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Experiments, Sample Spaces, and Events Definition of Probability Rules of Probability

7. Experiments, Sample Spaces, and Events Definition of Probability Rules of Probability Use of Counting Techniques in Probability Conditional Probability and Independent Events Bayes’ Theorem. Probability. 7.1. Experiments, Sample Spaces, and Events. Terminology. Experiment

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Experiments, Sample Spaces, and Events Definition of Probability Rules of Probability

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  1. 7 • Experiments, Sample Spaces, and Events • Definition of Probability • Rules of Probability • Use of Counting Techniques in Probability • Conditional Probability and Independent Events • Bayes’ Theorem Probability

  2. 7.1 Experiments, Sample Spaces, and Events

  3. Terminology Experiment • An experiment is an activity with observable results. • The results of an experiment are called outcomes of the experiment.

  4. Examples • Tossing a coin and observing whether it falls heads or tails • Rolling a die and observing which of the numbers 1, 2, 3, 4, 5, or 6 shows up • Testing a spark plug from a batch of 100 spark plugs and observing whether or not it is defective

  5. Terminology Sample Point, Sample Space, and Event • Sample point: An outcome of an experiment • Sample space: The set consisting of all possible sample points of an experiment • Event: A subset of a sample space of an experiment

  6. Example • Describe the sample space associated with the experiment of tossing a coin and observing whether it falls heads or tails. • What are the events of this experiment? Solution • The two outcomes are heads and tails, and the required sample space is given by S = {H, T} where H denotes the outcomeheads and T denotes the outcometails. • The events of the experiment, the subsets of S, are Ø, {H}, {T}, S Example 1, page 354

  7. Terminology Union of Two Events • The union of two events E and F is the event E F. • Thus, the event E  F contains the set of outcomes of Eand/orF.

  8. Terminology Intersection of Two Events • The intersection of two events E and F is the event E F. • Thus, the event E  F contains the set of outcomes common to EandF.

  9. Terminology Complement of an Event • The complement of an event E is the event Ec. • Thus, the event Ec is the set containingallthe outcomes in the sample space Sthat arenot in E.

  10. Example • Consider the experiment of rolling a die and observing the number that falls uppermost. • Let S = {1, 2, 3, 4, 5, 6} denote the sample space of the experiment and E = {2, 4, 6} and F = {1, 3} be events of this experiment. • Compute E F. Interpret your results. Solution • E F = {1, 2, 3, 4, 6} and is the event that the outcome of the experiment is a 1, a 2, a 3, a 4, or a 6. Example 2, page 355

  11. Example • Consider the experiment of rolling a die and observing the number that falls uppermost. • Let S = {1, 2, 3, 4, 5, 6} denote the sample space of the experiment and E = {2, 4, 6} and F = {1, 3} be events of this experiment. • Compute E F. Interpret your results. Solution • E F = Ø since there are no elements in common between the two sets E and F. Example 2, page 355

  12. Example • Consider the experiment of rolling a die and observing the number that falls uppermost. • Let S = {1, 2, 3, 4, 5, 6} denote the sample space of the experiment and E = {2, 4, 6} and F = {1, 3} be events of this experiment. • Compute Fc. Interpret your results. Solution • Fc = {2, 4, 5, 6} is precisely the event that the event Fdoes not occur. Example 2, page 355

  13. Terminology Mutually Exclusive Events • E and F are mutually exclusive if E F = Ø

  14. Example • An experiment consists of tossing a coin three times and observing the resulting sequence of heads and tails. • Describe the sample spaceS of the experiment. • Determine the eventE that exactly two heads appear. • Determine the eventF that at least one head appears. Example 3, page 356

  15. Example Solution • As the tree diagram demonstrates, the sample space is S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} Example 2, page 355

  16. Example Solution • Scanning the sample space obtained S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} we see that the outcomes in which exactly two heads appear are given by the event E = {HHT, HTH, THH} • We can also see that the outcomes in which at least one head appears are given by the event F = {HHH, HHT, HTH, HTT, THH, THT, TTH} Example 2, page 355

  17. Applied Example: Movie Attendance • The manager of a local cinema records the number of patrons attending a first-run movie screening. • The theatre has a seating capacity of 500. • Determine an appropriate sample space for this experiment. • Describe the eventE that fewer than50 people attend the screening. • Describe the eventF that the theatre is more than half full at the screening. Solution • The number of patrons at the screening could run from 0 to 500. Therefore, a sample space for this experiment is S = {0, 1, 2, 3, …, 500} Applied Example 5, page 357

  18. Applied Example: Movie Attendance • The manager of a local cinema records the number of patrons attending a first-run movie screening. • The theatre has a seating capacity of 500. • Determine an appropriate sample space for this experiment. • Describe the eventE that fewer than50 people attend the screening. • Describe the eventF that the theatre is more than half full at the screening. Solution • The eventE that fewer than50 people attend the screening can be described as E = {0, 1, 2, 3, …, 49} Applied Example 5, page 357

  19. Applied Example: Movie Attendance • The manager of a local cinema records the number of patrons attending a first-run movie screening. • The theatre has a seating capacity of 500. • Determine an appropriate sample space for this experiment. • Describe the eventE that fewer than50 people attend the screening. • Describe the eventF that the theatre is more than half full at the screening. Solution • The eventF that the theatre is more than half full at the screening can be described as F = {251, 252, …, 500} Applied Example 5, page 357

  20. 7.2 Definition of Probability

  21. Probability of an Event in a Uniform Sample Space • If S = {s1, s2, … , sn} is the sample space for an experiment in which the outcomes are equally likely, then we assign the probabilities to each of the outcomes s1,s2,… , sn.

  22. Example • A fair die is rolled, and the number that falls uppermost is observed. • Determine the probability distribution for the experiment. Solution • The sample space for the experiment is S = {1, 2, 3, 4, 5, 6} and the simple events are accordingly given by the sets {1}, {2}, {3}, {4}, {5}, and {6} • Since the die is assumed to be fair, the six outcomes are equally likely. • We therefore assign a probability of 1/6 to each of the simple events. Example 1, page 364

  23. Example • A fair die is thrown, and the number that falls uppermost is observed. • Determine the probability distribution for the experiment. Solution • Thus, the probability distribution of these simple events is: Example 1, page 364

  24. Finding the Probability of Event E • Determine a sample spaceS associated with the experiment. • Assign probabilities to the simple events of S. • If E = {s1, s2, … , sn} where {s1}, {s2}, {s3}, … , {sn} are simple events, then P(E) = P(s1) + P(s2) + P(s3) + ··· + P(sn) If E is the empty set, Ø, then P(E) = 0.

  25. Applied Example: Rolling Dice • A pair of fair dice is rolled. • Calculate the probability that the two dice show the same number. • Calculate the probability that the sum of the numbers of the two dice is 6. Applied Example 3, page 365

  26. Applied Example: Rolling Dice Solution • The sample spaceS of the experiment has 36outcomes S = {(1, 1), (1, 2), … , (6, 5), (6, 6)} • Both dice are fair, making each of the 36 outcomes equally likely, so we assign the probability of 1/36 to each simple event. • The event that the two dice show the same number is E = {(1, 1), (2, 2) , (3, 3), (4, 4), (5, 5), (6, 6)} • Therefore, the probability that the two dice show the same number is given by Six terms Applied Example 3, page 365

  27. Applied Example: Rolling Dice Solution • The event that the sum of the numbers of the two dice is 6 is given by E6 = {(1, 5), (2, 4) , (3, 3), (4, 2), (5, 1)} • Therefore, the probability that the sum of the numbers on the two dice is 6 is given by Applied Example 3, page 365

  28. 7.3 Property 1.P(E)  0 for every E. Property 2.P(S) = 1. Property 3. If E and F are mutually exclusive (E  F = Ø), then P(E  F) = P(E) + P(F) Property 4.If E and F are any two events of an experiment, then P(E  F) = P(E) + P(F)– P(E F) Property 5.If E is an event of an experiment and Ec denotes the complement of E, then P(Ec) = 1 – P(E) Rules of Probability

  29. Properties of the Probability Function Property 1.P(E)  0 for every E. Property 2.P(S) = 1. Property 3. If E and F are mutually exclusive (E  F = Ø), then P(E  F) = P(E) + P(F)

  30. Applied Example: SAT Verbal Scores • The superintendent of a metropolitan school district has estimated the probabilities associated with the SAT verbal scores of students from that district. The results are shown in the table below. • If a student is selected at random, find the probability that his or her SAT verbal score will be • More than 400. • Less than or equal to 500. • Greater than 400 but less than or equal to 600. Applied Example 1, page 372

  31. Applied Example: SAT Verbal Scores Solution • Let A, B, C, D, E, and F denote, respectively, the mutually exclusiveevents listed in the table below. • The probability that the student’s score will be more than400 is given by P(D  C  B  A) = P(D) + P(C) + P(B) + P(A) = .23 + .19 + .07 + .01 = .50 Applied Example 1, page 372

  32. Applied Example: SAT Verbal Scores Solution • Let A, B, C, D, E, and F denote, respectively, the mutually exclusiveevents listed in the table below. • The probability that the student’s score will be less than or equal to500 is given by P(D  E  F) = P(D) + P(E) + P(F) = .23 + .31 + .19 = .73 Applied Example 1, page 372

  33. Applied Example: SAT Verbal Scores Solution • Let A, B, C, D, E, and F denote, respectively, the mutually exclusiveevents listed in the table below. • The probability that the student’s score will be greater than400 but less than or equal to600 is given by P(C  D) = P(C) + P(D) = .19 + .23 = .42 Applied Example 1, page 372

  34. Properties of the Probability Function Addition Rule Property 4.If E and F are any two events of an experiment, then P(E  F) = P(E) + P(F)– P(E F)

  35. Example • A card is drawn from a shuffled deck of 52playing cards. • What is the probability that it is an ace or a spade? Solution • Let E denote the event that the card drawn is an ace, and let F denote the event that the card drawn is a spade. • Then, • Note that E and F are not mutually exclusive events: • E F is the event that the card drawn is an ace of spades. • Consequently, Example 2, page 373

  36. Example • A card is drawn from a shuffled deck of 52playing cards. • What is the probability that it is an ace or a spade? Solution • The event that a card drawn is an ace or a spade is E  F, with probability given by Example 2, page 373

  37. Applied Example: Quality Control • The quality-control department of Vista Vision, manufacturer of the Pulsar plasma TV, has determined from records obtained from the company’s service centers that 3% of the sets sold experience video problems, 1% experience audio problems, and 0.1% experience both video and audio problems before the expiration of the warranty. • Find the probability that a plasma TV purchased by a consumer will experience video or audio problems before the warranty expires. Applied Example 3, page 373

  38. Applied Example: Quality Control Solution • Let E denote the event that a plasma TV purchased will experience video problems within the warranty period, and let F denote the event that a plasma TV purchased will experience audio problems within the warranty period. • Then, P(E) = .03 P(F) = .01 P(E F) = .001 • The event that a plasma TV purchased will experience video or audio problems before the warranty expires is E  F, and the probability of this event is given by Applied Example 3, page 373

  39. Properties of the Probability Function Rule of Complements Property 5.If E is an event of an experiment and Ec denotes the complement of E, then P(Ec) = 1 – P(E)

  40. Applied Example: Warranties • What is the probability that a Pulsar plasma TV (from the last example) bought by a consumer will not experience video or audio problems before the warranty expires? Solution • Let E denote the event that a plasma TV bought by a consumer will experience video or audio problems before the warranty expires. • Then, the event that the plasma TV will not experience either problem before the warranty expires is given by Ec, with probability Applied Example 4, page 375

  41. 7.4 Use of Counting Techniques in Probability

  42. Computing the Probability of an Event in a Uniform Sample Space • Let S be a uniform sample space and let E be any event. Then,

  43. Example • An unbiased coin is tossed six times. • Find the probability that the coin will land heads • Exactlythree times. • At mostthree times. • On the first and the last toss. Example 1, page 381

  44. Example Solution • Each outcome of the experiment may be represented as a sequence of heads and tails. • Using the generalized multiplication principle, we see that the number of outcomes of this experiment is 26, or 64. • Let E denote the event that the coin lands headsexactlythree times. • Since there are C(6, 3) ways this can occur, we see that the required probability is Example 1, page 381

  45. Example Solution • Let F denote the event that the coin lands headsat mostthree times. • Then, n(F) is given by the sum of the number of ways the coin lands headszero times (no heads), exactly once, exactly twice, and exactly three times. That is, • Thus, the required probability is Example 1, page 381

  46. Example Solution • Let H denote the event that the coin lands heads on the first and the last toss. • Then, • Therefore, the required probability is Example 1, page 381

  47. 7.5 Conditional Probability and Independent Events

  48. Conditional Probability of an Event • If A and B are events in an experiment and P(A)  0, then the conditional probability that the event B will occur given that the event A has already occurred is

  49. Example • A pair of fair dice is rolled. What is the probability that the sum of the numbers falling uppermost is 7 if it is known that exactlyone of the numbers is a 5? Solution • Let A denote the event that exactlyone of the numbers is a 5 and let B denote the event that the sum of the numbers falling uppermost is 7. Thus, so Example 2, page 390

  50. Example • A pair of fair dice is rolled. What is the probability that the sum of the numbers falling uppermost is 7 if it is known that exactlyone of the numbers is a 5? Solution • Since the dice are fair, each outcome of the experiment is equally likely; therefore, • Thus, the probability that the sum of the numbers falling uppermost is 7 given that exactlyone of the numbers is a 5 is given by Example 2, page 390

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