ECEN 303: Homework 3 Solutions

Transcription

1 Problem 1: ECEN 0: Homework Solutions Let A be the event that the dog was lost in forest A and A c be the event that the dog was lost in forest B. Let D n be the event that the dog dies on the nth day. Let F n be the event that the dog is found on the nth day. Let S n be the event that Oscar searches forest A on nth day and S c n be the event that he searches forest B on day n. We know that PrA 0. PrA c 0.6 PrD n+1 D c n, F c n N N + PrF n A, S n, F c n PrF n B, S c n, F c n a We want to find PrF 1 S 1 PrF 1 S c 1 Now PrF 1 S 1 PrF 1 S 1, A PrA + PrF 1 S 1, A c PrA c Similarly PrF 1 S1 c PrF 1 S1, c A PrA + PrF 1 S1, c A c PrA c Thus PrF 1 S 1 > PrF 1 S1 c, so Oscar should search in forest A. b Using Baye s Rule PrA S 1, F c 1 PrA PrF c 1 A, S 1 PrA PrF c 1 A, S 1 + PrA c PrF c 1 A c, S c Now PrS 1 F 1 PrS 1 F 1 PrF 1 PrS 1 F 1 PrS 1 F 1 A PrA PrS 1 PrF 1 S 1, A

2 and PrS c 1 F 1 PrS c 1 F 1 A c PrA c PrS c 1 PrF 1 S c 1, A c Using, PrF 1 PrF 1 S 1 + PrF 1 S1 c, we have 0.05 PrS 1 F d Probability that the Oscar will find a live dog for the first time on the second day is PrF D c F c 1 S 1, S PrD c F, F c 1, S 1, S PrF F c 1, S 1, S PrF c 1 S 1, S PrD F c 1 c PrF F1 c, S PrF1 c S 1 [ PrD F c 1 c PrF c F1 c, S, A PrA F1 c, S 1 ] [ ] + PrF F1 c, S, A c PrA c F1 c, S 1 0 PrF1 c S 1, A PrA + PrF1 c S 1, A c PrA c [0.5 0.] [ ] 0.05 e Probability that Oscar does not find a dead dog at the end of the second day is Pr F D c F1 c, S 1, S 1 PrF D F1 c, S 1, S f Probability that he found a live dog is 1 PrD F, F c 1, S 1, S PrF F c 1, S 1, S 1 PrD F c 1 PrF F c 1, S [0.5 0.] 0.97 PrD c F, F c, F c, F c 1, S 1, S, S, S c PrD c D c D c F, F c, F c, F c 1, S 1, S, S, S c PrD c D c, D c, F, F c, F c, F c 1, S 1, S, S, S c PrD c D c, F, F c, F c, F c 1, S 1, S, S, S c PrD c F, F c, F c, F c 1, S 1, S, S, S c PrD c D c, F c PrD c D c, F c PrD c F c g Let L be the event that he searches twice in forest A and twice in forest B. Then PrD c F, F c, F c, F c 1, L PrD c D c D c F, F c, F c, F c 1, L PrD c D c, D c, F, F c, F c, F c 1, L PrD c D c, F, F c, F c, F c 1, L PrD c F, F c, F c, F c 1, L PrD c D c, F c PrD c D c, F c PrD c F c

3 Problem : a Let P the event that exactly one parent went and R be the event that Rover, the oldest dog, went. We want to find PrR P Number of ways exactly one parent and rover can go Number of ways exactly one parent goes Now if exactly one parent and rover are going, we can choose the parent in two ways and the remaining three positons can be filled by any of the remaining two dogs or the four kids. The number of combinations of this is 6 0 If exactly one parent is going, we can choose the parent in two ways. Both dogs and cats cannot go together, so we have two cases: either the dogs go or the cats go. Suppose the pet is a dog. So we have to choose four positions with three dogs or four kids such that at least one dog is chosen. This can be done in + total number of ways 1 number of ways of not choosing any dog Suppose the pet is cat, the number of ways we can fill the remaining four positions is + 1 number of ways of not choosing any cat total number of ways Thus, the total number of ways exactly one parent can go is Hence, Problem : PrR P Number of ways of distributing cards to players, each getting cards is! ; ; ;!!!! a Number of ways for one player to receive spades is 9 1 ; ; Choose the player Distribute the remaining 9 cards between players

4 Thus, Pr One of the players receive all spades 1 9 ;; ;;; b Number of ways for two of players to receive spades such that each one has at least one spade is Choose the players 9 Choose more cards [ ] 6 Distribute the 6 cards spades and others just chosen between the two players such that each player gets at least 1 spade 6 Thus, Distribute the remaining 6 cards between the remaining players Pr Two players receive all spades such that each one has atleast one spade 9 [ 6 ] 6 ;;; c Number of ways in which each player gets one ace is 8 1; 1; 1; 1 }{{! } Number of ways of distributing the aces between players Number of ways to distribute the remaining 8 cards between players Thus d Pr Each player gets one ace 8 1;1;1;1! ;;; Pr Void in at least one suit Pr Void in exactly one suit + Pr Void in exactly two suits + Pr Void in exactly three suits 0 Further, number of ways to draw hands that are void in exactly one suit [ ] 6 1 Number of ways to choose the suit Number of ways to draw cards from 6 cards that is not void in any suit

5 and number of ways to draw hands that are void in exactly two suits 1 Thus, Problem : Choose the two suits in which we are void [ 6 ] Pr The hand drawn is void in at least one suit 1 + a Let A be the event that 1 is transmitted and A c be the event that 0 is transmitted. Let B n be the event that the n th bit we receive is 1. Then using the law of Total Probability and Baye s rule we have p 1 PrA B 1 9 PrB 1 A PrA 1 ε1 p PrB 1 A PrA + PrB 1 A c PrA c 1 ε1 p + εp. b Note that B 1 and B are conditionally independent given A. Now, p PrA B 1, B PrB A, B 1 PrA B 1 PrB B 1 PrB A, B 1 PrA B 1 PrB A, B 1 PrA B 1 + PrB A c, B 1 PrA c B 1 PrB A, B 1 PrA B 1 PrB A PrA B 1 + PrB A c PrA c B 1 1 εp 1 1 εp 1 + ε1 p 1 c Using the same argument as in part, we get p n 1 εp n 1 1 εp n 1 + ε1 p n 1 1 ε n p 1 ε n p 0 + ε n 1 p 0 p 0 n p 0 + ε 1 ε 1 p0 where p 0 1 p. As n, we have 1 if ε < 1 p n p 0 if ε 1 0 if ε > 1 These results meet the intuitions. When ε < 1/, it means the observations are positively correlated with the bit transmitted, thus knowing the observations B n, n 1,... will increase the conditional probability of A given the observations, until it hits 1. When ε 1/, it means the observations are independent to the bit transmitted, thus given a bunch of observations B n, n 1,... won t change the conditional probability of A given the observations, which is the same as the prior probability 5

6 p. When ε > 1/, it means the observations are negatively correlated with the bit transmitted, thus knowing the observations B n, n 1,... will decrease the conditional probability of A given the observations, until it hits 0. c By part c, to get p n 0.99, the following condition must be true: There are three cases p 0 p n n 0.99 p 0 + ε 1 ε 1 p0 p 0 log 991 p 0 n log ε 1 ε 1 when ε < 1/, we need n log p 0 log[991 p 0 ], log ε log[1 ε] The intuition is the following. In this case, knowing the observations B n, n 1,... will monotonously increase p n, until it hits 1. So we expect after the number of observations n exceeds a threshold, p n can be no less than when ε 1/, the right hand side of 1 is 0. Thus for 1 to be true, the left hand side must be no less than 0, which implies p If p , then for any n, p n exceeds 0.99; otherwise no p n exceeds The intuition is that when ε 1/, the observations are independent to the bit transmitted. Thus the conditional probability of A given the observations won t change as we get more observations. Consequently, the only possible way to have p n > 0.99 is the prior probability p When ε > 1/, we need the following to be true n log p 0 log[991 p 0 ] log ε log[1 ε] Since n is no less than 1, the above inequality further requires log p 0 log[991 p 0 ] log ε log[1 ε] p 0 99ε ε. The intuition is that when ε > 1/, knowing the observations B n, n 1,,... will monotonously decrease p n, until it hits 0. Thus the only possible way for p n 0.99 is that p 0 is large enough so as to make p n 0.99 for n less than a certain threshold. Problem 5: a Let X be the event that A can communicate with B and Y be the event that exactly five links have failed. We are asked to find PrX Y. PrX Y PrX Y PrY 6

7 Note that X Y is the event that exactly 5 links fail and A can communicate with B. This is only possible if the following sets of links fail: either adgef or bchef. The probabilities of each of these possibilities is p 5 1 p. Thus, PrX Y p 5 1 p + p 5 1 p p 5 1 p Furthermore, we can calculate the probability of 5 links failing as 8 PrY p 5 1 p 5 Thus, PrX Y p5 1 p 8 p 5 1 p 1 8 b Let C be the event that either g or h but not both are working. PrC Y PrC Y PrY Note that C Y consists of all the possibilities such that exactly one gate out of g and h is working while the other are not working and exactly four gates out of the remaining six a,b,c,d,e,f are not working while the remaining two are working. Thus, 6 PrC Y p1 p p 1 p 1 1 p1 p 6 p 1 p PrC B p 5 1 p 8 c Let D be the event that a, d and h have failed. Given this the only way A can communicate with B is if gates b, b, f and g are working. Let this event be E. As the gates fail independently, events E and D are independent. Thus PrE D PrE 1 p 7