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A Solution Manual for: A First Course In Probability by Sheldon M. Ross. John L. Weatherwax∗ February 7, 2012 Introduction Here you’ll ﬁnd some notes that I wrote up as I worked through this excellent book. I’ve worked hard to make these notes as good as I can, but I have no illusions that they are perfect. If you feel that that there is a better way to accomplish or explain an exercise or derivation presented in these notes; or that one or more of the explanations is unclear, incomplete, or misleading, please tell me. If you ﬁnd an error of any kind – technical, grammatical, typographical, whatever – please tell me that, too. I’ll gladly add to the acknowledgments in later printings the name of the ﬁrst person to bring each problem to my attention. Acknowledgements Special thanks to (most recent comments are listed ﬁrst): Mark Chamness, Dale Peterson, Doug Edmunds, Marlene Miller, John Williams (several contributions to chapter 4), Timothy Alsobrooks, Konstantinos Stouras, William Howell, Robert Futyma, Waldo Arriagada, Atul Narang, Andrew Jones, Vincent Frost, and Gerardo Robert for helping improve these notes and solutions. It should be noted that Marlene Miller made several helpful suggestions on most of the material in Chapter 3. Her algebraic use of event “set” notation to solve probability problems has opened my eyes to this powerful technique. It is a tool that I wish to become more proﬁcient with. All comments (no matter how small) are much appreciated. In fact, if you ﬁnd these notes useful I would appreciate a contribution in the form of a solution to a problem that is not yet ∗wax@alum.mit.edu 1

worked in these notes. Sort of a “take a penny, leave a penny” type of approach. Remember: pay it forward. Miscellaneous Problems The Crazy Passenger Problem 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 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 N (1) + 1 N (0) + p(N − k, 1) N−1 1 N Xk=2 p(N − k, 1) . + 1 N N−1 Xk=2 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 arrangements 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 arrangements (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 arrangements for this problem so we see that p(3, 1) = 3 6 = 1 2 . If we hypothesis that p(N, 1) = 1 formulation above gives 2 for all N, placing this assumption into the recursive p(N, 1) = 1 N + 1 N 1 2 = 1 2 . N−1 Xk=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 total license plates. 262 · 105 = 67600000 , 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 · 7 · 6 = 19656000 , 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 possible unique band assignments. 2! · 2! = 4 ,

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) Part (a): Since we assume that each person is unique, the total number of ordering is given by 6! = 720. Part (b): We have 3! orderings of each group of the three boys and girls. Since we can put these groups of boys and girls in 2! diﬀerent ways (either the boys ﬁrst or the girls ﬁrst) we have possible orderings. (2!) · (3!) · (3!) = 2 · 6 · 6 = 72 , Part (c): If the boys must sit together we have 3! = 6 ways to arrange the block of boys. This block of boys can be placed either at the ends or in between any of the individual 3! orderings of the girls. This gives four locations where our block of boys can be placed we have possible orderings. 4 · (3!) · (3!) = 144 , Part (d): The only way that no two people of the same sex can sit together is to have the two groups interleaved. Now there are 3! ways to arrange each group of girls and boys, and to interleave we have two diﬀerent choices for interleaving. For example with three boys and girls we could have thus we have possible arrangements. g1b1g2b2g3b3 vs. b1g1b2g2b3g3 , 2 · 3! · 3! = 2 · 62 = 72 ,

Problem 8 (counting arrangements of letters) Part (a): Since “Fluke” has ﬁve unique letters we have 5! = 120 possible arrangements. Part (b): Since “Propose” has seven letters of which four (the “o”’s and the “p”’s) repeat we have arrangements. 7! 2! · 2! = 1260 , Part (c): Now “Mississippi” has eleven characters with the “i” repeated four times, the “s” repeated four times and the “p” repeated two times, so we have possible rearranges. 11! 4! · 4! · 2! = 34650 , Part (d): “Arrange” has seven characters with a double “a” and a double “r” so it has 7! 2! · 2! = 1260 , diﬀerent arrangements. Problem 9 (counting colored blocks) Assuming each block is unique we have 12! arrangements, but since the six black and the four red blocks are not distinguishable we have 12! 6! · 4! = 27720 , possible arrangements. Problem 10 (seating people in a row) Part (a): We have 8! = 40320 possible seating arrangements. Part (b): We have 6! ways to place the people (not including A and B). We have 2! ways to order A and B. Once the pair of A and B is determined, they can be placed in between any ordering of the other six. For example, any of the “x”’s in the expression below could be replaced with the A B pair x P1 x P2 x P3 x P4 x P5x P6 x .

Giving seven possible locations for the A,B pair. Thus the total number of orderings is given by 2! · 6! · 7 = 10080 . Part (c): To place the men and women according to the given rules, the men and women must be interleaved. We have 4! ways to arrange the men and 4! ways to arrange the women. We can start our sequence of eight people with a woman or a man (giving two possible choices). We thus have possible arrangements. 2 · 4! · 4! = 1152 , Part (d): Since the ﬁve men must sit next to each other their ordering can be speciﬁed in 5! = 120 ways. This block of men can be placed in between any of the three women, or at the end of the block of women, who can be ordered in 3! ways. Since there are four positions we can place the block of men we have possible arrangements. 5! · 4 · 3! = 2880 , Part (e): The four couple have 2! orderings within each pair, and then 4! orderings of the pairs giving a total of total orderings. (2!)4 · 4! = 384 , Problem 11 (counting arrangements of books) Part (a): We have (3 + 2 + 1)! = 6! = 720 arrangements. Part (b): The mathematics books can be arranged in 2! ways and the novels in 3! ways. Then the block ordering of mathematics, novels, and chemistry books can be arranged in 3! ways resulting in possible arrangements. (3!) · (2!) · (3!) = 72 , Part (c): The number of ways to arrange the novels is given by 3! = 6 and the other three books can be arranged in 3! ways with the blocks of novels in any of the four positions in between giving possible arrangements. 4 · (3!) · (3!) = 144 ,

Problem 12 (counting awards) Part (a): We have 30 students to choose from for the ﬁrst award, and 30 students to choose from for the second award, etc. So the total number of diﬀerent outcomes is given by 305 = 24300000 Part (b): We have 30 students to choose from for the ﬁrst award, 29 students to choose from for the second award, etc. So the total number of diﬀerent outcomes is given by 30 · 29 · 28 · 27 · 26 = 17100720 Problem 13 (counting handshakes) With 20 people the number of pairs is given by 2 = 190 . 20 Problem 14 (counting poker hands) A deck of cards has four suits with thirteen cards each giving in total 52 cards. From these 52 cards we need to select ﬁve to form a poker hand thus we have 52 5 = 2598960 , unique poker hands. Problem 15 (pairings in dancing) We must ﬁrst choose ﬁve women from ten in 10 in 12 5 possible ways, and ﬁve men from 12 5 ways. Once these groups are chosen then we have 5! pairings of the men and women. Thus in total we will have 10 5 12 5 5! = 252 · 792 · 120 = 23950080 , possible pairings.

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