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Additive Rule Review

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:.

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Additive Rule Review

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  1. 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: EGR 252.001 Spring 2008

  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 FMEA data indicates the probabilities that if a short circuit occurs, it will be at any one of several locations is as follows: EGR 252.001 Spring 2008

  3. 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 EGR 252.001 Spring 2008

  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 2008

  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 2008

  6. 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) EGR 252.001 Spring 2008

  7. Independence Example • A quality engineer collected the following data on defects: • 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. EGR 252.001 Spring 2008

  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 2008

  9. Serial and Parallel Systems • 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 1 2 EGR 252.001 Spring 2008

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