Review of Probability

1. Review of Probability Jake Blanchard Spring 2010 Uncertainty Analysis for Engineers

2. Introduction • Interpretations of Probability • Classical – If an event can occur in N equally likely and different ways, and if n of these have an attribute A, then the probability of the occurrence of A, denoted Pr(A), is defined as n/N • Example: the probability of drawing an Ace from a full deck of cards is 1/13 (4/52) Uncertainty Analysis for Engineers

3. Introduction • Interpretations of Probability • Frequency (empirical) – If an experiment is conducted N times, and a particular attribute A occurs n times, then the limit of n/N as N becomes large is defined as the probability of A • Example: If, in the past, 73 cars out of 10,000 are defective (coming from a particular factory), then the probability that a car will be defective is 0.0073 • This interpretation is the most common among statisticians Uncertainty Analysis for Engineers

4. Introduction • Interpretations of Probability • Subjective – Pr(A) is a measure of the degree of belief one holds in a specified proposition A • Broader than other interpretations • Probability is directly related to the odds one would wager on a specific proposition Uncertainty Analysis for Engineers

5. Set Theory • Set=collection of distinct objects • Union: C1=AB • Intersection: C2=AB=AB • Complement (“not”): C3=A • =null set • I=entire set • m(A)=number of elements in set A Uncertainty Analysis for Engineers

6. Identities • A   =A • A  I=I • A   = A =  • AI=A • A  A=A • AA=A •  =I • AA=  • A  A=I Uncertainty Analysis for Engineers

7. Example • The EZ Company employs 10 non-professional employees: 3 assemblers, 5 machinists, and 2 clerks • m(A)=3; m(M)=5; m(C)=2; m(I)=10 • Q=set of workers who are both a machinist and an assembler • Q=AM=Z; m(Q)=0 • F=all factory workers; F=A  M • m(F)=m(A  M)=8 Uncertainty Analysis for Engineers

8. Example (continued) • Suppose the company employs 8 engineers, 3 supervisors, and 2 employees who are both an engineer and a supervisor m(E  S) =m(E)+m(S)-m(ES) =10+5-2=13 Uncertainty Analysis for Engineers

9. Probability • Rolling dice • m(I)=6 • A=rolling a 2 • Pr(A)=m(A)/m(I)=1/6 Uncertainty Analysis for Engineers

10. Definition • Two events are independent if the occurrence of one does not change the probability of occurrence of the other Uncertainty Analysis for Engineers

11. Probability Laws • Pr(A)=1-Pr(A) • If A and B are independent, then • Pr(A and B)=Pr(AB)=Pr(A)Pr(B) • If A and B are mutually exclusive [m(AB)=0] then • Pr(A or B)=Pr(A  B)=Pr(A)+Pr(B) • In general • Pr(A and/or B) • =Pr(A  B) • =Pr(A)+Pr(B)-Pr(AB) Uncertainty Analysis for Engineers

12. One More Probability Law Pr(A and/or B and/or C) =Pr(A  B  C) =Pr(A)+Pr(B)+Pr(C)-Pr(AB)-Pr(AC)-Pr(BC)+Pr(ABC) Uncertainty Analysis for Engineers

13. Example • Consider a 3-stage process, as diagrammed below • Our goal is to find the probability of success of the entire operation, assuming all individual probabilities are independent • Branches represent parallel redundancy, so success in a stage requires success of, for example, A or B Pr(A)=0.9 Pr(D)=0.9 D A C E Pr(E)=0.9 B Pr(C)=0.95 F Pr(B)=0.8 Uncertainty Analysis for Engineers Pr(F)=0.5

14. Solution • Pr(S)=Pr(I)Pr(II)Pr(III) • where I, II, and III represent the three stages • Pr(I)=Pr(A)+Pr(B)-Pr(AB) • =0.9+0.8-0.9*0.8=0.98 (success requires A or B to succeed) • Pr(II)=Pr(C)=0.95 • Pr(III)=Pr(D)+Pr(E)+Pr(F)-Pr(DE)-Pr(EF)-Pr(DF)+Pr(DEF) • =0.9+0.9+0.5-0.9*0.9-0.9*0.5-0.9*0.5+0.9*0.9*0.5=0.995 • So, Pr(S)=0.98*0.95*0.995=0.926 Uncertainty Analysis for Engineers

15. Another Example • What if 2 events are not independent? • Consider the system below, where event G is in both stages G G H J Uncertainty Analysis for Engineers

16. Conditional Probability • Conditional probability [Pr(B|A)] of an event B with respect to some other event A is the probability that B will occur, given that A has taken place • For our example, Pr(II|I) represents the probability of successful operation of stage II, given successful operation of stage I • Once A has occurred, then A replaces I as the sample space of interest, so the size of AB relative to the new set is given by m(AB)/m(A) Uncertainty Analysis for Engineers

17. Cond. Probability (cont.) • Pr(B|A)=[m(AB)/m(I)]/[m(A)/m(I)] • =Pr(AB)/Pr(A) • Or • Pr(A and B)=Pr(AB)=Pr(A)Pr(B|A) • Extending… • Pr(A and B and C)=Pr(ABC)= Pr(A)Pr(B|A)Pr(C|AB) • Pr(C|AB) is probability of C, given that A and B have occurred Uncertainty Analysis for Engineers

18. Example • 75% of transistors come from vendor 1 and 25% from vendor 2 • 99% of supply from vendor 1 and 90% of supply from vendor 2 are acceptable • If we randomly pick a transistor, what is the probability that it came from vendor 1 and is defective? • Also, what is the probability that the transistor is defective, irrespective of the vendor Uncertainty Analysis for Engineers

19. Solution • A1=transistor from vendor 1 • A2=transistor from vendor 2 • B1=good transistor • B2=bad transistor • Pr(A1)=0.75; Pr(A2)=0.25 • Pr(B1|A1)=0.99; Pr(B2|A1)=0.01 • Pr(B1|A2)=0.90; Pr(B2|A2)=0.10 Uncertainty Analysis for Engineers

20. Solution (cont.) • Pr(A1B2)=Pr(A1) Pr(B2|A1)=0.75*0.01=0.0075 • Pr(A2B2)=Pr(A2) Pr(B2|A2)=0.25*0.1=0.025 • Pr(B2)=0.0075+0.025=0.0325 Uncertainty Analysis for Engineers

21. A Generalization • If B depends on a series of previous events (Ai) then Uncertainty Analysis for Engineers

22. Example (2.25) • Hurricanes: C1=Category 1, C2=Category 2, etc. • P(C1)=.35, P(C2)=.25, P(C3)=.14, P(C4)=.05, P(C5)=.01 • D=Damage; P(D|C1)=.05, P(D|C2)=.1, P(D|C3)=.25, P(D|C4)=.6, P(D|C5)=1.0 • What is probability of damage? • P(D)=P(D|C1)P(C1)+ P(D|C2)P(C2)+ P(D|C3)P(C3)+ P(D|C4)P(C4)+ P(D|C5)P(C5)=0.1175 Uncertainty Analysis for Engineers