Chapter 2:Probability Lecture 2 Recall: Lecture 1 Probability Basics Definitions Symbols Recall: Lecture 1 Graphics Decision Trees Venn Diagrams Recall: Lecture 1 Calculations nPr nCr Multiplication Rule JMB Ch2 Lecture 2 EGR 252 JMB 2019

Additive Rule/ Contingency Table Experiment: Draw 1 card from a standard 52 card deck. Record Value (A, 2, 3, …Q,K) & Color The probabilities associated with drawing an ace and with drawing a black card are shown in the following contingency table: Event A = ace Event B = black card Therefore the probability of drawing an ace or a black card is: P(A U B) = P(A) + P(B) – P (A ∩ B) = 4/52 + 26/52 – 2/52 = 28/52 Type Color Total Red Black Ace 2 4 Non-Ace 24 48 26 52 JMB Ch2 Lecture 2 EGR 252 JMB 2019

Short Circuit Example - Data An appliance manufacturer has learned of an increased incidence of short circuits and fires in a line of ranges sold over a 5 month period. A review of the defect data indicates the probabilities that if a short circuit occurs, it will be at any one of several locations is as follows: The sum of the probabilities equals _____ Location P House Junction (HJ) 0.46 Oven/MW junction (OM) 0.14 Thermostat (T) 0.09 Oven coil (OC) 0.24 Electronic controls (EC) 0.07 .46+.14+.09+.24+.07 = 1 JMB Ch2 Lecture 2 EGR 252 JMB 2019

Short Circuit Example - Probabilities If we are told that the probabilities represent mutually exclusive events, we can calculate the following: The probability that the short circuit does not occur at the house junction is P(HJ’) = 1 - P(HJ) = 1 – 0.46 = 0.54 The probability that the short circuit occurs at either the Oven/MW junction or the oven coil is P(OM U OC) = P(OM)+P(OC) = 0.14 + 0.24 = 0.38 JMB Ch2 Lecture 2 EGR 252 JMB 2019

Conditional Probability The conditional probability of B given A is denoted by P(B|A) and is calculated by P(B|A) = P(B ∩ A) / P(A) Example: S = {1,2,3,4,5,6,7,8,9,11} Event A = number greater than 6 P(A) = 4/10 Event B = odd number P(B) = 6/10 (B∩A) = {7, 9, 11} P (B∩A) = 3/10 P(B|A) = P(B ∩ A) / P(A) = (3/10) / (4/10) = 3/4 In general, the conditional probability of B given A is denoted by P(B|A) and is given by P(B|A) = P(B ∩ A) / P(A) JMB Ch2 Lecture 2 EGR 252 JMB 2019

Multiplicative Rule If in an experiment the events A and B can both occur, then P(B ∩ A) = P(A) * P(B|A) Previous Example: S = {1,2,3,4,5,6,7,8,9,11} Event A = number greater than 6 P(A) = 4/10 Event B = odd number P(B) = 6/10 P(B|A) = 3/4 (calculated in previous slide) P(B∩A) = P(A)*P(B|A) = (4/10)*(3/4) = 3/10 P(B|A) = P(B ∩ A) / P(A) JMB Ch2 Lecture 2 EGR 252 JMB 2019

Independence Definitions If the conditional probabilities P(A|B) and P(B|A) exist, the events A and B are independent if and only if P(A|B) = P(A) or P(B|A) = P(B) Two events A and B are independent if and only if P (A ∩ B) = P(A) P(B) In general, the conditional probability of B given A is denoted by P(B|A) and is given by P(B|A) = P(B ∩ A) / P(A) JMB Ch2 Lecture 2 EGR 252 JMB 2019

Independence Example A quality engineer collected the following data on 100 defective items produced by a manufacturer in the southeast: Problem/Shift Electrical Mechanical Other Day 20 15 25 Night 10 What is the probability that the defective items were associated with the day shift? P(Day) = (20+15+25) / 100 = .60 or 60% What was the relative frequency of defectives categorized as electrical? (20 + 10) / 100 P(Electrical) = .30 Are Electrical and Day independent? P(E ∩ D) = 20 / 100 = .20 P(D) P(E) = (.60) (.30) = .18 Since .20 ≠.18, Day and Electrical are not independent. JMB Ch2 Lecture 2 EGR 252 JMB 2019

Serial and Parallel Systems For increased safety and reliability, systems are often designed with redundancies. A typical system might look like the following: Principles: NOTE: if components are in serial (e.g., A & B), all must work in order for the system to work. If components are in parallel, the system works if any of the components work. If components are in serial (e.g., A & B), all must work in order for the system to work. If components are in parallel, the system works if any of the components work. JMB Ch2 Lecture 2 EGR 252 JMB 2019

Serial and Parallel Systems 1 What is the probability that: Segment 1 works? A and B in series P(A∩B) = P(A) * P(B) = (0.95)(0.90) = 0.855 Segment 2 works? C and D in parallel will work unless both C and D do not function 1 – P(C’) * P(D’) = 1 – (0.12) * (0.15) = 1 - 0.018 = 0.982 The entire system works? Segment 1, Segment 2 and E in series P(Segment1) * P(Segment2) * P(E) = 0.855*0.982*0.97 = 0.814 2 Segment 1: P(A∩B) = P(A)P(B) = (.95)(.9) = .855 Segment 2: 1 – P(C’)P(D’) = 1 – (.12)(.15) = 1-.018 = 0.982 Entire system: P(1)P(2)P(E) = .855*.982*.97 =.814 JMB Ch2 Lecture 2 EGR 252 JMB 2019

Study Suggestions Examples Exercises 2.15 page 46 2.21 page 50 In general, the conditional probability of B given A is denoted by P(B|A) and is given by P(B|A) = P(B ∩ A) / P(A) JMB Ch2 Lecture 2 EGR 252 JMB 2019