# hw2_sol - EE 351K Probability, Statistics, and Random...

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EE 351K Probability, Statistics, and Random Processes SPRING 2008 Instructor: Sanjay Shakkottai shakkott@ece.utexas.edu TA: Rony El Haddad rony.elhaddad@ece.utexas.edu Homework 2 - Solution Problem 1 A new test has been developed to determine whether a given student is overstressed. This test is 85% accurate if the student is not overstressed, but only 80% accurate if the student is overstressed. It is known that 40% of all students are over-stressed. Given that a particular student tests negative for stress, what is the probability that the test result is correct? Solution : Let A be the event that the student is not overstressed, and let A c be the event that the student is in fact overstressed. Now let B be the event that the test results indicate that the student is not overstressed. The desired probability, P ( A | B ) , is found by Bayes’ rule: P ( A | B ) = P ( A ) P ( B | A ) P ( A ) P ( B | A ) + P ( A c ) P ( B | A c ) = 0 . 6 0 . 85 0 . 6 0 . 85 + 0 . 4 0 . 2 = 0 . 8644 . Problem 2 A parking lot consists of a single row containing n parking spaces ( n 2) . Mary arrives when all spaces are free. Tom is the next person to arrive. Each person makes an equally likely choice among all available spaces at the time of arrival. Describe the sample space. Obtain P ( A ) , the probability the parking spaces selected by Mary and Tom are at most 3 spaces apart. Solution : For convenience, we will number each of the parking spaces. We will draw a sequential probability tree to illustrate the sample space: Mary 1 2 N 1/N 1/N 1/N Tom 2 N 1/(N-1) 1/(N-1) 1 3 N 1/(N-1) 1/(N-1) 1/(N-1) 1 N-1 1/(N-1) 1/(N-1) Figure 1: Problem 2 Mary can choose any of the n parking spaces. She has a probability of 1 /n of selecting any particular space. Tom can choose any of the remaining n 1 spaces and has a probability of 1 / ( n 1) of choosing any particular space (other than the one Mary chose). There are n ( n 1) leaves on the tree, and each leaf is equally likely to occur. When we look at the leaves on the branches where Mary does not choose spaces 1 , 2 , 3 ,n 2 ,n 1 , or n , we see that 6 leaves on each of these branches is in our event (three spaces on each side of Mary’s car). When Mary chooses spaces 3 or n 2 , there are five such leaves (two spaces on one side and three on the other side). When Mary chooses

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spaces 2 or n 1 , there are four such leaves (one on one side, three on the other). When Mary chooses spaces 1
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## This note was uploaded on 06/10/2008 for the course EE 351k taught by Professor Bard during the Spring '07 term at University of Texas at Austin.

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hw2_sol - EE 351K Probability, Statistics, and Random...

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