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Unformatted text preview: SIEO 3600 (IEOR Majors) Solution to Assignment #3 Introduction to Probability and Statistics February 9, 2010 Solution to Assignment #3 1. A group of 5 boys and 10 girls is lined up in random order  that is, each of the 15! is assumed to be equally likely. (a) What is the probability that the person in the 4th position is a boy? Solution: To calculate this probability first we need to calculate in how many different ways we can organize the 15 different kids in the line, regardless of their gender, which is equal to the permutation of 15 different objects (15!). We know that position 4 in the line is gonna be a boy, then we have 14 positions left to organize the other 14 kids (regardless of their gender), Then the total number of different ways that we can organize the other 14 kids is the permutation of 14 different objects (14!). Finally let denote the event gender of kid in position 4 as p 4 , and this event has to possible outcomes b(boy) or g(girls). Then the probability that a boy (any one of the 5 boys) is in position 4, is: P ( { p 4 = boy } ) = { p 4 = boy } 15! = 14!5 15! = 1 3 (b) What about the person in the 12th position? Solution: To answer this question you should realize that the probability that the kid in position i is a boy is independent of i . Then this probability is exactly the same that the calculate in (a). P ( { p 4 = boy } ) = P ( { p 12 = boy } ) = 1 3 (c) What is the probability that a particular boy is in the 3rd position? Solution: If we are speaking about the specific boy j ( j goes from 1 to 5), then we just need to consider the total number of ways to organize the 15 kids in a line, and the total number of ways to organize 14 kids in 14 empty positions given that boy j is in position 3. P ( { p 3 = boy j } ) =  p 3 = boy j  15! = 14! 15! = 1 15 2. A woman has n keys, of which one will open her door. If she tries the keys at random, discarding those that do not work, what is the probability that she will open the door on her k th try? What if she does not discard previously tried keys? 2 SIEO 3600, Solution to Assignment #3 Solution: case 1 The woman discard the keys that are not working. Let the K denote the time she gets the right key. Then, P ( K = 1) = 1 n , and notice that P ( K = 2  K 6 = 1) = 1 n 1 . Using conditional probability we have: P ( K = 2) = P ( { K = 2 }{ K 6 = 1 } ) = P ( K = 2  K 6 = 1) P ( K 6 = 1) = 1 n 1 1 1 n = 1 n Continuing in the same fashion we have: P ( K = k ) = P ( K = k  K 6 = k 1 ,...,K 6 = 1) P ( K 6 = k 1 ,...,K 6 = 1) = 1 n k +1 n k +1 n k n 2 n 1 n 1 n = 1 n case 2 The woman doesnt discard the tried keys. In this case K is a geometric random variable which we will learn next week, then the probability that we need to make k trials is given by: P ( K = k ) = n 1 n k 1 1 n ....
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This note was uploaded on 10/03/2011 for the course SIEO W3600 taught by Professor Yunanliu during the Spring '10 term at Columbia.
 Spring '10
 YUNANLIU

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