Presentation is loading. Please wait.

Presentation is loading. Please wait.

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.

Similar presentations


Presentation on theme: "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."— Presentation transcript:

1

2 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

3 Statistics for Business and Economics Chapter 3 Probability

4 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

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

6 3.5 Conditional Probability

7 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)

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

9 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

10 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

11 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

12 3.6 The Multiplicative Rule and Independent Events

13 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.

14 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)

15 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

16 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*

17 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)

18 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?

19 Solution* Using the multiplicative rule, the probabilities are:

20 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

21 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

22 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

23 3.7 Bayes’s Rule

24 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.

25 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?

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

27 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

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


Download ppt "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."

Similar presentations


Ads by Google