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ProbabilitySimulationThe imitation of chance behavior based on a model that accurately reflect the phenomenon under consideration.Box of simulationsA basketball player…

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ProbabilitySimulationThe imitation of chance behavior based on a model that accurately reflect the phenomenon under consideration. Box of simulations A basketball player makes 70% of her free throws in a long season. In a tournament game she shoots 5 free throws late in the game and misses 3 of them. The fans think she was nervous, but the missing may simply be by chance. How can we simulate this to see how likely it is that she misses 3 out of 5?Use your calculator to simulate a couple’s having children until they have a girl or until they have 4 children, which ever comes first. Use this simulation to estimate the probability that they will have a girl among their no-more-than-four children.Imagine a large group of families, each having 4 children. You want to use simulation to estimate the percent of families that have exactly 0 girls, exactly 1 girl, exactly 2 girls, exactly 3 girls, and exactly 4 girls. Assume that having a boy is equally likely as having a girl. Conduct 40 replications and share your results.Simulations: imitationUse to make judgments Confirm concepts: What happens to μ & σ when……..? Steps of a simulationState the problem or describe the random phenomenon. State any and all assumptions. Assign digits to represent outcomes (Correspondence). Simulate many repetitions (trials). State your conclusions. PracticeHomework: Page 398 Problems 6.3 and 6.4Problem 6.3Chance experiments and events: any activity or situation in which there is uncertainty about which possible outcome will result.Long run: the probability should approach the true probability. Random behavior gains some order only in long runs.Independent TrialsThe outcome of one trial must not influence the outcome of any other. Sample space: the collection of all possible outcomes.Event: any collection of outcomes from the sample space. The Event is a subset of the sample space.Probability Model:Simple event: consists of exactly one outcome.ProbabilitySample space for rolling two numbered cubes (6.35)The sum of each possible outcome is:Techniques to enumerate the outcomesTree Diagram Multiplication Principle: If you do one task in a number of ways and a second task in b number of ways, then both tasks can be done in axb number of ways.With replacement, without replacement!! Organized, systematic list Page 411 Problem 6.24Page 416-417Problems 6.29, 6.32, 6.33 (up to)Venn Diagrams are helpful for probabilities of multiple events.Sports you playComplement: outcomes that are not the event.Notation: Ac, A'Union: The event A or B (A U B)Intersection: The event A and B (A B)Mutually disjoint: two events that have no common outcomes.U Consider the experiment in which an automobile is selected and both the number of defective headlights (0, 1, or 2) and the number of defective tires (0, 1, 2, 3, or 4) are determined.A. Display possible outcomes.B. Let A be the event that at most one headlight is defective and B be the event that at most one tire is defective. What outcomes are in Ac? In A U B? In A intersect B?C. Let C denote the event that all 4 tires are defective. Are A and C disjoint? Are B and C disjoint? A family consisting of three people – P1, P2, and P3- belong to a medical clinic that always has a physician at each of three stations 1, 2, and 3. During a certain week, each member of the family visits the clinic exactly once and is randomly assigned to a station. One experimental outcome is (1, 2, 1), which means that P1 is assigned to station 1, P2 is assigned to station 2, and P3 is assigned to station 1.a. List all possible outcomes.b. List all outcomes in the event A, that all three people go to the same station.c. List all outcomes in the event B, that all three people go to different stations.d. List all outcomes in the event C, that no one goes to station 2.Identify the outcomes in the following events:BcC'A U BA intersect BA intersect CBACProbability RulesThe probability P(A) of event A satisfies 0 < P(A) < 1 If S is the total sample space in a probability model, then P(S) = 1 Two events A and B are disjoint (mutually exclusive) if they have no outcomes in common [P(A and B) = 0] and so can never occur simultaneously. If A and B are disjoint, P(A or B) = P(A) + P(B) The complement of any event A is the event that A does not occur. The complement rule states that P(Ac) = 1 - P(A) The probability of events in a sample space.These probabilities must be numbers between 0 and 1 and have a sum of 1. Rule 5: Multiplication RuleTwo events A and B are independent if knowing that one occurs does not change the probability that the other occurs. If A and B are independent, P(A and B) = P(A)P(B) This is the multiplication rule for independent events Addition Rule for Disjoint EventsIf events A, B, C, ……., are disjoint in the sense that no two have any outcomes in common, then P(A or B or C or …..) = P(A) + P(B) + P(C)… General Addition Rule for Unions of Two EventsFor any two events A and B P(A or B) = P(A) + P(B) – P(A and B) Equivalently P(A U B) = P(A) + P(B) – P(A ∩ B) Great place to see in a Venn diagram!!! PracticeSheet and 6.69 Homework: page 441, problem 6.70 Monty Hall, Lets Make a DealThe game show had three “doors”. Behind one door was a very nice prize, like a new car and behind the other two doors were prizes like goats or camels. The contestant picks a door and the host, Monty Hall, opens one of the remaining doors, the one he knows doesn’t hide the car. The contestant is given the option to switch doors. What is the probability of winning the car if the contestant stays with there first choice? If the decide to switch? General Multiplication Rule for Any Two EventsThe joint probability that event A and B both happen can be found by: P(A and B) = P(A)P(B|A) Equivalently P(A ∩ B) = P(A)P(B|A) Here (B|A) is the conditional probability that B occurs, given the information that A occurs. (Has Occurred) Conditional ProbabilityWhen P(A)>0, using ALGEBRA, the conditional probability of B, given A, is P(B|A) = However, just think about it, if you are given information that A has occurred, the sample space has reduced. Tree Diagram Revisited for a reason!!!!!Problems that have several STAGES! This is for problems that look like, sound like, taste like, feel like, and smell like this problem. At a certain gas station, 40% of the customers use regular unleaded gas, 35% use extra unleaded gas, and 25% use premium unleaded gas. Of those customers using regular gas, only 30% fill their tanks. Of those customers using extra unleaded, 60% fill their tanks, whereas of those using premium, 50% fill their tanks.a) What is the probability that the next customer will request extra unleaded and fill the tank?b) What is the probability that the next customer fills the tank?c) If the next customer fills the tank, what is the probability that regular gas is requested? Extra gas? Premium gas?Suppose a computer company has manufacturing plants in three states. 50% of their computers are manufactured in California, and 85% of these are desktops, 30% of computers are manufactured in Washington, and 40% of these are laptops, and 20% of computers are manufactured in Oregon, and 40% of these are desktops. All computers are first shipped to a distribution site in Nebraska before being sent out to stores. If you pick a computer at random from the Nebraska distribution center, what is the probability that it is a laptop?Given it was a laptop, what is the probability that it came from Oregon?Bayes’ TheoremBayes’ Theorem looks backwards in the tree diagram. Used to examine incidence of rare events.Let us first look at the problem when quite a few athletes use performance enhanced drugs and we have a pretty good test for detection.20 out of 100 athletes use the drug.The test is accurate 99% of the time in detecting the drug and individuals not using will show positive 2% of the time.What is the probability that a person who tests positive actually is using the drug? Let us look at the problem when only a few athletes use performance enhanced drugs and we have a pretty good test for detection.1 out of 300 athletes use the drug.The test is accurate 99% of the time in detecting the drug and individuals not using will show positive 2% of the time.What is the probability that a person who tests positive actually is using the drug?

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