This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 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 childs 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 theyre 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 colorfirst 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 theyre too useful for the rest of the problem, and since theyre trivial to calculate, you should probably always write them down. Heres the first tree: @ @ @ @ @ @ @ @ R R B G 6 / 3 9/30 1 5 / 3 * H H H H j * H H H H j * H H H H j Choc Choc Choc Clean Clean Clean 6 / 6 / 6 3 / 9 6 / 9 1 / 1 5 5 / 1 5 P ( Choc R ) = 1 5 P ( Clean R ) = 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 chocolatecovered marbles and the number of clean marbles. Obviously there are 19 chocolatecovered marbles ( 6 red, 3 blue, and 10 green), leaving 11 clean marbles ( red, 6 blue, and 5 green). This tree has the advantage that all the probabilities are already in lowest terms: SOLUTIONS @ @ @ @ @ @ R Choc Clean 1 9 / 3 1 1 / 3 R P ( R Choc ) = 1 5 B P ( B Choc ) = 1 10 G P ( G Choc ) = 1 3 @ @ @ @ R @ @ @ @ R R P ( R Clean ) = B P ( B Clean ) = 1 5 G P ( G Clean ) = 1 6 6 / 1 9 3/19...
View
Full
Document
This homework help was uploaded on 02/05/2008 for the course MATH 218 taught by Professor Haskell during the Fall '06 term at USC.
 Fall '06
 Haskell
 Math, Probability

Click to edit the document details