | Ace)

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For two events A and B, we have following probabilities: P(B A)=0.3, P(A c )=0.6, P(B c )= 0.8 Are events A and B mutually exclusive? Find P(A B). Thinking Challenge

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P(B A)=0.3, P(A c )=0.6, P(B c )= 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(A c )=1-0.6=0.4 P(B)=1- P(B c )=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 Solution*

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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)

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Thinking Challenge 1. P(C B) = 2. P(B D) = 3. P(A B) = Event CDTotal A 426 B Using the multiplicative rule, what’s the probability?

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Solution* Using the multiplicative rule, the probabilities are:

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Tree Diagram Experiment: Select 2 pens from 20 pens: 14 blue & 6 red. Don’t replace. Dependent! B R B R B R 6/20 5/19 14/19 14/20 6/19 13/19 P(R R)=P(R_1)P(R_2 R_1) =(6/20)(5/19) =3/38 P(R B)= P(R_1)P(B_2 R_1) =(6/20)(14/19) =21/95 P(B R)= P(B_1)P(R_2 B_1) =(14/20)(6/19) =21/95 P(B B)= P(B_1)P(B_2 B_1) =(14/20)(13/19) =91/190

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a)A and C, B and C Since A C is empty space Since B C is empty space. _________________________________ b)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

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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

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3.7 Bayes’s Rule

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Given k mutually exclusive and exhaustive events B 1, B 1,... B k, such that P(B 1 ) + P(B 2 ) + … + P(B k ) = 1, and an observed event A, then Bayes’s rule is useful for finding one conditional probability when other conditional probabilities are already known.

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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?

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Bayes’s Rule Example Factory II Factory I Defective Defective Good Good

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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

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

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