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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. Probability Concepts • Three Helpful Concepts in Understanding Probability: • Experiment • Sample Space • Event

  5. Experiment • An activity for which the outcome is uncertain is an experiment. • Example 5.1.1: Examples of experiments • Flipping a coin • Rolling two dice • Taking an exam • Observing the number of arrivals at a drive-up window over a 5-minute period

  6. Sample Space • The list of all possible outcomes of an experiment is called the sample space. • Example 5.1.2: Example of sample space • Flipping a coin twice results in one of four possible outcomes. These possible outcomes are HH, HT, TH, TT. Therefore, sample space = {HH, HT, TH, TT}. • If there are n outcomes of an experiment, sample space lists all n outcomes.

  7. Event • An event consists of one or more possible outcomes of the experiment. • It is usually denoted by a capital letter. • Example 5.1.3: Examples of experiments and some corresponding events • Experiment: Rolling two dice; events: A = rolling a total of 7, B = rolling a total greater than 8, C = rolling two 4s. • Experiment: Taking an exam; events: A = pass, B = fail. • Experiment: Observing the number of arrivals at a drive-up window over a 5-minute period; events: A0 = no arrivals, A1 = seven arrivals, etc.

  8. Probability • A numerical measure of the chance OR likelihood that a particular event will occur. • The probability that event A will occur is written P(A). • The probability of any event ranges from 0 to 1, inclusive. • P(A) = 0 means event A will never occur. • P(A) = 1 means event A must occur.

  9. 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

  10. 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

  11. 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

  12. 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

  13. 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

  14. 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

  15. 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

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

  17. 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”

  18. 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

  19. 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

  20. 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)

  21. 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)

  22. 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

  23. 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

  24. 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

  25. 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

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

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

  28. 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

  29. 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

  30. 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

  31. 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

  32. 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

  33. 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

  34. 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

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

  36. 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

  37. 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

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

  39. 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

  40. 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

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

  42. 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

  43. 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

  44. 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

  45. 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%)

  46. 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

  47. Applying the Two Rules Components in Parallel • P(system fails) = P(A fails and B fails and C fails) • This is P(A fails) · P(B fails) · P(C fails) • = (.02)(.02)(.02) = .000008 (8 in a million chance) • This is called built-in redundancy

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