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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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**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

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**Contents Conditional Probability**

The Multiplicative Rule and Independent Events Bayes’s Rule

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**Conditional Probability**

3.5 Conditional Probability

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**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)

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**Conditional Probability Using Venn Diagram**

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

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**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

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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

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**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

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**The Multiplicative Rule and Independent Events**

3.6 The Multiplicative Rule and Independent Events

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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.

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**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)

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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).

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**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)= =0.48

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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 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) =

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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! R P(R R)=P(R_1)P(R_2R_1) =(6/20)(5/19) =3/38 5/19 R 6/20 P(R B)= P(R_1)P(B_2R_1) =(6/20)(14/19) =21/95 14/19 B R 6/19 P(B R)= P(B_1)P(R_2B_1) =(14/20)(6/19) =21/95 14/20 B 13/19 B P(B B)= P(B_1)P(B_2B_1) =(14/20)(13/19) =91/190

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A and C, B and C Since AC is empty space Since BC is empty space. _________________________________ If P(AB)=P(A)P(B) then they are independent. 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 If 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)= =0.65

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**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(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+) = =0.995

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3.7 Bayes’s Rule

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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.

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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 Defective 0.02 Factory I 0 .6 0.98 Good Defective**

0.01 0 .4 Factory II 0.99 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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