2006%20fall%20solutions - FALL 2006 FINAL EXAM SOLUTIONS...

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FALL 2006 FINAL EXAM SOLUTIONS MATH 218, DECEMBER 13, 2006 This is a solution set for the Math 218 final common exam held from 2:00 to 4:00 PM on Wednesday, December 13, 2006. There were ten problems, weighted equally (for a total of 200 points). These solu- tions are rather more full and complete than anything we could have expected from any student. Problem 1 (20 pts) . 30 marbles are sitting on a table: 6 red, 9 blue, and 15 green. All of the red marbles, 3 of the blue marbles, and 10 of the green marbles are covered in chocolate, because a child played with them while eating a piece of pie. The remaining marbles were out of the child’s reach and remained clean. (a) Draw a probability tree describing the situation. Be sure to include the labels of events, probabilities, and conditional probabilities. (b) If a randomly selected marble is covered in chocolate, what is the probability that it is red? (c) If a randomly selected marble is red, what is the probability that it is covered in chocolate? (d) Are the events “red” and “covered in chocolate” indepen- dent? Solution. (a) The marbles are partitioned in two ways: according to color, and according to whether they’re covered in chocolate or are clean. There are, accordingly, two ways you can draw the tree: in the first, the first level of the tree is based on color, and the second level based on chocolate/clean; in the second, the first level of the tree is based on chocolate/clean, and the second is based on color. The two ways are about equally easy. Our first tree uses the color-first approach. Of course, you were instructed to reduce fractions to lowest terms, so instead of “ 10/15 ”, for example, as a conditional probability in the tree, you should have written “ 2/3 ”. We have left the fractions unreduced to make it more obvious where they came from. Also, the calculation of the probabilities at the right edge was not, strictly speaking, asked for; but they’re too useful for the rest of the problem, and since they’re trivial to calculate, you should probably always write them down. Here’s the first tree: R B G 6/30 9/30 15/30 Choc Choc Choc Clean Clean Clean 6/6 0/6 3/9 6/9 10/15 5/15 P ( Choc R ) = 1 5 P ( Clean R ) = 0 P ( Choc B ) = 1 10 P ( Clean B ) = 1 5 P ( Choc G ) = 1 3 P ( Clean G ) = 1 6 In the second possible tree description, we have to count the number of chocolate-covered marbles and the number of clean marbles. Obviously there are 19 chocolate-covered marbles ( 6 red, 3 blue, and 10 green), leaving 11 clean marbles ( 0 red, 6 blue, and 5 green). This tree has the advantage that all the probabilities are already in lowest terms:
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SOLUTIONS Choc Clean 19/30 11/30 R P ( R Choc ) = 1 5 B P ( B Choc ) = 1 10 G P ( G Choc ) = 1 3 R P ( R Clean ) = 0 B P ( B Clean ) = 1 5 G P ( G Clean ) = 1 6 6/19 3/19 10/19 0 6/11 5/11 (b) P ( R | Choc ) = P ( R Choc ) P ( Choc ) = P ( R Choc ) P ( R Choc ) + P ( B Choc ) + P ( G Choc ) , which from the tree in part (a) is P ( R | Choc ) = 1/5 1/5 + 1/10 + 1/3 = 6 19 0.316.
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