Exam (elaborations) TEST BANK FOR First Course In Probability 9th Edition By Sheldon M. Ross. John L. (Solution Manual)-Converted

The following is known as the “crazy passenger problem” and is stated as follows. A line of 100 airline passengers is waiting to board the plane. They each hold a ticket to one of the 100 seats on that flight. (For convenience, let’s say that the k-th passenger in line has a ticket for the seat number k.) Unfortunately, the first person in line is crazy, and will ignore the seat number on their ticket, picking a random seat to occupy. All the other passengers are quite normal, and will go to their proper seat unless it is already occupied. If it is occupied, they will then find a free seat to sit in, at random. What is the probability that the last (100th) person to board the plane will sit in their proper seat (#100)? If one tries to solve this problem with conditional probability it becomes very difficult. We begin by considering the following cases if the first passenger sits in seat number 1, then all  1 the remaining passengers will be in their correct seats and certainly the #100’th will also. If he sits in the last seat #100, then certainly the last passenger cannot sit there (in fact he will end up in seat #1). If he sits in any of the 98 seats between seats #1 and #100, say seat k, then all the passengers with seat numbers 2, 3, . . . , k−1 will have empty seats and be able to sit in their respective seats. When the passenger with seat number k enters he will have as possible seating choices seat #1, one of the seats k + 1, k +2, . . . , 99, or seat #100. Thus the options available to this passenger are the same options available to the first passenger. That is if he sits in seat #1 the remaining passengers with seat labels k+1, k+2, . . . , 100 can sit in their assigned seats and passenger #100 can sit in his seat, or he can sit in seat #100 in which case the passenger #100 is blocked, or finally he can sit in one of the seats between seat k and seat #99. The only difference is that this k-th passenger has fewer choices for the “middle” seats. This k passenger effectively becomes a new “crazy” passenger. From this argument we begin to see a recursive structure. To fully specify this recursive structure lets generalize this problem a bit an assume that there are N total seats (rather than just 100). Thus at each stage of placing a k-th crazy passenger we can choose from • seat #1 and the last or N-th passenger will then be able to sit in their assigned seat, since all intermediate passenger’s seats are unoccupied. • seat # N and the last or N-th passenger will be unable to sit in their assigned seat. • any seat before the N-th and after the k-th. Where the k-th passenger’s seat is taken by a crazy passenger from the previous step. In this case there are N −1−(k+1)+1 = N − k − 1 “middle” seat choices. If we let p(n, 1) be the probability that given one crazy passenger and n total seats to select from that the last passenger sits in his seat. From the argument above we have a recursive structure give by p(N, 1) = 1 N (1) + 1 N (0) + 1 N NX−1 k=2 p(N − k, 1) = 1 N + 1 N NX−1 k=2 p(N − k, 1) . where the first term is where the first passenger picks the first seat (where the N will sit correctly with probability one), the second term is when the first passenger sits in the N-th seat (where the N will sit correctly with probability zero), and the remaining terms represent the first passenger sitting at position k, which will then require repeating this problem with the k-th passenger choosing among N − k + 1 seats. To solve this recursion relation we consider some special cases and then apply the principle of mathematical induction to prove it. Lets take N = 2. Then there are only two possible arraignments of passengers (1, 2) and (2, 1) of which one (the first) corresponds to the second passenger sitting in his assigned seat. This gives p(2, 1) = 1 2 . If N = 3, then from the 3! = 6 possible choices for seating arraignments (1, 2, 3) (1, 3, 2) (2, 3, 1) (2, 1, 3) (3, 1, 2) (3, 2, 1) Only (1, 2, 3) (2, 1, 3) (3, 2, 1) correspond to admissible seating arraignments for this problem so we see that p(3, 1) = 3 6 = 1 2 . If we hypothesis that p(N, 1) = 1 2 for all N, placing this assumption into the recursive formulation above gives p(N, 1) = 1 N + 1 N NX−1 k=2 1 2 = 1 2 . Verifying that indeed this constant value satisfies our recursion relationship. Chapter 1 (Combinatorial Analysis) Chapter 1: Problems Problem 1 (counting license plates) Part (a): In each of the first two places we can put any of the 26 letters giving 262 possible letter combinations for the first two characters. Since the five other characters in the license plate must be numbers, we have 105 possible five digit letters their specification giving a total of 262 · 105 = , total license plates. Part (b): If we can’t repeat a letter or a number in the specification of a license plate then the number of license plates becomes 26 · 25 · 10 · 9 · 8 = , total license plates. Problem 2 (counting die rolls) We have six possible outcomes for each of the die rolls giving 64 = 1296 possible total outcomes for all four rolls. Problem 3 (assigning workers to jobs) Since each job is different and each worker is unique we have 20! different pairings. Problem 4 (creating a band) If each boy can play each instrument we can have 4! = 24 ordering. If Jay and Jack can play only two instruments then we will assign the instruments they play first with 2! possible orderings. The other two boys can be assigned the remaining instruments in 2! ways and thus we have 2! · 2! = 4 , possible unique band assignments. Problem 5 (counting telephone area codes) In the first specification of this problem we can have 9 − 2 + 1 = 8 possible choices for the first digit in an area code. For the second digit there are two possible choices. For the third digit there are 9 possible choices. So in total we have 8 · 2 · 9 = 144 , possible area codes. In the second specification of this problem, if we must start our area codes with the digit “four” we will only have 2 · 9 = 18 area codes. Problem 6 (counting kittens) The traveler would meet 74 = 2401 kittens. Problem 7 (arranging boys and girls)