finalexamsoloutions

# finalexamsoloutions - Math 310-1 Final Exam Fall 2011 1(30...

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Math 310-1 Final Exam Fall 2011 1. (30 points): A random variable X is defined in each of the following parts. For each part, identify the name of the distribution of X and determine its density function f X . You do not need to justify your answers. (a) A box contains 5 red and 18 blue balls. Pick 4 balls picked at random without replacement. Let X be the number of red balls picked. (b) A box contains 5 red and 18 blue balls. Pick 4 balls picked at random with replacement. Let X be the number of red balls picked. (c) A woman has 4 keys in her pocket, of which exactly one will open her door. She chooses one key at random from her pocket. If the key does not work, she drops it on the ground and chooses another key from her pocket. Let X be the number of keys on the ground after she opens the door. (d) Consider the woman from part (c) with the 4 keys in her pocket. This time, if a key does not work, she puts it back into her pocket with the other keys. Let X be the number of attempts before the correct key is chosen. (e) Let Y be a normal random variable with mean 12 and variance 9 . Define X = 4 3 ( Y - 6) . Solution: (a) X Hypergeometric with density function is given by f X ( k ) = 5 k 18 4 - k 23 4 for k = 0 , 1 , 2 , 3 , 4 . (b) X Binomial with n = 4 and p = 5 / 23 . The density function is given by f X ( k ) = 4 k 5 23 k 1 - 5 23 4 - k = 4 k 5 23 k 18 23 4 - k for k = 0 , 1 , 2 , 3 , 4 . (c) X Uniform on { 0 , 1 , 2 , 3 } . The density function is given by f X ( k ) = 1 / 4 for k = 0 , 1 , 2 , 3 . The best way to see this is to directly compute each probability P ( X = k ) . (d) X Geometric with parameter p = 1 / 4 . The density function is given by f X ( k ) = 1 4 1 - 1 4 k = 1 4 3 4 k for k = 0 , 1 , 2 , . . . (e) X Normal with mean μ = 8 and variance σ 2 = 16 . The density is given by f X ( x ) = 1 4 2 π exp - ( x - 8) 2 32 for all real x . Page 1 of 8

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Math 310-1 Final Exam Fall 2011 2. (15 points): A bag contains 5 dice. One die has the numbers 2-3-3-4-5-6 marked on its sides. The other four dice are standard fair dice. One die is picked at random from the bag and rolled twice. Given that both rolls are 3, what is the probability the die is fair? Solution: Let A be the event that the first two rolls both show 3, and let F be the event that the die is fair. Then, Bayes’ Law gives us that P ( F | A ) = P ( A | F ) P ( F ) P ( A | F ) P ( F ) + P ( A | F c ) P ( F c ) = (1 / 6)(1 / 6)(4 / 5) (1 / 6)(1 / 6)(4 / 5) + (2 / 6)(2 / 6)(1 / 5) = 1 2 .
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