Additive Rule/ Contingency Table

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: EGR 252.001 Spring 2010

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 _____ EGR 252.001 Spring 2010

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) = 0.14 + 0.24 = 0.38 EGR 252.001 Spring 2010

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 EGR 252.001 Spring 2010

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 EGR 252.001 Spring 2010

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) EGR 252.001 Spring 2010

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) = (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. EGR 252.001 Spring 2010

8. Serial and Parallel Systems • For increased safety and reliability, systems are often designed with redundancies. A typical system might look like the following: 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. EGR 252.001 Spring 2010

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) = 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 1 2 EGR 252.001 Spring 2010