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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

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7.1 Experiments, Sample Spaces, and Events

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TerminologyExperiment An experiment is an activity with observable results. The results of an experiment are called outcomes of the experiment.

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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

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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

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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 outcome heads and T denotes the outcome tails. The events of the experiment, the subsets of S, are Ø, {H}, {T}, S Example 1, page 354

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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 E and/or F.

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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 E and F.

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Terminology Complement of an Event The complement of an event E is the event E c. Thus, the event E c is the set containing all the outcomes in the sample space S that are not in E.

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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

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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

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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 F c. Interpret your results. Solution F c = {2, 4, 5, 6} is precisely the event that the event F does not occur. Example 2, page 355

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Terminology Mutually Exclusive Events E and F are mutually exclusive if E F = Ø

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Example An experiment consists of tossing a coin three times and observing the resulting sequence of heads and tails. ✦ Describe the sample space S of the experiment. ✦ Determine the event E that exactly two heads appear. ✦ Determine the event F that at least one head appears. Example 3, page 356

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

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ExampleSolution 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

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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 event E that fewer than 50 people attend the screening. Describe the event F 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

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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 event E that fewer than 50 people attend the screening. Describe the event F that the theatre is more than half full at the screening. Solution The event E that fewer than 50 people attend the screening can be described as E = {0, 1, 2, 3, …, 49} Applied Example 5, page 357

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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 event E that fewer than 50 people attend the screening. Describe the event F that the theatre is more than half full at the screening. Solution The event F that the theatre is more than half full at the screening can be described as F = {251, 252, …, 500} Applied Example 5, page 357

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7.2 Definition of Probability

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Probability of an Event in a Uniform Sample Space If S = {s 1, s 2, …, s n } is the sample space for an experiment in which the outcomes are equally likely, then we assign the probabilities to each of the outcomes s 1, s 2, …, s n.

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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

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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: SimpleeventProbability {1}1/6 {2}1/6 {3}1/6 {4}1/6 {5}1/6 {6}1/6 Example 1, page 364

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Finding the Probability of Event E 1.Determine a sample space S associated with the experiment. 2.Assign probabilities to the simple events of S. 3.If E = {s 1, s 2, …, s n } where {s 1 }, {s 2 }, {s 3 }, …, {s n } are simple events, then P(E) = P(s 1 ) + P(s 2 ) + P(s 3 ) + ··· + P(s n ) If E is the empty set, Ø, then P(E) = 0.

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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

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Applied Example: Rolling Dice Solution The sample space S of the experiment has 36 outcomes 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

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Applied Example: Rolling Dice Solution The event that the sum of the numbers of the two dice is 6 is given by E 6 = {(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

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Rules of Probability 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) 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) P(E F) = P(E) + P(F) – P(E F) Property 5.If E is an event of an experiment and E c denotes the complement of E, then P(E c ) = 1 – P(E) P(E c ) = 1 – P(E)

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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) P(E F) = P(E) + P(F)

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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. Score, x Probability x > < x < x < x < x x Applied Example: SAT Verbal Scores Applied Example 1, page 372

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Applied Example: SAT Verbal Scores Solution Let A, B, C, D, E, and F denote, respectively, the mutually exclusive events listed in the table below. The probability that the student’s score will be more than 400 is given by P(D C B A)= P(D) + P(C) + P(B) + P(A) = =.50 Score, x Probability x > < x < x < x < x x AB C D E F Applied Example 1, page 372

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Applied Example: SAT Verbal Scores Solution Let A, B, C, D, E, and F denote, respectively, the mutually exclusive events listed in the table below. The probability that the student’s score will be less than or equal to 500 is given by P(D E F)= P(D) + P(E) + P(F) = =.73 Score, x Probability x > < x < x < x < x x AB C D E F Applied Example 1, page 372

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Applied Example: SAT Verbal Scores Solution Let A, B, C, D, E, and F denote, respectively, the mutually exclusive events listed in the table below. The probability that the student’s score will be greater than 400 but less than or equal to 600 is given by P(C D)= P(C) + P(D) = =.42 Score, x Probability x > < x < x < x < x x AB C D E F Applied Example 1, page 372

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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)

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Example A card is drawn from a shuffled deck of 52 playing 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

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Example A card is drawn from a shuffled deck of 52 playing 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

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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

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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 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

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

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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 E c, with probability Applied Example 4, page 375

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7.4 Use of Counting Techniques in Probability

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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,

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Example An unbiased coin is tossed six times. Find the probability that the coin will land heads ✦ Exactly three times. ✦ At most three times. ✦ On the first and the last toss. Example 1, page 381

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ExampleSolution 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 2 6, or 64. Let E denote the event that the coin lands heads exactly three times. Since there are C(6, 3) ways this can occur, we see that the required probability is Example 1, page 381

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ExampleSolution Let F denote the event that the coin lands heads at most three times. Then, n(F) is given by the sum of the number of ways the coin lands heads zero times (no heads), exactly once, exactly twice, and exactly three times. That is, Thus, the required probability is Example 1, page 381

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ExampleSolution 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

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7.5 Conditional Probability and Independent Events

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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

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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 exactly one of the numbers is a 5? Solution Let A denote the event that exactly one 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

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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 exactly one 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 exactly one of the numbers is a 5 is given by Example 2, page 390

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Applied Example: Color Blindness In a test conducted by the U.S. Army, it was found that of 1000 new recruits (600 men and 400 women), 50 of the men and 4 of the women were red-green color-blind. Given that a recruit selected at random from this group is red-green color-blind, what is the probability that the recruit is a male? Solution Let C denote the event that a randomly selected subject is red-green color-blind, and let M denote the event that the subject is a male recruit. Applied Example 3, page 390

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Applied Example: Color Blindness In a test conducted by the U.S. Army, it was found that of 1000 new recruits (600 men and 400 women), 50 of the men and 4 of the women were red-green color-blind. Given that a recruit selected at random from this group is red-green color-blind, what is the probability that the recruit is a male? Solution Since 54 out of 1000 subjects are color-blind, we see that Furthermore, since 50 of the subjects are colorblind and male, we see that Applied Example 3, page 390

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Applied Example: Color Blindness In a test conducted by the U.S. Army, it was found that of 1000 new recruits (600 men and 400 women), 50 of the men and 4 of the women were red-green color-blind. Given that a recruit selected at random from this group is red-green color-blind, what is the probability that the recruit is a male? Solution Therefore, the probability that a subject is male given that the subject is red-green color-blind is given by Applied Example 3, page 390

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Product Rule

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Example Two cards are drawn without replacement from a well- shuffled deck of 52 playing cards. What is the probability that the first card drawn is an ace and the second card drawn is a face card? Solution Let A denote the event that the first card drawn is an ace, and let F denote the event that the second card drawn is a face card. Then, Example 5, page 392

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Example Two cards are drawn without replacement from a well- shuffled deck of 52 playing cards. What is the probability that the first card drawn is an ace and the second card drawn is a face card? Solution After drawing the first card, there are 51 cards left in the deck, of which 12 are face cards. Therefore, the probability of drawing a face card given that the first card drawn was an ace is given by Example 5, page 392

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Example Two cards are drawn without replacement from a well- shuffled deck of 52 playing cards. What is the probability that the first card drawn is an ace and the second card drawn is a face card? Solution By the product rule, the probability that the first card drawn is an ace and the second card drawn is a face card is given by Example 5, page 392

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Independent Events If A and B are independent events, then

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Test for the Independence of Two Events Two events A and B are independent if and only if

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Example Consider the experiment consisting of tossing a fair coin twice and observing the outcomes. Show that obtaining heads on the first toss and tails on the second toss are independent events. Solution Let A denote the event that the outcome of the first toss is a head, and let B denote the event that the outcome of the second toss is a tail. The sample space of the experiment is so Example 8, page 396

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Example Consider the experiment consisting of tossing a fair coin twice and observing the outcomes. Show that obtaining heads in the first toss and tails in the second toss are independent events. Solution Next, we compute and observe that the test for independent events is satisfied: Example 8, page 396

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Applied Example: Medical Survey A survey conducted by an independent agency for the National Lung Society found that, of 2000 women, 680 were heavy smokers and 50 had emphysema. Of those who had emphysema, 42 were heavy smokers. Using the data in this survey, determine whether the events being a heavy smoker and having emphysema are independent events. Solution Let A denote the event that a woman is a heavy smoker, and let B denote the event that a woman has emphysema. Then, the probabilities that a woman is a heavy smoker, has emphysema, or both are given by, respectively, Applied Example 9, page 396

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Applied Example: Medical Survey A survey conducted by an independent agency for the National Lung Society found that, of 2000 women, 680 were heavy smokers and 50 had emphysema. Of those who had emphysema, 42 were heavy smokers. Using the data in this survey, determine whether the events being a heavy smoker and having emphysema are independent events. Solution Next, we see that the test for independent events is not satisfied: so and conclude that A and B are not independent events. Applied Example 9, page 396

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7.6 Bayes’ Theorem

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Let A 1, A 2, …, A n be a partition of a sample space S, and let E be an event of the experiment such that P(E) 0 and P(A i ) 0 for 1 ≤ i ≤ n Then the a posteriori probability is given by

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Applied Example: Quality Control The panels for the Pulsar widescreen LCD HDTVs are manufactured in three locations and then shipped to the main plant of Vista Vision for final assembly. Plants A, B, and C supply 50%, 30%, and 20%, respectively, of the panels used by Vista Vision. The quality-control department of the company has determined that 1% of the panels produced by plant A are defective, whereas 2% of the panels produced by plants B and C are defective. If a Pulsar widescreen TV is selected at random and the panel is found to be defective, what is the probability that the panel was manufactured in plant C? Applied Example 1, page 404

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Applied Example: Quality Control Solution Let A, B, and C denote the event that the set chosen has a panel manufactured in plant A, plant B, or plant C, respectively. Also, let D denote the event that a set has a defective panel. We can draw a tree diagram with this information: Applied Example 1, page 404

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Applied Example: Quality Control Solution Next, using Bayes’ theorem, we find that the required a posteriori probability is given by Applied Example 1, page 404

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End of Chapter

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