Review of Probability Jake Blanchard Spring 2010 Uncertainty Analysis for Engineers1

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 Engineers2

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 ◦ This interpretation is the most common among statisticians Uncertainty Analysis for Engineers3

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 Engineers4

Set Theory Set=collection of distinct objects Union: C 1 =A  B Intersection: C 2 =A  B=AB Complement (“not”): C 3 =A  =null set I=entire set m(A)=number of elements in set A Uncertainty Analysis for Engineers5

Identities A   =A A  I=I A   = A  =  A I=A A  A=A A A=A  =I A A=  A  A=I Uncertainty Analysis for Engineers6

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 Engineers7

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 Engineers8

Probability Rolling dice m(I)=6 A=rolling a 2 Pr(A)=m(A)/m(I)=1/6 Uncertainty Analysis for Engineers9

Definition Two events are independent if the occurrence of one does not change the probability of occurrence of the other Uncertainty Analysis for Engineers10

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 Engineers11

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 Engineers12

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 Uncertainty Analysis for Engineers13 A B C F E D Pr(C)=0. 95 Pr(B)=0.8 Pr(A)=0. 9 Pr(D)=0. 9 Pr(F)=0.5 Pr(E)=0.9

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) = *0.8=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*0.5=0.995 So, Pr(S)=0.98*0.95*0.995=0.926 Uncertainty Analysis for Engineers14

Another Example What if 2 events are not independent? Consider the system below, where event G is in both stages Uncertainty Analysis for Engineers15 G H J G

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 Engineers16

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 Engineers17

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 Engineers18

Solution A 1 =transistor from vendor 1 A 2 =transistor from vendor 2 B 1 =good transistor B 2 =bad transistor Pr(A 1 )=0.75; Pr(A 2 )=0.25 Pr(B 1 |A 1 )=0.99; Pr(B 2 |A 1 )=0.01 Pr(B 1 |A 2 )=0.90; Pr(B 2 |A 2 )=0.10 Uncertainty Analysis for Engineers19

Solution (cont.) Pr(A 1 B 2 )=Pr(A 1 ) Pr(B 2 |A 1 )=0.75*0.01= Pr(A 2 B 2 )=Pr(A 2 ) Pr(B 2 |A 2 )=0.25*0.1=0.025 Pr(B 2 )= = Uncertainty Analysis for Engineers20

A Generalization If B depends on a series of previous events (A i ) then Uncertainty Analysis for Engineers21

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)= Uncertainty Analysis for Engineers22