Additive Rule Review Experiment: Draw 1 Card. Note Kind, Color & Suit. Probabilities associated with drawing an ace and with drawing a black card are shown in the following contingency table: Therefore the probability of drawing an ace or a black card is given by: Type Color Total Red Black Ace 2 4 Non-Ace 24 48 26 52 JMB Chapter 2 Lecture 2 EGR 252.001 Spring 2008

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 FMEA data indicates the probabilities that if a short circuit occurs, it will be at any one of several locations is as follows: Location P House Junction 0.46 Oven/MW junction 0.14 Thermostat 0.09 Oven coil 0.24 Electronic controls 0.07 JMB Chapter 2 Lecture 2 EGR 252.001 Spring 2008

Short Circuit Example - Probabilities The probability that the short circuit does not occur at the house junction is … P(H’) = 1 – 0.46 = 0.54 The probability that the short circuit occurs at either the Oven/MW junction or the oven coil is … P(J U C) = P(J) + P(C) = 0.14 + 0.24 = 0.38 P(H’) = 1 – 0.46 = 0.54 P(J U C) = P(J) + P(C) = 0.14 + 0.24 = 0.38 P(E ∩ T) = P(E)*P(T) = 0.09 * 0.07 = 0.0063 JMB Chapter 2 Lecture 2 EGR 252.001 Spring 2008

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 P(C’|2) = P(C’∩D)/P(2) = ((1-.88)*0.85)/0.982 = 0.1039 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) (Definition 2.9, pg. 48) JMB Chapter 2 Lecture 2 EGR 252.001 Spring 2008

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(C’|2) = P(C’∩D)/P(2) = ((1-.88)*0.85)/0.982 = 0.1039 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) (Definition 2.9, pg. 48) JMB Chapter 2 Lecture 2 EGR 252.001 Spring 2008

Independence If in an experiment 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) P(C’|2) = P(C’∩D)/P(2) = ((1-.88)*0.85)/0.982 = 0.1039 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) (Definition 2.9, pg. 48) JMB Chapter 2 Lecture 2 EGR 252.001 Spring 2008

Independence Example A quality engineer collected the following data on defects: Electrical Mechanical Other Day 20 15 25 Night 10 What is the likelihood that defects were associated with the day shift? P(Day) = (20+15+25) / 100 = .60 What was the relative frequency of electrical defects? (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 Chapter 2 Lecture 2 EGR 252.001 Spring 2008

Serial and Parallel Systems For increased safety and reliability, systems are often designed with redundancies. A typical system might look like the following: 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 Chapter 2 Lecture 2 EGR 252.001 Spring 2008

Serial and Parallel Systems 1 What is the probability that: Segment 1 works? P(A∩B) = P(A)P(B) = (.95)(.9) = .855 Segment 2 works? 1 – P(C’)P(D’) = 1 – (.12)(.15) = 1-.018 = 0.982 The entire system works? P(Segment1)P(Segment2)P(E) = .855*.982*.97 =.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 Chapter 2 Lecture 2 EGR 252.001 Spring 2008