Example for calculating your final grade for this course

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*80+0.20*70+0.20*88+0.15*95+.25*85=83.1 • In 4 point scale=3.0

3. Statistics for Business and Economics Chapter 3 Probability

4. 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(AB)≠0 c) Ac B = {E3, E6} P(Ac B) = P(E3)+P(E6) = 0.2+0.3=0.50 d) AB = {E1, E2, E3, E4, E5, E6, E7} P(AB) =1-P(E8)= 1-0.03=0.97 e) AB = {E2, E4, E5} P(AB) =0.05+0.20+0.06=0.31 f) Ac  Bc = (AB)c = {E8} P(Ac  Bc) = P((AB)c )= 1- P(AB) =0.03

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

6. 3.5 Conditional Probability

7. 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)

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

9. Event 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 Table Experiment: Draw 1 Card. Note Kind & Color. Revised Sample Space

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

14. 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)

15. 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).

16. 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)=0.4+0.2-0.12=0.48

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. Event Event C D Total 4 2 6 A 1 3 4 B 5 5 10 Total Thinking Challenge • P(CB) = • P(BD) = • P(A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! 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 B 14/19 R P(BR)= P(B_1)P(R_2B_1) =(14/20)(6/19) =21/95 6/19 14/20 B 13/19 B P(BB)= P(B_1)P(B_2B_1) =(14/20)(13/19) =91/190

21. A and C, B and C Since AC is emptyspace Since BC is emptyspace. _________________________________ • If P(AB)=P(A)P(B) thentheyareindependent. 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 Ifwechecktheotherpairs, wefindthattheyare not independent, either. _________________________________ c) P(AB)=P(1)+P(2)+P(3)+P(4)=0.65 usingadditiverule; P(AB)=P(A)+P(B)- P(AB)=0.55+0.4-0.3=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+) = 0.9+0.95-0.855=0.995

23. 3.7 Bayes’s Rule

24. 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.

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 Defective 0.02 Factory I 0 .6 0.98 Good Defective 0.01 Factory II 0 .4 0.99 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)