MAT 210 Spring 2006 Exam 2 03/31/2006Name:________________________________________________ Class Code | | | | - Write your full name and code clearly on the first page of your solutions.- This is an open book/notes test. Work independently. Please do not write on the test. One page per questiononly.- 50 minutes. Total score is 100. Good luck!Part I: Probability, Conditional Probability, Bayes Theorem, 50 pts1. (25 pts, 5 pts each) You are going to toss a coin repeatedly in successive trials until you get a head for the first time, and then you are going to stop the experiment. Assuming that the coin is unbalanced with P(head) = 1/5, answer the following questions:(1) Draw a tree diagram that is used to determine the outcomes;(2) List at least 4 possible outcomes;(3) Express the event “at least 4 tosses are required” in terms of the outcomes;(4) Compute the probabilities P(“at least 4 tosses are required before the experiment terminates”);(5) Explain why the probability calculated in (4) is large.Geometric Distribution: In problems 3 and 4 below, the number of Bernoulli trials (n) is fixed while the number of defective items(or international requests) is the variable. One has dependent trials (selections without replacement) and the other has independent trials (selections with replacement; or without replacement but only “few” items are selected from a large number of products).In the first probability model given in problem 1, trails are independent and the number of trials required (until certain event occurs) is the variable. This model is the so-called Geometric distribution.2. (25 pts, 25/3 pts each, Bayes Theorem) A test for a certain disease is found to be 95% accurate, meaning that it will correctly diagnose the disease in 95 out of 100 people who have the ailment. The test is also 95% accurate for a negative result, meaning that it will correctly exclude the disease in 95 out of 100 people who do not have the ailment. For a certain segment of the population, the incidence of the disease is 4%. (1) If a person tests positive, find the probability that the person actually has the disease (Hint: define appropriate events and use Bayes Theorem);(2) Now suppose the incidence of the disease is 49%. Compute the probability that the person actually hasthe disease, given that the test is positive;(3) The probability you obtained in (1) is much smaller than 0.95, if your computation is correct. Hence, You can conclude that there is only a much smaller probability to claim “the person really has the disease” after knowing that “the test is positive”, though the test has 95% “correctness”. Explain this difference.Hint. Population: the segment of the populationExperiment: Determine whether or not a selected person has the diseaseOutcomes: D, Dc (where D and Dc are defined below)Sample space: S = {D, Dc}Define necessary events: D = the selected person has the disease;Dc = the selected person does not have the disease. Rate of disease = 4%  P(D) = 4% , P(Dc) = 96% E = Testing result shows positive for the selected person Ec = Testing result shows negative for the selected person Correctness of testing: if the selected person does have the disease, the testing will show positive with probability 95%, i.e., (about) 95 positive out of 100 persons who do have the disease. if the selected person does NOT have the disease, the testing will show negative with probability 95%, i.e., (about) 95 positive out of 100 persons who do NOT have the disease.In other words, P(E | D) = 95% and P(Ec | Dc) = 95% Want P(D | E) = ?Part II: Discrete Variables and their Distributions, 50 points (Read Hints below)3. (25 pts, 25/6 pts each) A certain disease has an infection rate 1/5 in a rural area of 40 people. A sample of size 10 is taken at random without replacement. Let X be the number of infected people in the sample. (1) What is the distribution of X? Give the name of the distribution and values of the parameters; (2) List all the possible values of X;(3) Find the probability that exactly 2 individuals in the sample are infected by the disease;(4) Compute the mean - and variance -2 of X using the formulas for this distribution;(5) Explain the formula for -; (6) Interpret the probability you calculated in (3).4. (25 pts, 5 pts each) Of all airline flight requests received by a certain discount ticket broker, 10% are for international travels and 90% are for domestic flights. Let X be the number of international flight requests of the next 20 customers. Assuming independence of successive requests, answer the following questions.(1) What is the distribution of X? Give the name of the distribution and values of the parameters; (2) List all the possible values of X;(3) Compute the probability that P(X - 5); (4) Compute the mean and variance of X (using formulas).(5) Interpret the mean you computed in (4).About Binomial and HyperGeometric Distributions: If trials are independent (or approximately independent), the resulting distribution is Binomial; otherwise,it is HyperGeometric.For HyperGeometric distribution, Mean = n * [a / (a + b)] Variance = n * [a / (a + b)] * [b / (a + b)] * [(a + b – n) / (a + b –