Additive Rule/ Contingency Table Experiment: Draw 1 card from a standard 52 card deck. Record Value (A-K), Color & Suit. Ø The probabilities associated with drawing an ace and with drawing a black card are shown in the following contingency table: Color Type Ace Red Black Total 2 2 4 Non-Ace 24 24 48 Total 26 26 52 Ø Event A = ace Event B = black card Ø Therefore the probability of drawing an ace or a black card is: JMB Ch 2 Lecture 2 EGR 252 2014 Slide 1
Short Circuit Example - Data q 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: 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 q The sum of the probabilities equals _____ JMB Ch 2 Lecture 2 EGR 252 2014 Slide 2
Short Circuit Example - Probabilities q If we are told that the probabilities represent mutually exclusive events, we can calculate the following: q The probability that the short circuit does not occur at the house junction is P(HJ’) = 1 - P(HJ) = 1 – 0. 46 = 0. 54 q 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 Ch 2 Lecture 2 EGR 252 2014 Slide 3
Conditional Probability q 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) q Example: Ø S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 11} Ø Event A = number greater than 6 Ø Event B = odd number Ø (B∩A) = {7, 9, 11} P(A) = 4/10 P(B) = 6/10 P (B∩A) = 3/10 q P(B|A) = P(B ∩ A) / P(A) = (3/10) / (4/10) = 3/4 JMB Ch 2 Lecture 2 EGR 252 2014 Slide 4
Multiplicative Rule q If in an experiment the events A and B can both occur, then P(B ∩ A) = P(A) * P(B|A) q Previous Example: Ø S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 11} Ø Event A = number greater than 6 P(A) = 4/10 Ø Event B = odd number P(B) = 6/10 Ø P(B|A) = 3/4 (calculated in previous slide) q P(B∩A) = P(A)*P(B|A) = (4/10)*(3/4) = 3/10 JMB Ch 2 Lecture 2 EGR 252 2014 Slide 5
Independence Definitions q 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) q Two events A and B are independent if and only if P (A ∩ B) = P(A) P(B) JMB Ch 2 Lecture 2 EGR 252 2014 Slide 6
Independence Example q A quality engineer collected the following data on 100 defective items produced by a manufacturer in the southeast: Problem/Shift 20 15 25 10 20 10 What is the probability that the defective items were associated with the day shift? Ø P(Day) = (20+15+25) / 100 =. 60 or 60% What was the relative frequency of defectives categorized as electrical? Ø q Other Night q Mechanical Day q Electrical (20 + 10) / 100 P(Electrical) =. 30 Are Electrical and Day independent? Ø Ø P(E ∩ D) = 20 / 100 =. 20 P(D) P(E) = (. 60) (. 30) =. 18 Since. 20 ≠. 18, Day and Electrical are not independent. JMB Ch 2 Lecture 2 EGR 252 2014 Slide 7
Serial and Parallel Systems q For increased safety and reliability, systems are often designed with redundancies. A typical system might look like the following: q Principles: 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 Ch 2 Lecture 2 EGR 252 2014 Slide 8
Serial and Parallel Systems q What is the probability that: Ø Segment 1 works? Ø A and B in series P(A∩B) = P(A) * P(B) = (0. 95)(0. 9) = 0. 855 1 2 Ø Segment 2 works? Ø 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 Ø The entire system works? Ø Segment 1, Segment 2 and E in series P(Segment 1) * P(Segment 2) * P(E) = 0. 855*0. 982*0. 97 =0. 814 JMB Ch 2 Lecture 2 EGR 252 2014 Slide 9
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