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Additive Rule/ Contingency Table

Learn about probabilities, contingency tables, conditional probability, independence, serial vs. parallel systems, and random variables with examples and calculations.

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Additive Rule/ Contingency Table

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  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 2015

  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 2015

  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 2015

  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 2015

  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 2015

  6. Independence Definitions • 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) • Two events A and B are independent if and only if P (A ∩ B) = P(A) P(B) EGR 252 2015

  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 2015

  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. EGR 252 2015

  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 2015

  10. Chapter 3: Random Variables and Probability Distributions • Definition and nomenclature • A random variable is a function that associates a real number with each element in the sample space. • We use an uppercasel letter such as X to denote the random variable. • We use a lowercase letter such as x for one of its values. • Example: Consider a random variable Y which takes on all values y for which y > 5. EGR 252 2015

  11. Defining Probabilities: Random Variables • Examples: • Out of 100 heart catheterization procedures performed at a local hospital each year, the probability that more than five of them will result in complications is P(X > 5) • Drywall anchors are sold in packs of 50 at the local hardware store. The probability that no more than 3 will be defective is P(Y < 3) EGR 252 2015

  12. Discrete Random Variables • Pr. 2.51 P.59 (Modified) A box contains 500 envelopes (75 have $100, 150 have $25, 275 have $10) • Assume someone spends $75 to buy 3 envelopes. The sample space describing the presence of $10 bills (H) vs. bills that are not $10 (N) is: • S = {NNN, NNH, NHN, HNN, NHH, HNH, HHN, HHH} • The random variable associated with this situation, X, reflects the outcome of the experiment • X is the number of envelopes that contain $10 • X = {0, 1, 2, 3} • Why no more than 3? Why 0? EGR 252 2015

  13. Discrete Probability Distributions 1 • The probability that the envelope contains a $10 bill is 275/500 or .55 • What is the probability that there are no $10 bills in the group? P(X = 0) =(1-0.55) * (1-0.55) *(1-0.55) = 0.091125 P(X = 1) = 3 * (0.55)*(1-0.55)* (1-0.55) = 0.334125 • Why 3 for the X = 1 case? • Three items in the sample space for X = 1 • NNH NHN HNN EGR 252 2015

  14. Probabilities per of Sample Space Outcomes NNN P(X=0)= (1-0.55)* (1-0.55)* (1-0.55)= 0.091125 P(X=0) = 0.091125 NNH P(X=1)= (1-0.55)* (1-0.55)* (0.55)= 0.111375 NHN P(X=1)= (1-0.55)* (0.55) *(1-0.55)= 0.111375 HNN P(X=1)= (0.55)* (1-0.55)* (1-0.55)= 0.111375 P(X=1) = 0.111375 + 0.111375 + 0.111375 = 0.334125 Add all outcome probabilities for each X = 1 to get the total probability of drawing an ONE $10 envelope. NHH P(X=2)= (1-0.55)* (0.55)* (0.55)= 0.136125 HHN P(X=2)= (0.55)* (0.55)* (1-0.55)= 0.136125 HNH P(X=2)= (0.55)* (1-0.55)* (0.55)= 0.136125 P(X=2) = 0.136125 + 0.136125 + 0.136125 = 0.408375 HHH P(X=3) =(0.55)* (0.55)* (0.55)= 0.166375 P(X=3) = 0.166375 A probability was calculated for all outcomes in the sample space S = {NNN, NNH, NHN, HNN, NHH, HNH, HHN, HHH} EGR 252 2015

  15. Discrete Probability Distributions 2 P(X = 0) = 0.091125 P(X = 1) = 0.334125 P(X = 2) = 0.408375 P(X = 3) = 0.166375 • The probability distribution associated with the number of $10 bills is given by: EGR 252 2015

  16. Another View • The probability histogram EGR 252 2015

  17. Another Discrete Probability Example • Given: • A shipment consists of 8 computers • 3 of the 8 are defective • Experiment: Randomly select 2 computers • Definition: random variable X = # of defective computers selected • What is the probability distribution for X? • Possible values for X: X = 0 X =1 X = 2 • Let’s start with P(X=0) [0 defectives and 2 nondefectives are selected] Recall that P = specified target / all possible (all ways to get 0 out of 3 defectives) ∩ (all ways to get 2 out of 5 nondefectives) (all ways to choose 2 out of 8 computers) (all ways to choose 2 out of 8 computers) EGR 252 2015

  18. Discrete Probability Example • What is the probability distribution for X? • Possible values for X: X = 0 X =1 X = 2 • Let’s calculate P(X=1) [1 defective and 1 nondefective are selected] (all ways to get 1 out of 3 defectives) ∩ (all ways to get 1 out of 5 nondefectives) (all ways to choose 2 out of 8 computers) (all ways to choose 2 out of 8 computers) EGR 252 2015

  19. Discrete Probability Distributions • The discrete probability distribution function (pdf) • f(x) = P(X = x) ≥ 0 • Σxf(x) = 1 • The cumulative distribution,F(x) • F(x) = P(X ≤ x) = Σt ≤ xf(t) • Note the importance of case: F not same as f EGR 252 2015

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