JMB Ch2 Lecture 2 vSp2012 EGR Slide 1 Additive Rule/ Contingency Table Experiment: Draw 1 card from a standard 52 card deck. Record Value (A-K), Color & Suit.  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: Type Color Total RedBlack Ace224 Non-Ace24 48 Total26 52

JMB Ch2 Lecture 2 vSp2012 EGR Slide 2 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 _____ LocationP 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

JMB Ch2 Lecture 2 vSp2012 EGR Slide 3 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) = = 0.38

JMB Ch2 Lecture 2 vSp2012 EGR Slide 4 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

JMB Ch2 Lecture 2 vSp2012 EGR Slide 5 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

JMB Ch2 Lecture 2 vSp2012 EGR Slide 6 Independence Definitions  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)

JMB Ch2 Lecture 2 vSp2012 EGR Slide 7 Independence Example  A quality engineer collected the following data on 100 defective items produced by a manufacturer in the southeast:  What is the probability that the defective items were associated with the day shift?  P(Day) = ( ) / 100 =.60 or 60%  What was the relative frequency of defectives categorized as electrical?  ( ) / 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. Problem/ShiftElectricalMechanicalOther Day Night102010

JMB Ch2 Lecture 2 vSp2012 EGR Slide 8 Serial and Parallel Systems  For increased safety and reliability, systems are often designed with redundancies. A typical system might look like the following:  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 Ch2 Lecture 2 vSp2012 EGR Slide 9 Serial and Parallel Systems  What is the probability that:  Segment 1 works?  A and B in series P(A∩B) = P(A) * P(B) = (0.95)(0.9) =  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) = =  The entire system works?  Segment 1, Segment 2 and E in series P(Segment1) * P(Segment2) * P(E) = 0.855*0.982*0.97 =