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Chapter 5 Probability Concepts

KVANLI PAVUR KEELING. Chapter 5 Probability Concepts. Chapter Objectives. At the completion of this chapter, you should be able to answer the following questions: • What is meant by the term "probability"? • How would each of these techniques be used

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Chapter 5 Probability Concepts

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  1. KVANLI PAVUR KEELING Chapter 5Probability Concepts

  2. Chapter Objectives • At the completion of this chapter, you should be able to answer the following questions: • What is meant by the term "probability"? • How would each of these techniques be used to generate a probability: ∙ classical approach ∙ relative frequency approach ∙ subjective probability approach

  3. Chapter Objectives - Continued • At the completion of this chapter, you should be able to answer the following questions: • What is meant by the term joint probability? • How are multiple events handled? • How do you compute probabilities using the additive rule? The multiplicative rule?

  4. Let’s Make a Deal Curtain #2 Curtain #3 Curtain #1 Behind two curtains sits a goat and behind one curtain sits a new Corvette Remember: The host of the show knows what is behind each curtain

  5. Three Questions • Question #1: Should you trade (does it make any difference)? • Question #2: If you’re a curtain trader, what is the probability of you getting the Corvette? • Question #3: If you’re not a curtain trader (you’ll stay with your original pick), what is this probability?

  6. Question #3 • For this situation, all the curtains are closed and you’ll need to pick the curtain with the Corvette behind it • Since one curtain out of three has the Corvette, your chances of getting the Corvette are 1 out of 3 • We say that this probability is 1/3

  7. Questions #1 and #2 • What happens if, on your initial pick (all the curtains are closed), you pick a goat curtain? • What must be behind the remaining curtain? The curtain you picked – has a goat behind it The remaining curtain

  8. Questions #1 and #2 The curtain you picked – has a goat behind it Conclusion: If you’re a curtain trader (you will trade curtains when given the offer), all you need to do on the initial pick (with all curtains closed) is pick a goat curtain.

  9. Questions #1 and #2 • What are the chances of you picking a goat curtain on your initial pick? • Since two of the three curtains are goat curtains, your chances of picking a goat curtain are 2 out of 3 • We say that this probability is 2/3 • A curtain trader is twice as likely to win the Corvette as a non-curtain trader

  10. Summary • This was the solution proposed by Marilyn Vos Savant • And she’s correct! • Many readers wrote to her chewing her out for offering such a lame answer • A sample of these comments is shown on the next two slides

  11. Reader Comments • I have been a faithful reader of your column, and I have not, until now, had any reason to doubt you. However, in this matter (for which I do have expertise), your answer is clearly at odds with the truth. James Rauff, Ph.D.MillikinUniversity • May I suggest that you obtain and refer to a standard textbook on probability before you try to answer a question of this type again? Charles Reid, Ph.D.University of Florida • You are the goat! Glenn Calkins Western State College

  12. More Comments • You are utterly incorrect about the game show question, and I hope this controversy will call some public attention to the serious national crisis in mathematical education. If you can admit your error, you will have contributed constructively towards the solution of a deplorable situation. How many irate mathematicians are needed to get you to change your mind? E. Ray Bobo, Ph.D.Georgetown University • I am in shock that after being corrected by at least three mathematicians, you still do not see your mistake. Kent FordDickinson State University

  13. Coming up with a Probability • Make it up (called a subjective probability) P(Dow-Jones goes up more than 1000 points next month) is .1

  14. Classical Definition of a Probability 2. Classical definition of a probability Example: Draw a single card from a deck containing 52 cards, 13 cards of each suit P(draw a six) is 4/52 = .077 • Have n equally likely outcomes • m of these satisfy the condition • The probability is m/n 8% chance of drawing a six 52 4

  15. Relative Frequency Definition 3. The relative frequency definition of a probability is based on past observations Example: Your last 200 customers • P(next customer is a female) is 140 / 200 = .7 • Based on these observations, there is a 70% chance that the next customer is a female 140 females 60 males

  16. Using a Contingency Table • A group of 400 undergraduate students is classified • according to their gender and year in school • One student is selected at random

  17. Marginal Probabilities • The chances that the person selected is a female is: P(F) = 150/400 = .375 • The chances that the person selected is a junior is: P(JR) = 120/400 = .3 There is a 30% chance of this event

  18. Joint Probabilities There is a 20% chance of selecting a male and a junior • What is P(M and JR)? • P(M and JR) = 80/400 = .2 • What is P(SO and F)? • P(SO and F) = 60/400 = .15

  19. More on Joint Probabilities • What is P(FR and JR)? • This probability is zero since a student cannot be • both a freshman and a junior • The two events, FR and JR, are said to be mutually • exclusive – they cannot both occur

  20. Finding an OR Probability • What is P(M or JR)? • P(M or JR) is (250 + 40)/400 = 290/400 = .725

  21. Finding an OR Probability • In general, when finding “or” probabilities, the • boxes that qualify have a “t” shape • Sometimes, the “t” is upside down • Sometimes, it doesn’t quite look like a “t”

  22. OR Probabilities for Mutually Exclusive Events • P(FR or SO) is (80 + 140)/400 = 220/400 = .55 • (80 + 140)/400 = 80/400 + 140/400 = P(FR) + P(SO) • Rule: When two events are mutually exclusive, • the two probabilities can be added

  23. Complement of an Event • What is the probability of not selecting a junior? • This is written P(JR) and is equal to 1 – P(JR) • P(JR) = 1 – 120/400 = .7 • JR is called the complement of JR

  24. Conditional Probability • Suppose you know (someone tells you) that the • person selected is a male • What are the chances this person is a freshman? • This is called a conditional probability and is • written P(FR|M)

  25. Conditional Probability • There are 250 males and we are told that the person • selected is one of these • 50 of these 250 males are freshmen • P(FR|M) = 50/250 = .2 • 20% of the males are freshmen (but not the other • way around)

  26. Independent Events • Note that P(FR) is 80/400 = .2 • Here, P(FR|M) is the same as P(FR) • When the occurrence of one event has no effect • on the occurrence of another event, these events • are said to be independent

  27. Two Other Events • Are the events female (F) and Junior (JR) mutually • exclusive? • Can both events occur? Yes, there are 40 • people who are female and juniors. • These two events are not mutually exclusive

  28. Events F and JR • Are the events female (F) and Junior (JR) independent? • P(F|JR) is 40/120 = .33 since 40 of the juniors are female • P(F) is 150/400 = .375 since 150 of the students are female • Since these two probabilities are not the same, these two • events are not independent

  29. Events M and SR These must be the same • Are the events male (M) and senior (SR) independent? • P(M|SR) is 40/60 = .67 since 40 of the 60 seniors are male • P(M) is 250/400 = .625 since 250 of the 400 students are male • Since these two probabilities are not the same, these two • events are not independent

  30. Drawing a Card Let A = drawing an ace B = drawing a diamond Are these two events mutually exclusive? No, since both events can occur (an ace of diamonds)

  31. Drawing a Card A = drawing an ace and B = drawing a diamond Are these two events independent? P(A|B) = 1/13 since 1 of the 13 diamonds is an ace P(A) = 4/52 = 1/13 since 4 of the 52 cards are aces These two events are independent

  32. Another Example • A = The Dow-Jones goes up more than 50 points tomorrow • B = It rains in Dallas tomorrow • Are these events mutually exclusive? • No, because both events can occur

  33. Raining in Dallas • A = The Dow-Jones goes up more than 50 points tomorrow and B = It rains in Dallas tomorrow • Are these two events independent? • They are independent since P(A|B) is the same as P(A) • It’s not necessary to determine these two probabilities since raining in Dallas clearly has no effect on the Dow-Jones going up more than 50 points • And certainly, the Dow going up more than 50 points has no effect on whether it rains in Dallas

  34. The Oil Wells Example • A = Oil well #1 hits oil B = Oil well #2 hits oil • Are these events mutually exclusive? • No, since both events could occur • Are events A and B independent? • It depends

  35. The Oil Wells Example • It depends on the location of these wells • If wells #1 and #2 are right next to each other then well #1 hitting oil could very well have an effect on well #2 hitting oil (this probability just went up) • In this case, events A = well #1 hitting and B = well #2 hitting are not independent

  36. The Oil Wells Example • If well #1 is in Oklahoma and well #2 is in Alaska, then well #1 hitting oil would most likely have no effect on well #2 hitting oil • P(B|A) is the same as P(B) • In this case, events A = well #1 hitting and B = well #2 hitting are independent

  37. The Dreaded Word Problems • You’ll be given three (always three) pieces of information • You can solve these by setting up a contingency table with two rows and two columns that satisfies these three conditions

  38. Example 5.4 • This example is in the textbook • A certain community has a morning paper and an evening paper • The three pieces of information: • 20% of the people take the morning paper P(M) = .2 • 30% of the people take the evening paper P(E) = .3 • 10% of the people take both P(M and E) = .1 This is a joint probability

  39. Example 5.4 Table M M 30 E 10 20 E 10 60 70 Any value can go here 20 80 100

  40. Example 5.4 Solution Want to know: The probability a person takes the morning or evening paper: P(M or E) 2. What percentage of the morning subscribers take the evening paper? This is P(E|M) This is a conditional probability 1. P(M or E) is 40/100 = .4 So, 40% of the people take one paper or the other 2. P(E|M) is 10/20 = .5 So, 50% of the morning subscribers take the evening paper

  41. Example 5.6 • This example is also in the textbook • You are examining the banks in a certain northeastern state • The three pieces of information: • 5% of the banks will fail P(fail) = .05 • 90% of the banks are insured P(ins) = .90 • 3% of the insured banks will fail P(fail|ins) = .03 This is a conditional probability

  42. Example 5.6 Table ins ins 50 fail 27 23 fail 873 77 950 Remember: Any value can go here 900 100 1000

  43. Example 5.6 Solution Want to know: 1. The probability a bank will fail and is insured: P(fail and ins) 2. The probability a bank will fail or is insured: P(fail or ins) 3. What percentage of the failed banks are insured? 1. P(fail and ins) is 27/1000 = .027 2. P(fail or ins) is (900 + 23)/1000 = .923 This is a conditional probability 3. P(ins|fail) is 27/50 = .54 4. P(ins) is 900/1000 = .9 These two events are not independent

  44. Example 5.6 Revised • The three pieces of information are now: • 5% of the banks will fail P(fail) = .05 • 90% of the banks are insured P(ins) = .90 • The two events, fail and insured, are independent • As we will see shortly, this means that you can multiply the two probabilities to obtain the joint probability • P(fail and ins) is P(fail) · P(ins) = (.05)(.90) = .045

  45. Revised Example 5.6 Table ins ins 50 fail 45 5 fail 855 95 950 900 100 1000

  46. Two Rules • P(A and B) = P(A) · P(B) provided A and B are independent • This means that events A and B don’t affect each other • This also applies to more than two events • P(A and B and C) = P(A) · P(B) · P(C) provided these three events are independent

  47. Second Rule • P(A or B) = P(A) + P(B) provided A and B are mutually exclusive • This means that events A and B cannot both occur • This also applies to more than two events • P(A or B or C) = P(A) + P(B) + P(C) provided these three events are mutually exclusive

  48. Applying the Two Rules Components in Series The system There is a 2% chance of failure for each of the three components A B C The system fails if one or more components fail

  49. Applying the Two Rules Components in Series • P(system fails) = 1 – P(system doesn’t fail) • This is 1 – P(A works and B works and C works) • 1 – [P(A works) · P(B works) · P(C works)] • 1- [(.98) ·(.98)·(.98)] = .059 (roughly 6%)

  50. Applying the Two Rules Components in Parallel There is a 2% chance of failure for each of the three components The system A B C The system fails if all three components fail

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