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Welcome to BUAD 310

Welcome to BUAD 310. Instructor: Kam Hamidieh Lecture 5, Wednesday January 29, 2014. Agenda & Announcement. Today : Continue with Chapters 7 (Read all of it.) Start Chapter 8 (Read all of it.) HW 1 is due today by 5 PM. No extensions will be given.

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Welcome to BUAD 310

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  1. Welcome to BUAD 310 Instructor: Kam Hamidieh Lecture 5, Wednesday January 29, 2014

  2. Agenda & Announcement • Today: • Continue with Chapters 7 (Read all of it.) • Start Chapter 8 (Read all of it.) • HW 1 is due today by 5 PM. No extensions will be given. • HW 2 will be posted soon and due on Wednesday February 12th, 5 pm. • Marshall peer tutoring:http://students.marshall.usc.edu/undergrad/marshall-peer-tutoring-program/ BUAD 310 - Kam Hamidieh

  3. FYI • “The Age of Big Data” NY Times, February 2012 • Do a search on “Big Data” at:www.google.com/trends BUAD 310 - Kam Hamidieh

  4. From Last Time • Scatterplots: • Two dimensional plot of two variables • Used to assess relationship between two variables: nature (linear/nonlinear), strength, direction • Probability model has two parts: • List of possible outcomes of a random phenomenon (or an experiment) • Their probabilities • Law of large number: justifies our intuition about probabilities in the long run BUAD 310 - Kam Hamidieh

  5. From Last Time • Sample Space = exhaustive listing of simple events from an experiment • Events = combination of simple events • Equal likely events: P(A) = (count simple events in A) / (Total count in S) • Complementary events: Can’t happen at the same time, together give S • Mutually exclusive events: Can’t happen at the same time BUAD 310 - Kam Hamidieh

  6. Essential Rules of Probability • For any events A and B defined on S, • 0  P(A)  1 • P( S ) = 1 ( P(empty set )=0 ) • If A and B are disjoint, then P(A or B) = P(A  B) = P(A) + P(B) • Using the above, you can also show that:P(A) + P(AC) = 1. BUAD 310 - Kam Hamidieh

  7. Examples Our experiment is to roll a fair die once. The sample space is S = {1,2,3,4,5,6}. Consider the following events: A = {1,2,3} and AC = {4,5,6} B = {Odd outcome} = {1,3,5} C = {6} Then: P(A) + P(AC) = P({1,2,3}) + P({4,5,6}) = 1/3 + 1/3 = 1 P(B  C) = P({1,3,5,6}) = 4/6 ?=? P(B) + P(C) = 3/6 + 1/6 BUAD 310 - Kam Hamidieh

  8. In Class Exercise 1 (http://www.eia.gov/forecasts/steo/uncertainty/index.cfm?src=Markets-f1) The graph below show the probabilities of crude oil exceeding a certain level in April 2014 at the current time of early January 2014. (Ignore blue line.) By eyeballing: P(Oil ≥ 90) ≈ 0.90, P(Oil ≥ 100) ≈ 0.40, P(Oil ≥ 110) ≈ 0.07 You work for a hedge fund that speculates by physically buying and selling crude oil. In Jan (now), your fund bought 100,000 barrels of oil for $96, and then went into a contract to sell the oil in April. Your boss asks you “Hey, what are the chances we’ll lose money on this deal? I need the answer in the next five minutes!!!” What should you say? After you give him the correct answer he says “Hey, what are the chances we’ll make a million bucks or more here?” What would you say? BUAD 310 - Kam Hamidieh

  9. Examples • Some states are banning the use of cell phone while driving. • One study classified the accidents by the day of the week when they occurred. (D.A. Redelmeirer and R.J. Tibshirani, “Association between cellular-telephone calls and more vehicle collisions” New England Journal of Medicine, 336, 1997, pages 453-458. Here are the results: • All probabilities are between 0 and 1, and add up to 1. • The days and the probabilities together make up our model. BUAD 310 - Kam Hamidieh

  10. Example Continued • What is the probability that an accident occurs on a weekend?P(Saturday  Sunday) = P(Saturday) + P(Sunday) = 0.02 + 0.03 = 0.05 • What is the probability that an accident occurs on a weekday? (There are two ways to do this)P(Weekday) = 1 – P(weekend) = 1 – 0.05 = 0.95 BUAD 310 - Kam Hamidieh

  11. Addition Rule • What if you want to find P(A  B) but A and B are not mutually exclusive? • Then the general rule is:P(A  B) = P(A) + P(B) – P(A  B) (WHY?) BUAD 310 - Kam Hamidieh

  12. Example Our experiment is to toss a fair die once. The sample space is S = {1,2,3,4,5,6}. Consider the following events: A = {1,2,3} B = {Odd outcome} = {1,3,5} Note P(A  B) = P({1,3}) = 2/6 Then:P(A  B) = P({1,2,3,5}) = 4/6 which is alsoP(A) + P(B) - P(A  B) = 3/6 + 3/6 – 2/6 = 4/6 BUAD 310 - Kam Hamidieh

  13. Example (set up first!) BUAD 310 - Kam Hamidieh

  14. Example What is the probability of randomly drawing either an ace or a heart from a deck of 52 playing cards? There are 4 aces in the pack and 13 hearts. However, 1 card is both an ace and a heart. Thus: P(ace or heart) = P(ace) + P(heart) – P(ace and heart) = 4/52 + 13/52 - 1/52 = 16/52 ≈ .3 BUAD 310 - Kam Hamidieh

  15. Independent/Dependent Events • Two events are independent of each other if knowing that one will occur (or has occurred) will not change the probability of that the other occurs. • Two events are dependent if knowing that one will occur (or has occurred) changes the probability that the other occurs. BUAD 310 - Kam Hamidieh

  16. Probabilities for Independent Events Probabilities for Independent Events • If A and B are independent then P(A  B) = P(A) × P(B) WHY? BUAD 310 - Kam Hamidieh

  17. Example • Throw a fair coin twice: S = {HH, HT, TH, TT}. • Are the events A = {first toss is a head} = {HH, HT}B = {second toss is a tail} = {HT, TT} independent? In other words, P(A  B) = P(A) × P(B)? • Let’s see…Note A  B = {HT} soP(A  B) = 1/4P(A) × P(B) = ½ × ½ = ¼ soA and B are independent! BUAD 310 - Kam Hamidieh

  18. Example • Throw a fair coin twice: S = {HH, HT, TH, TT}. • Are the events A = {HH}, and B = {TT} independent? In other words, P(A  B) = P(A) × P(B)? • Let’s see…Note A  B = {HH}  {TT} = {} = Ø, soP(A  B) = P(Ø) = 0 butP(A) × P(B) = P({HH}) × P({TT}) = ¼ × ¼ = 1/16 soObviously 0 ≠ 1/16!A and B are not independent! BUAD 310 - Kam Hamidieh

  19. Independence vs Mutually Exclusive • When you want to check to see if A and B are disjoint events, you have to see if A  B = {} = Ø. You ask can A and B happen at the same time? • When you want to check to see if A and B are independent events, you have to see if P(A  B) = P(A) × P(B). You ask given one of the events has occurred, does it change the probability of the other event? • Note: • When two events are mutually exclusive and one happens, it turns the probability of the other one to 0. • When two events are independent and one happens, it leaves the probability of the other one alone. BUAD 310 - Kam Hamidieh

  20. In Class Exercise 2 Suppose you are given the following information: P(A) = 0.2, P(B) = 0.3, P(C) = 0.4, P(A and C) = 0.08, P(A and B) = 0.1 Answer: • Are A and B independent? • Are A and C independent? • Find P( A or C ) • Suppose you had not been told that P(A and C) = 0.08 but you were told that A and C are independent events. Would you be able to find P(A or C)? BUAD 310 - Kam Hamidieh

  21. Motivating Conditional Probability • Toss a fair die once: S = {1,2,3,4,5,6}. • Let A = { 1 } = { toss a 1 } • What is P(A)? • Suppose the die is tossed but I don’t tell you what value showed up. All I tell you is that the number is odd. • Given that a odd number has been tossed, would P(A) remain the same? Should you make any adjustments to this probability? BUAD 310 - Kam Hamidieh

  22. Conditional Probability • The conditional probability of the event A, given that the event B occurs, is the long-run frequency with which event A occurs when circumstances are such that B also occurs. • Note given B has happened, the sample space has been restricted to B. • Notation: P(A|B) The given part. This say Bhas happened or will happen. The event we want to find the probability for, in this case A. Just a separator. This vertical line does not mean division. BUAD 310 - Kam Hamidieh

  23. Rules for Conditional Probabilities • For conditional probabilities:P(A|B) = P(A  B) / P(B) • Example: • Toss a fair die once: S = {1,2,3,4,5,6}. • Consider the following events: • B = {1,2,3} • A = {Odd outcome} = {1,3,5} • Note P(A  B) = P({1,3}) = 2/6 • P(A|B) = P(A  B) / P(B) = (2/6) / (3/6) = 2/3 BUAD 310 - Kam Hamidieh

  24. More on Conditional Probabilities • In general P(A|B) ≠ P(B|A) • Just like regular probabilities: 0  P(A|B)  1 • Conditional probabilities will obey all the probability rules we have learned. • Technical: P(A|B) is defined only when P(B) > 0. • If A and B are independent then: P(A|B) = P(A) • Now we can explain this: P(A  B) = P(A) × P(B), for A and B independent. BUAD 310 - Kam Hamidieh

  25. Dependent Events and Multiplication Rule • Suppose A and B are dependent (not independent!) and you want to find: P(A  B) = ? • Use conditional probabilities to come up with the multiplication rule: P(A  B) = P(A) × P(B|A) = P(B) × P(A|B) • HOW? • In words: The probability that both A and B occur is equal to the probability that A occurs multiplied by the conditional probability of B given A occurs. BUAD 310 - Kam Hamidieh

  26. Example • When measuring the quality of a product coming out of a manufacturing process, it is not possible to test every item. • Generally a small “sample” out of a batch is taken and assumed to represent the entire batch. • Quality analysts must assess the chance a sample represents the entire batch. BUAD 310 - Kam Hamidieh

  27. Example Continued Suppose a box contains two defective components and two good ones. The inspector randomly draws two components. What is the probability that both are good? ( The answer is not 1/2 ) Draw 1 Draw 2 The chance of first drawing a good component is 2/4…. The chance of second drawing a good component is 1/3. P( first draw is good and second draw is good) = ? A = first draw is good, B = second draw is good P( A and B) = P(A) × P(B|A) = (2/4) × (1/3) = 1/6 BUAD 310 - Kam Hamidieh

  28. In Class Exercise 3 A fire insurance company insures a building that they estimate would have a 0.70 probability of being completely destroyed given it catches fire. They have accurate estimates that a building has only a 0.02 probability of experiencing a fire. What is the that a building catches fire and is completely destroyed by fire? Hint: apply multiplication rule. BUAD 310 - Kam Hamidieh

  29. Next Time • Chapter 8: More probability! • Start Chapter 9. BUAD 310 - Kam Hamidieh

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