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Unformatted text preview: Midterm 2009 Answers Question 1: Question 1: A business school graduate schedules 7 interviews in a given week. Suppose that the probability that an individual interview results in a job offer is 0.40. Assume that the outcomes of the interviews are independent of one another. Let X be the number of job offers that this person has at the end of the week. Write out one expression for P(X=x) where x may be any arbitrary positive integer between 1 and 7: 1 ≤ x ≤ 7 . Explain each part of the expression and why this makes sense. Midterm 2009 Answers Question 1 Answer: The number of job offers at the end of the week is a Binomial Random variable with parameters p=0.40 and n=7. That is, we are looking at the sum of “successes” in n=7 independent trials. Midterm 2009 Answers Question 1 Answer: P(X=x) is given by the Probability Distribution Function for a Binomial Random Variable: 7 7 ( ) (1 ) for 0 , 1 , 2, ... , or, given that p=0.40, n=7 ( ) (0.4) (1 0.4) for 0 , 1 , 2, ... , n x x x n x x x P X x C p p x n P X x C x n = = × ×  = = = × ×  = Midterm 2009 Answers 7 7 7 7 ( ) (0.4) (1 0.4) for 0 , 1 , 2, ... , Explanation: There are two parts of this equation: and (0.4) (1 0.4) We have 7 independent trials. We want to fin x x x x x x P X x C x n C = = × ×  = ×  7 d the probability that x of these trials experience a success and nx do not experience a success. The expression (0.4) (1 0.4) gives the probability of one particular sequence of outcomes that resu x x ×  7 lts in x successes and 7x failures. E.g., if x=2, then the probability of getting the sequence {1 0 0 0 1 0 0} is (0.4)(0.6)(0.6)(0.6)(0.4)(0.6)(0.6)=(0.4) (1 0.4) . However, this is only one x x ×  7 very specific way that we can get two successes (successes on trials 1 and 5). The total number of ways that we can get x successes in 7 trials is given by , so the probability of observing any one x C 7 7 of these outcomes is given by the sum of the probabilities of all of these events, which is given by (0.4) (1 0.4) x x x C × ×  Midterm 2009 Answers Question 2: Question 2: Consider the following estimates from the Sheboygan Fire Department: The probability that a house fire is due to arson is 0.10. The probability that the Fire Department correctly blames a house fire on arson is 0.82 (probability that they blame it on arson, given that the fire was actually caused by arson). The probability that the Fire Department incorrectly blames a house fire on arson is 0.05 (probability that they blame it on arson, given that it was not caused by arson) a) What is the probability that a house fire was actually caused by arson if the Fire Department blames the fire on arson?...
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This note was uploaded on 04/07/2012 for the course STATS V31.0018.0 taught by Professor Thom during the Spring '10 term at NYU.
 Spring '10
 Thom
 Probability

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