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Probability Essentials Chapter 3

Probability Essentials Chapter 3

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Probability Essentials Chapter 3

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  1. Probability EssentialsChapter 3 “A pinch of probably is worth a pound of perhaps.” James Thurber, American Humorist MGMT 242 Fall, 1998

  2. Goals for Chapter 3 • Understand • Two different interpretations of probability--Frequency and Subjective • Rules for using probability • Numerical Scale • Addition Rule--probability of “either A or B” • Multiplication Rule--probability of “A and B” • Statistical Independence • Conditional Probability • Bayes’ Rule MGMT 242 Fall, 1998

  3. Interpretation of Probability • Historical Perspective-- • origins: Pascal & Fermat, early 17th century • purpose: to determine best wager for gambling, from which comes odds, expectation. • Frequency Interpretation-- • example: what is probability of drawing the ace of diamonds from a complete, well shuffled deck? Ans: 1/52; • example: what is the probability of drawing any ace from a complete, well shuffled deck? Ans: 4/52 = 1/13 • example: if 50.6 % of babies born during the last 50 years in Union County were male, what is the probability that the next baby will be male? Ans: 50.6 %. • Probability of event A = N(A) / N • N(A) is number of times event A can occur in total number of trials, N. MGMT 242 Fall, 1998

  4. Interpretation of Probability--cont. • Subjective Interpretation-- • based on estimate (experience, intuition, ...) of likelihood that event will occur • example: stockbroker says probability is 30% of Microsoft gaining $50 / share during next six months; • example: after diagnostic tests, physician says the probability that patient has malignant tumor is 40%. • More Examples • An economist estimates that the probability that the inflation rate will exceed 3% next year is 85% (Subjective) • 25% of employees in a firm are over 50 years of age--if the firm downsizes, the probability that a discharged employee will be over 50 should be 25% , if employees selected for downsizing are randomly chosen. (Frequency) MGMT 242 Fall, 1998

  5. Probability Scale • If the probability of an event A occurring is P(A), then 0 is less than or equal to P(A) and P(A) is less than or equal to 1 [ 0 P(A)  1]. • P(A) = 0: the event A will not happen. • P(A) = 1: the event A is sure to happen. • Example: if A is “the sun will rise in the east tomorrow”, then P(A) = 1 (or very close). • Example; if A is “the sun will rise in the west tomorrow”, then P(A) = 0 (or very close). MGMT 242 Fall, 1998

  6. Complement of Events • AC, the complement of an event A, comprises all the events that can occur, other than A; • example: suppose A is the event that an employee over 50 is downsized. AC is the event that employees 50 years old or younger are downsized; • example: A is the event that Microsoft stock will rise $50 / share or more during the next 6 months; AC is the event Microsoft will rise less than $50 / share (or fall) during the next 6 months. • P(A) + P(AC) = 1, i.e. Either the event or its complement must occur • This relation is useful--sometimes P(AC) is easier to calculate than P(A). MGMT 242 Fall, 1998

  7. Complement of Event--examples • 5 out of 1000 Widgets have defective paint (on average). What is the probability that a Widget will not have defective paint cover? • Answer: let A be event that a Widget has a defective paint cover; P(A) = 0.005 (5 in 1000). Then we have P(A) + P(AC) = 1, or P(AC) = 1 - P(A) = 1 - 0.005=0.995 MGMT 242 Fall, 1998

  8. Addition Law for Probabilities--Mutually Exclusive Events • Consider two mutually exclusive events, A & B • example: . event A--you receive an A in MGMT 242, event B--you receive a B • Then the probability of either A or B occurring, P(A or B) = P(A) + P(B); • A or B denoted as AB (“union of A and B”) • example: in previous years, I’ve given 15% A’s in undergraduate courses, 25% B’s. What is the probability that an average student in MGMT 242 will receive either an A or a B? P(A) = 0.15, P(B) = 0.25, so P(A) + P(B) = 0.40 MGMT 242 Fall, 1998

  9. Addition Law for Probabilities--General Case • Suppose that both events A & B can occur in a trial • example: event A--downsized employee is over 50, event B--downsized employee is female • Then P(A or B) = P(A) + P(B) - P(A and B); • A and B means that both A and B occur, e.g. the downsized employee is over 50 and female. A and B is denoted as AB, (“intersection of A and B”) • The term -P(A and B) is included so that the intersection, AB, isn’t counted twice (see Venn diagram illustration) • To calculate P(A and B) we have to use data directly or learn about Multiplication Law and Statistical Independence. MGMT 242 Fall, 1998

  10. Addition Law for Probabilities--General Case--Example • Consider Table below; total number of employees is 1000 : event A:employee is female, event B: employee is over 50 • P(A) = 350 / 1000 = 0.35; P(B) = 250 / 1000 = 0.25; P(A and B) = 50 / 1000 = 0.05 • Thus the Probability that an employee is either over 50 or female is P(A or B) = 0.35 + 0.25 - 0.05 = 0.55 MGMT 242 Fall, 1998

  11. Conditional Probabilities • P (A | B) denotes the conditional probability, probability of event A if event B occurs • Example from previous table, what is the probability that an employee is female if the employee is over 50? • P (A | B) = 50 / 250 = 0.20 or 20% • Formally, P(A | B ) given by relation P(A and B) = P(A | B) P(B) • Example: P (A and B) = 50 / 1000; P (A | B) = 50 /250; P(B) = 250 / 1000 for A, female; B, over 50. MGMT 242 Fall, 1998

  12. Multiplication Law for Probabilities • Suppose we have two events, A (e.g. being a female employee) and B (e.g. being over 50). Then P(A and B) = P(A) P(B) if and only if the events A and B are “statistically independent” • previous example: P(A) = 350/ 1000 = 0.35, P(B) = 250/1000 = 0.25; then if A and B are independent events, P(A and B) = 0.35 x 0.25 = 0.0875; • however, P(A and B) = 0.05 (from Table). Therefore, being a female employee and being an employee over 50 are not statistically independent events. MGMT 242 Fall, 1998

  13. Multiplication Law for Probabilities-Examples • A manufacturer of widgets finds that 5 out of 1000 widgets have a defective paint cover (event A) and that 2 out of 1000 widgets are not within length specs (Event B). If the two defects (A and B) are statistically independent, what is the probability (1) that any widget will have both defects (A and B); (2) have no defects; (3) have either defect A or defect B or both. • (1) P(A and B) = P(A)P(B) = .005 x .002 = .00001 • (2) P(ACand BC) = 0.995 x 0.998 = 1 -.005 - .002 -.00001=0.99299 • (3) P(A or B) = P(A) + P(B) - P(A and B) = 0.005 + 0.002 - 0.0001=.00701 MGMT 242 Fall, 1998

  14. Multiplication Law for Probabilities-Examples • A manufacturer of widgets finds that 5 out of 1000 widgets have a defective paint cover (event A). What is the probability that (1) two widgets in a row will have defective paint covers; (2) one widget out of the two (either the first or the second) will have a defective paint cover; (3) that neither of these two widgets will be defective? (A1: first widget is defective; A2: second widget is defective). • (1) P(A1 and A2) = P(A1) P(A2) = (.005)2; • (2) P(A1 or A2) - P(A1 and A2) = 2 x 0.995 x 0.005 • (3) P(A1C and A2C) = (0.995)2 = 1 - 2x0.995 x 0.005 - (.005)2 = = 1 - 0.005 x[2x(1 - 0.005) + 0.005]=(1-.005)2 MGMT 242 Fall, 1998

  15. Bayes’ Theorem--Example • Bayes’ Theorem is best approached by an example. Suppose we have a diagnostic test (for example, mammography for breast cancer). The test gives 99% “true positive” results, that is 99% of the people who have the disease give a positive test and 20% “false positive” results, that is 20% of the people who don’t have the disease give a positive result. Suppose you’re a woman in an age group and with family background such that there’s a 5 % probability (in the absence of a test) that you might get breast cancer next year. You test positive; what is the probability that you have breast cancer? MGMT 242 Fall, 1998

  16. Bayes’ Theorem--Example (cont.) • To answer this question consider the calculation below; we take 100,000 women for whom the prior probability of breast cancer is 5%, so that 5000 of these women have breast cancer. Of these 5000, 0.99 x 5000 = 4950 test positive. Of the 100,000 women, 95,000 don’t have breast cancer. Of these 95,000, 0.20 = 19,000 give a false positive test. The total number testing positive is 19,000+4950 = 23,950. The conditional probability that if you test positive you then have the disease is P (D+ | T+) = 4950 / 23,950 = 0.2067 or 21% MGMT 242 Fall, 1998

  17. Bayes’ Theorem • Bayes’ Theorem deals with conditional probabilities--given a “posterior” event, what is the probability of the “prior” event. • Suppose we have prior events B1, B2, … (e.g. B1: have cancer, B2: don’t have cancer). Also, suppose we know conditional probabilities P(A |B1), P(A |B2), … (e.g. A is testing positive for cancer). • Then P (B1 | A) = P(A | B1) P(B1) P(A | Bi) P(Bi) Note that the summation term in the denominator is simply P(A). MGMT 242 Fall, 1998