Probability Concepts
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Probability Concepts

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classical, empirical, and subjective

classical, empirical, and subjective
approaches to probability.conditional probability and joint
probability.Bayes’ theorem.

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Probability Concepts Probability Concepts Document Transcript

  • 1 McGraw-Hill/Irwin Copyright © 2010 by The McGraw-Hill Companies, Inc. All rights reserved. A Survey of Probability Concepts Chapter 5 5-2 GOALS 1. Define probability. 2. Describe the classical, empirical, and subjective approaches to probability. 3. Explain the terms experiment, event, outcome, permutations, and combinations. 4. Define the terms conditional probability and joint probability. 5. Calculate probabilities using the rules of addition and rules of multiplication. 6. Apply a tree diagram to organize and compute probabilities. 7. Calculate a probability using Bayes’ theorem. 5-3 Probability, Experiment, Outcome, Event, Sample Spce: Defined PROBABILITY A value between zero and one, inclusive, describing the relative possibility (chance or likelihood) an event will occur.  An experiment is a process that leads to the occurrence of one and only one of several possible observations.  An outcome is the particular result of an experiment.  An event is the collection of one or more outcomes of an experiment.  Sample space is the collection of all possible events of an experiment 5-4 Mutually Exclusive Events and Collectively Exhaustive Events  Events are mutually exclusive if the occurrence of any one event means that none of the others can occur at the same time.  Events are collectively exhaustive if at least one of the events must occur when an experiment is conducted.  The sum of all collectively exhaustive and mutually exclusive events is 1.0 (or 100%)  Events are independent if the occurrence of one event does not affect the occurrence of another. collectively exhaustive and mutually exclusive events 5-5 Assigning Probabilities Three approaches to assigning probabilities Classical Empirical Subjective 5-6 Classical Probability Consider an experiment of rolling a six-sided die. What is the probability of the event “an even number of spots appear face up”? The possible outcomes are: There are three “favorable” outcomes (a two, a four, and a six) in the collection of six equally likely possible outcomes. Assumptions: Each outcome is equally likely Probability of success is based on the prior knowledge of the process Limitations Can only work under orderly situation Can only be applied if the events are equally likely
  • 2 5-7 Empirical Probability The empirical approach to probability is based on what is called the law of large numbers. The key to establishing probabilities empirically is that more observations will provide a more accurate estimate of the probability. EXAMPLE: On February 1, 2003, the Space Shuttle Columbia exploded. This was the second disaster in 113 space missions for NASA. On the basis of this information, what is the probability that a future mission is successfully completed? 98.0= 113 111 = = flightsofnumberTotal flightssuccessfulofNumber flightsuccessfulaofyProbabilit Limitations If the number of outcomes is not sufficiently larger It past occurrences are scarce 5-8 Law of Large Numbers Suppose we toss a fair coin. The result of each toss is either a head or a tail. If we toss the coin a great number of times, the probability of the outcome of heads will approach 0.5. The following table reports the results of an experiment of flipping a fair coin 1, 10, 50, 100, 500, 1,000 and 10,000 times and then computing the relative frequency of heads 5-9 Subjective Probability - Example  If there is little or no past experience or information on which to base a probability, it may be arrived at subjectively.  Due to differences in individual judgment, the probability of the same event might be different to different people  Illustrations of subjective probability are: 1. Estimating the likelihood the New England Patriots will play in the Super Bowl next year. 2. Estimating the likelihood you will be married before the age of 30. 3. Estimating the likelihood the U.S. budget deficit will be reduced by half in the next 10 years. 5-10 Summary of Types of Probability 5-11 Rules of Addition Rules of Addition  Special Rule of Addition - If two events A and B are mutually exclusive, the probability of one or the other event’s occurring equals the sum of their probabilities. P(A or B) = P(A) + P(B)  The General Rule of Addition - If A and B are two events that are not mutually exclusive, then P(A or B) is given by the following formula: P(A or B) = P(A) + P(B) - P(A and B) EXAMPLE: An automatic Shaw machine fills plastic bags with a mixture of beans, broccoli, and other vegetables. Most of the bags contain the correct weight, but because of the variation in the size of the beans and other vegetables, a package might be underweight or overweight.Acheck of 4,000 packages filled in the past month revealed: What is the probability that a particular package will be either underweight or overweight? P(A or C) = P(A) + P(C) = .025 + .075 = .10 Union= Or = U Intersection= and= ∩ 5-12 Special Addition Rule – Example:1 New England Commuter Airways recently supplied the following information on their commuter flights from Boston to New York: Arrival Frequency Early 100 Late 75 On Time 800 Canceled 25 Total 1000
  • 3 5-13 Special Addition Rule – Example:1 The probability that a flight is either early or late is: P(A or B) = P(A) + P(B) = 0.10 + 0.075 = 0.175 If A is the event that a flight arrives early, then P(A) = 100/1000 = 0.10 If B is the event that a flight arrives late, then P(B) = 75/1000 = 0.075 5-14 General Addition Rule – Example:1 What is the probability that a card chosen at random from a standard deck of cards will be either a king or a heart? P(A or B) = P(A) + P(B) - P(A ∩ B) = 4/52 + 13/52 - 1/52 = 16/52, or 0.3077 5-15 General Addition Rule –Example:2 In a sample of 500 students, 320 said they had a stereo, 175 said they had a TV, and 100 said they had both. 5 said they had neither. If a student is selected at random, what is the probability that the student has only a stereo or TV? What is the probability that the student has both a stereo and TV? P(ST or TV) = P(ST) + P(TV) - P(ST ∩ TV) = 320/500 + 175/500 – 100/500 = 0.79. P(ST and TV) = 100/500 = 0.20 5-16 The Complement Rule The complement rule is used to determine the probability of an event occurring by subtracting the probability of the event not occurring from 1. P(A) + P(~A) = 1 or P(A) = 1 - P(~A). EXAMPLE An automatic Shaw machine fills plastic bags with a mixture of beans, broccoli, and other vegetables. Most of the bags contain the correct weight, but because of the variation in the size of the beans and other vegetables, a package might be underweight or overweight. Use the complement rule to show the probability of a satisfactory bag is .900 P(B) = 1 - P(~B) = 1 – P(A or C) = 1 – [P(A) + P(C)] = 1 – [.025 + .075] = 1 - .10 = .90 5-17 If D is the event that a flight is canceled, then P(D) = 25/1000 = 0.025. Recall Airways example. Use the complement rule to find the probability of an early (A) or a late (B) flight If C is the event that a flight arrives on time, then P(C) = 800/1000 = 0.8 The Complement Rule-Example 5-18 P(C) =0.8 P(D) =0.025 ~(C or D) = (A or B) 0.175 P(A or B) = 1 - P(C or D) = 1 - [0.8 +0.025] = 0.175 The Complement Rule-Example
  • 4 5-19 The General Rule of Addition and Joint Probability The Venn Diagram shows the result of a survey of 200 tourists who visited Florida during the year. The survey revealed that 120 went to Disney World, 100 went to Busch Gardens and 60 visited both. What is the probability a selected person visited either Disney World or Busch Gardens? P(Disney or Busch) = P(Disney) + P(Busch) - P(both Disney and Busch) = 120/200 + 100/200 – 60/200 = .60 + .50 –.30= .80 JOINT PROBABILITY Aprobability that measures the likelihood two or more events will happen concurrently. 5-20 Special Rules of Multiplication • The special rule of multiplication requires that two events A and B are independent. • Two events A and B are independent if the occurrence of one has no effect on the probability of the occurrence of the other. • This rule is written: P(A and B) = P(A)P(B) EXAMPLE A survey by the American Automobile Association (AAA) revealed 60% of its members made airline reservations last year. Two members are selected at random. Since the number of AAA members is very large, we can assume that R1 and R2 are independent. What is the probability both made airline reservations last year? SOLUTION: The probability the first member made an airline reservation last year is .60, written as P(R1) = .60 The probability that the second member selected made a reservation is also .60, so P(R2) = .60. Since the number of AAA members is very large, you may assume that R1 and R2 are independent. P(R1 and R2) = P(R1)P(R2) = (.60)(.60) = .36 5-21 The health department routinely conducts two independent inspection of each restaurant, with the restaurant passing only if both inspectors pass it. Inspector A is very experienced and hence, passes only 2% of restaurants that actually do have health code violations. Inspector B is less experienced and passes 7% of restaurants with violations. What is the probability that (a) Inspector A passes a restaurant, given that inspector B has found a violation? (b)Inspector B passes a restaurant with a violation, given that inspector A passes it? (c)The health department passes a restaurant with a violation? Special Multiplication Rule (Levin & Rubin) 5-22 a) P(A/B) = P(A) = 0.02 b) P(B/A) = P(B) = 0.07 c) P(A,B) = P(A)×P(B) =(0.02)×(0.07)=0.0014 Special Multiplication Rule (Levin & Rubin) SOLUTION: 5-23 Mr. Bill Borde has just launched a publicity campaign for a new restaurant in a town. Bill has just installed four billboards on a highway outside of town, and he knows from experience the probabilities that each will be noticed by a randomly chosen motorist. The probability of the first billboard’s being noticed by a motorist is 0.75. The probability of the second’s being noticed is 0.82, the third has a probability of 0.87 of being noticed, and the probability of the fourth sign’s being noticed is 0.9. Assuming that the event that a motorist notices any particular billboard is independent of whether or not he notices the others, what is the probability that (a) All four billboards will be noticed by a randomly chosen motorist? (b) The first and fourth, but not the second & third billboards will be noticed? (c) Exactly one of the billboards will be noticed? (d) None of the billboards will be noticed? (e) The third and fourth billboards won’t be noticed? Special Multiplication Rule(Levin & Rubin) 5-24 Special Multiplication Rule(Levin & Rubin) Given us, P(A)=0.75, P(B)=0.82, P(C)=0.87 and P(D)=0.90 P(~A)=0.25, P(~B)=0.18, P(~C)=0.13 and P(~D)=0.10 (a) P(A,B,C,D)=P(A).P(B).P(C).P(D)= (.75)(.82)(.87)(.90)=0.4815 (b) P(A, ~B, ~C,D)=P(A).P(~B).P(~C).P(D) = (.75)(.18)(.13)(.90)=0.015795 (c) P(A,~B,~C,~D) +P(~A,B,~C,~D) +P(~A,~B,C,~D) +P(~A,~B,~C,D) =(.75)(.18)(.13)(.10) + .25)(.82)(.13)(.10)+(.25)(.18)(.87)(.10) + (.25)(.18)(.13)(.90) =0.001755+0.002665+0.003915+0.005265=0.0136 (d) P(~A, ~B,~C, ~D)= P(~A).P(~B).P(~C).P(~D) =(0.25)(0.18)(0.13)(0.10)= 0.000585 (e) P(~C, ~D)= P(~C).P(~D)= (0.13)(0.10)=0.013 SOLUTION:
  • 5 5-25 Suppose you are tossing a coin which is not fair. It has P(H)=0.8 and P(T)=0.2. Find the probability that 3 heads will appear in 3 successive tosses? P(H1, H2, H3,) = P(H1) × P(H2) × P(H3) = (0.80) ×(0.80) ×(0.80) = 0.512 Special Multiplication Rule (Levin & Rubin) SOLUTION: 5-26 Suppose you are tossing a coin which is fair. It has P(H)=0.5 and P(T)=0.5. What is the probability of (a) at least two heads of a fair coin on 3 tosses? and (b) one tail on 3 tosses? (a) Outcomes satisfying conditions are: (H1, H2, H3),(H1,H2,T3), P(H1,T2,H3), P(T1,H2,T3) The probability of each outcome is =(0.5)×(0.5)×(0.5)= 0.125 The required probability will be= 4× 0.125=0.50 Special Multiplication Rule (Levin & Rubin) SOLUTION: 5-27 Special Multiplication Rule (Levin & Rubin) b) Possible outcomes of 7 events except for the event (H1,H2, H3) Thus, the probability is = 7x0.125 = 0.875 Or The probability is= 1-0.125 = 0.875 5-28 Toss-1 Toss-2 Toss-3 Outcome Probability Outcomes Probability Outcomes Probability H1,H2,H3 0.125 H1,H2,T3 0.125 H1,H2 0.25 H1,T2,H3 0.125 H1 0.5 H1,T2 0.25 H1,T2,T3 0.125 T1 0.5 T1,H2 0.25 T1,H2,H3 0.125 T1,H2 0.25 T1,H2,T3 0.125 T1,T2,H3 0.125 T1,T2,T3 0.125 Special Multiplication Rule (Levin & Rubin) 5-29 Chris owns two stocks, IBM and GE. The probability that IBM stock will increase in value next year is 0.5 and the probability that GE stock will increase in value next year is 0.7. Assume the two stocks are independent. What is the probability that both stocks will increase in value next year? P(IBM and GE) = P(IBM) × P(GE) =(0.5)×(0.7) =0.35 Special Multiplication Rule 5-30 Special Multiplication Rule What is the probability that at least one of these stocks increases in value in the next year? This means that either can increase or both Outcomes satisfying conditions are: (IBM and ~GE), (~IBM and GE) and (IBM and GE) The probabilities of these outcomes are: =(0.5)×(1-0.7)+ (1-0.5)×(0.7)+(0.5)×(0.7) =0.15 + 0.35 + 0.35 =0.85 SOLUTION:
  • 6 5-31 General Rules of Multiplication The general rule of multiplication is used to find the joint probability that two events are not independent and they will occur under conditions of statistical dependence. EXAMPLE A golfer has 12 golf shirts in his closet. Suppose 9 of these shirts are white and the others blue. He gets dressed in the dark, so he just grabs a shirt and puts it on. He plays golf two days in a row and does not do laundry. What is the likelihood both shirts selected are white? The event that the first shirt selected is white is W1. The probability is P(W1) = 9/12 The event that the second shirt (W2 )selected is also white. The conditional probability that the second shirt selected is white, given that the first shirt selected is also white, is P(W2 | W1) = 8/11. To determine the probability of 2 white shirts being selected we use formula: P(AB) = P(A) P(B|A) P(W1 and W2) = P(W1)P(W2 |W1) = (9/12)(8/11) = 0.55 )A(P )BandA(P =)A/B(P=obabilityPrlConditiona 5-32 At a soup kitchen, a social worker gathers the following data. Of those visiting the kitchen, 59% are men, 32% are alcoholics, and 21% are male alcoholics. What is the probability that a random male visitor to the kitchen is an alcoholic? Given us: P(M) = 0.59, P(A) = 0.32 and P(M and A) = 0.21 General Multiplication Rule (Levin and Rubin) 356.0= 0.59 .210 = )M(P )AandM(P = P(M) M)andP(A =P(A/M) SOLUTION: 5-33 If a hurricane forms in the eastern half of the Gulf of Mexico, there is a 76% chance that it will strike the western coast of Florida. From the data gathered over the past 50 years, it has been determined that the probability of a hurricane’s occurring in this area in a given year is 0.85. (a) What is the probability that a hurricane will occur in the Eastern Gulf of Mexico and strike Florida this year? (b) If a hurricane in the eastern Gulf of Mexico is seeded (induced to rain by addition of chemicals from aircraft), its probability of striking Florida’s West Coast is reduced by one-fourth. If it is decided to seed any hurricane in the eastern gulf, what is the new value for the probability in part (a)? General Multiplication Rule (Levin and Rubin) 5-34 Given us, P(S/H)= 0.76 and P(H) = 0.85 We have to find out P(H and S)=? P(H and S) = P(S and H) = P(S/H)×P(H) =0.76×0.85=0.646 If seeded chemicals, the revised probability is: P(H and S) = P(S and H) = P(S/H)×P(H) =0.57×0.85=0.4845 General Multiplication Rule(Levin and Rubin) SOLUTION: 5-35 Contingency Tables A CONTINGENCY TABLE is a table used to classify sample observations according to two or more identifiable characteristic EXAMPLE: A sample of executives were surveyed about their loyalty to their company. One of the questions was, “If you were given an offer by another company equal to or slightly better than your present position, would you remain with the company or take the other position?” The responses of the 200 executives in the survey were cross- classified with their length of service with the company. What is the probability of randomly selecting an executive who is loyal to the company (would remain) and who has more than 10 years of service? Event A1 happens if a randomly selected executive will remain with the company despite an equal or slightly better offer from another company. Since there are 120 executives out of the 200 in the survey who would remain with the company P(A1) = 120/200, or .60. Event B4 happens if a randomly selected executive has more than 10 years of service with the company. Thus, P(B4| A1) is the conditional probability that an executive with more than 10 years of service would remain with the company. Of the 120 executives who would remain 75 have more than 10 years of service, so P(B4| A1) = 75/120. 5-36 The Dean of the School of Business at Owens University collected the following information about undergraduate students in her college: Major Male Female Total Accounting 170 110 280 Finance 120 100 220 Marketing 160 70 230 Management 150 120 270 Total 600 400 1000 Contingency Tables
  • 7 5-37 P(A|F) = P(A and F)/P(F) = [110/1000]/[400/1000] = 0.275 If a student is selected at random, what is the probability that the student is a female (F) accounting major (A)? P(F and A) = 110/1000 = 0.11 Given that the student is a female, what is the probability that she is an accounting major? Contingency Tables SOLUTION: 5-38 Tree Diagrams A tree diagram is useful for portraying conditional and joint probabilities. It is particularly useful for analyzing business decisions involving several stages. A tree diagram is a graph that is helpful in organizing calculations that involve several stages. Each segment in the tree is one stage of the problem. The branches of a tree diagram are weighted by probabilities. 5-39 Bayes’ Theorem A revised probability or posterior probability which is based on additional information The basic formula for conditional probability under statistical dependence: P(B/A) = P(B and A)/P(A) is called Bayes’ Theorem 5-40 Bayes’ Theorem Example-1 We have equal numbers of two types of deformed (biased or weighted) dice in a bowl. On half of them, ace (or one dot) comes up 40% of the time; therefore, P(ace) = 0.40. On the other half, ace comes up 70% of the time; P(ace) = 0.7. Let us call the former type-1 and the latter type-2. One die is drawn, rolled once, and comes up ace. (a) What is the probability that it is type-1 or type-2 dice? (b) Assume that the same die is rolled a second time and again comes up ace. What is the further revised probability that the die is type-1 or type-2? 5-41 Bayes’ Theorem Example-1 Elementary Event (E) P(E) P(Ace/E) P(Ace and E) Type-I 0.5 0.4 0.4x0.5=0.20 Type-II 0.5 0.7 0.7x0.5=0.35 P(E)=1.00 P(Ace)=0.55 364.0=55.0/20.0= P(Ace) Ace)andP(T =/Ace)P(T 1 1 636.0=55.0/35.0= P(Ace) Ace)andP(T =/Ace)P(T 2 2 5-42 Bayes’ Theorem Example-1 246.0=325.0/080.0= P(2Aces) 2Aces)andP(T =/2Aces)P(T 1 1 Event (E) P(E) P(Ace/E) P(2Aces/E) P(2 Aces, E) Type-I 0.5 0.4 0.16 0.16x0.5=0.080 Type-II 0.5 0.7 0.49 0.47x0.5=0.245 P(E)=1.00 P(2Aces)=0.325 754.0=325.0/245.0= P(2Aces) 2Aces)andP(T =/2Aces)P(T 1 2
  • 8 5-43 Bayes’ Theorem Example-2 EconOcon is planning its company picnic. The only thing that will cancel the picnic is a thunderstorm. The Weather Service has predicted dry conditions with probability 0.2, moist conditions with probability 0.45, and wet conditions with probability 0.35. If the probability of a thunderstorm given dry conditions is 0.3, given moist conditions is 0.6, and given wet conditions is 0.8, what is the probability of a thunderstorm? If we know the picnic was indeed canceled, what is the probability moist conditions were in effect? 5-44 Bayes’ Theorem Example-2 Event (E) P(E) P(Th/E) P(Th and E) Dry 0.20 0.3 0.3x0.2=0.06 Moist 0.45 0.6 0.6x0.45=0.27 Wet 0.35 0.8 0.8x0.35=0.28 P(T)=0.61 443.0= 61.0 27.0 = )Th(P )ThandM(P =)Th/M(P 5-45 Bayes’ Theorem Example-3 We A doctor has decided to prescribe two new drugs to 200 heart patients as follows: 50 get drug A, 50 get drug B, 100 get both. The 200 patients were chosen so that each had an 80% chance of having a heart attack if given neither drug. Drug A reduces the probability of a heart attack by 35%, drug B reduces the probability by 20%, and the two drugs, when taken together, work independently. If a randomly selected patient in the program has a heart attack, what is the probability that the patient was given both drugs? 5-46 Bayes’ Theorem Example-3 Event P(E) P (H|E) P (H and E) P (E|H) Drug-A 0.25 (0.8) (0.65) = 0.520 0.130 0.130/0.498 = 0.2610 Drug-B 0.25 (0.8) (0.80) = 0.640 0.160 0.160/0.498 = 0.3213 Both 0.50 (0.8)(0.65)(0.80) = 0.416 0.208 0.208/0.498 = 0.4177 P (H) = 0.498 Thus, P(A and B)| P(H) = 0.208/0.498= 0.417 5-47 Bayes’ Theorem Example-4 A physical therapist at Enormous State University knows that the football team will play 40% of its game on artificial turf this season. He also Knows that a football player’s chances of incurring a knee injury are 50% higher if he is playing on artificial turf instead of grass. If a player’s probability of knee injury on artificial turf is 0.42 what is the probability that (a) A randomly selected football player incurs a knee injury. (b) A randomly selected football player with a knee injury the incurred the injury playing on grass? 5-48 Bayes’ Theorem Example-4 Elementary Event (E) P(E) P(I/E) P(I and E) Artificial (A) 0.40 0.42 0.168 Grass (G) 0.60 0.28 0.168 1.00 P(I) = 0.336 (b) P(G/I) = P(G and I)/P(I) = (0.168)/0.336 = 0.50
  • 9 5-49 Bayes’ Theorem Example-5 A baseball team has been using an automatic pitching machine. If the machine is correctly set-up, it will pitch strikes 85 percent of the time. If incorrectly set-up, it will pitch strikes only 35 percent of the time. Past experience indicates that 75% of the set-ups of the machine are correctly done. (a) After the machine has been set-up at batting practice one day, it throws three strikes on the first three pitches. What is the revised probability that the set-up has been done correctly? (b) In the case of pitching machine, we might find the five pitches to be: strike, ball, strike, strike, strike. What is the revised probability that the set-up has been done correctly? 5-50 Bayes’ Theorem Example-5 Event (E) P(E) P(Strike/E) P(3 Strikes/E) P(3 Strikes and E) Correct 0.75 0.85 .85x.85x.85=.6141 .6141x.75=.4606 Incorrect 0.25 0.35 .35x.35x.35=.0429 .0429x.25=.0107 P(E)=1.0 P(3 Strikes)=.4713 9773.= 4713. 4606. = )Strikes3(P )Strikes3andCorrect(P =)strikes3/Correct(P 0227.= 4713. 0107. = )Strikes3(P )Strikes3andIncorrect(P =)strikes3/Incorrect(P 5-51 Bayes’ Theorem Example-5 Event (E) P(E) P(Strike/E) P(SBSSS/E) P(SBSSS,E) Correct 0.75 0.85 .85x.15x.85x.85x.85=.07830 .07830x.75=.05873 Incorrect 0.25 0.35 .35x.65x.35x.35x.35=.00975 .00975x.25=.00244 P(E)=1.0 P(SBSSS)=.06117 9601.= 06117. 05873. = )SBSSS(P )SBSSSandCorrect(P =)SBSSS/Correct(P 0399.= 06117. 00244. = )SBSSS(P )SBSSSandIncorrect(P =)SBSSS/Incorrect(P 5-52 Permutation and Combination A permutation is any arrangement of r objects selected from n possible objects. The order of arrangement is important in permutations. EXAMPLE Suppose that in addition to selecting the group, he must also rank each of the players in that starting lineup according to their ability. A combination is the number of ways to choose r objects from a group of n objects without regard to order. EXAMPLE There are 12 players on the Carolina Forest High School basketball team. Coach Thompson must pick five players among the twelve on the team to comprise the starting lineup. How many different groups are possible? 792 )!512(!5 !12 512   C040,95 )!512( !12 512   P