Summer 2017 June 16, Written Homework 03

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1 CS1800 Discrete Structures Prof. Schnyder Summer 2017 June 16, 2017 Assigned: Fr 16 June 2017 Due: Tu 20 June 2017 Instructions: Written Homework 0 The assignment has to be uploaded to Blackboard by the due date. NO assignment will be accepted after 11:59pm on that day. We expect that you will study with friends and often work out problem solutions together, but you must write up your own solutions, in your own words. Cheating will not be tolerated. Professors, TAs, and peer tutors will be available to answer questions but will not do your homework for you. One of our course goals is to teach you how to think on your own. Provided your submission is in one single file, you may turn in work to Blackboard that is either handwritten and scanned, written in a word processor such as Word, or typeset in LaTeX. In the case of handwritten work, we may deduct points if the scan is upside down or the work is illegible. To get full credit, show INTERMEDIATE steps leading to your answers, throughout. Problem 1 [12 pts (4,4,4]: Permutations and Combinations i. How many strings can you write using 7 A s and 4 B s, so that no two B s are consecutive? Hint: Insert the B s between the A s Consider the string A A A A A A A. We need to select a set of 4 blanks where we will write the B s. The answer is thus ( 8 4 = 70. ii. You own 5 songs by Adele, 4 by Lady Gaga, and by Katie Perry. How many playlists can be formed that consist of Adele, 2 Gaga, and 2 Perry songs if 1. Repetitions are allowed (the same song may be played multiple times We need to fill 7 positions. Tasks are: Select Adele positions out of 7. - fill Adele positions - select 2 Gaga positions out of 4 - fill Gaga positions - select 2 Perry positions out of 2 - fill Perry positions This can be performed in ( 7 2. Repetitions aren t allowed. 5 ( (2 2 2 = =, 780, 000 This differs from above count in the way we fill the positions. ( 7 ( ( = 5 (5 4 6 (4 1 ( 2 = 907, 200 Alternative solution: first select the songs. Can be done in ( ( 5 4 ( 2 2 = 180 ways. Then permute them in all 7! = 5040 possible ways. 1

2 Problem 2 [14pts (4, 4, 2, 4]: Combinations with repetitions A bowl of fruit contains apples, bananas, oranges, and pears (at least 12 of each kind. i. In how many ways can we choose 12 pieces? Stars and bars. Have bars and 12 stars, thus strings of length 15. Choosing positions for the bars can be done in ( 15 = 455 ways. ii. In how many ways can we choose 12 pieces, including at least 4 oranges? 4 stars must be allocated to the oranges. There remain 8 stars, so strings have length 11 and the answer is ( 11 = 165. iii. In how many ways can we choose 12 pieces, so that there are at most oranges? The complement of at most three is at least 4. Thus (good-guys-bad-guys = 290 iv. In how many ways can we choose 12 pieces, so that there are at most oranges and at most 2 pears? This is more difficult than the above questions. Let A be the set of strings with at most oranges and B be the set of strings with at most 2 pears. We want A B but first calculate the complement: A B = A B = A + B A B. A is the set of strings with at least 4 oranges, so A = 165, B is the set of strings with at least pears, so B = ( 12 = 220. Finally, A B is the set of strings with at least 4 oranges and pears, so A B = ( 8 = 56. Thus A B = = 29 and therefore A B = = 126. Problem [16 pts (4 pts each]: Giving Presents i. We have three prizes valued at $10, $100, $1000 to give away and no person should receive more than one prize. In how many ways can we select persons out of 11 to receive these prizes? Ordered selection. 11 ways to select the recipient of $10, then 10 ways for the $100, and 9 ways for the $1000 for a total of = 990. ii. We have three identical prizes to give away and no person should receive more than one prize. In how many ways can we select persons out of 11 to receive these prizes? This is an unordered selection, all that matters is who is chosen. We need a set of persons. There are ( 11 = 165 such sets. iii. We have three prizes valued at $10, $100, $1000 to give away and recipients may receive more than one prize. In how many ways can we distribute our three prizes among 11 persons? Ordered selection with repetitions. 11 choices for $10, 11 choices for $100, and 11 choices for $1000, for a total of 11 = 11 possibilities. 2

3 iv. We have three identical prizes to give away and recipients may receive more than one prize. In how many ways can we distribute our three prizes among 11 persons? Unordered selection, a person may be chosen multiple times: stars and bars. stars and 10 bars, so strings have length 1 and there are ( 1 = 286 possible selections. Problem 4 [16 pts (4 pts each]: Distribution of balls in urns In this problem, we have four distinguishable urns named A, B, C, D and four identical balls to be placed in these urns. We first calculate the number of different placements that are possible and then discuss the probabilities of these placements. i. In how many ways can one distribute the four identical balls among urns A, B, C, and D? This is a stars-and-bars problem with four stars (the balls and three bars (separating the urns. Each placement of balls is described by a string of four stars and three bars. The string length is 7 and there are ( 7 = 5 ways of selecting positions for the bars, thus 5 strings. Therefore, there are 5 possible placements of the balls in the urns. ii. Suppose that the four balls are randomly thrown at the urns and that each ball has an equal chance to land in any of the urns. What is the probability that each urn receives a ball? Hint: paint the balls with numbers 1, 2,, 4 to make them distinguishable. Then each throw of the balls can be represented by a string of four letters. For example, string B A D D indicates that ball 1 landed in urn B, ball 2 in urn A, and balls and 4 in urn D. The sample space consists of all possible strings and has cardinality 4 4. The event Ball in each urn consists of all permutations of the four letters. It has cardinality 4! and thus probability 4!/4 4 = iii. Is the probability of landing one ball in each urn equal to the ratio of the number of placements with one ball per urn (this is 1 over the number you calculated in part (i? What can you infer from this: are the placements equally likely? The ratio of placements with one ball per urn over all placements is 1/ , it is different from the probability of landing one ball in each urn. This means that the placement of balls in urns don t all have the same probability, they don t form what is called a uniform space. iv. You receive $1 for each ball landing in urn A, $2 for each ball in urn B, $ for each ball in urn, and $4 for each ball in urn D. Use linearity of expectation to calculate your expected gain. (Explain how and where linearity of expectation was used in your calculation. Let X be your gain and X A, X B, X C, X D be the numbers of balls in urns A, B, C, D respectively. Then X = X A + 2X B + X C + 4X D and thus, by linearity of expectation, E[X] = E[X A ]+2E[X B ]+E[X C ]+4E[X D ]. Furthermore, by symmetry, E[X A ] = E[X B ] = E[X C ] = E[X D ], so that E[X] = 10E[X A ]. Also, from X A + X B + X C + X D = 4 follows 4E[X A ] = E[X A ] + E[X B ] + E[X C ] + E[X D ] = E[X A + X B + X C + X D ] = 4 and thus E[X A ] = 1. The expected gain is thus $10.

4 Problem 5 [16 pts (4 pts each]: More cars A dealership sells cars manufactured by Chrysler, Ford, and GM. On its lot, 5% of the cars are Fords and 40% are GMs. The remainder are Chryslers. The cars are either sedans or SUVs, with GM s cars being 0% SUVs, Chrysler s being 50% SUVs, and Ford s being 80% SUVs. i. Draw a probability tree representing these breakdowns. 5/10 sedan 25/100 Chrysler 5/10 SUV Cars 5/100 Ford 2/10 sedan 8/10 SUV 40/100 GM 7/10 sedan /10 SUV ii. Calculate the probability that a randomly selected car is a SUV manufactured by GM. This is Pr[GM SUV] and can be read as product of the relative probabilities along 4 the lowest branch of the tree: = iii. Calculate the probability that a randomly selected car is a SUV. Add the products of the paths ending with SUV: = iv. Suppose that a randomly selected car is a SUV. What is the probability that it is a Chrysler? Pr[Chrysler SUV] = Pr[Chrysler SUV]. The numerator can be calculated from Pr[SUV] the path Cars Chrysler SUV; the denominator was calculated in the last question. Pr[Chrysler SUV] = = Thus Problem 6 [16 pts; (4,4,8]: Markov Chain Insurance An insurance policy is sold at three price levels: low, medium, high in the following way. (1 New customers pay the medium rate in their first year. (2 If a customer had no accident over the last year, then this customer moves to the next lower rate class (remains in the low class if already there. 1 4 Spri

5 medium ( If a customer had at least one accident over the last year, then that customer moves to the next higher rate category (remains in category high 0.9 if already 0.9 there. The insurance company estimates that a randomly selected customer has a 20% chance of accident over any year. low high 1. Draw a state diagram to0.9represent this process. 0.1 medium low high Write the corresponding transition matrix. from\ to low med high low medium high What, if any, is the long term distribution of customers among the three rate classes? System of linear equations π 1 = 0.8π π 2 π 2 = 0.2π π π = 0.2π π homogeneous system: 0 = 0.2π π 2 0 = 0.2π 1 π π 0 = 0.2π 2 0.8π the equations have sum 0, each of them is thus a combination of the other two. We thus only need to keep two equations, say the first and third, multiply by = 2π 1 +8π 2 0 = 2π 2 8π yields π 2 = 4π and π 1 = 4π 2 = 16π substituting these values into equation π 1 + π 2 + π = 1 then yields π = 1/21 and therefore: 5

6 (π 1, π 2, π = 1 21 (16, 4, 1 Problem 7 [10 pts (4, 6]: Induction i. Fill in the blanks. Let P (n be a statement about integers n. The Principle of Mathematical Induction asserts that to prove P (n true for all integers n 2, it suffices to establish that both 1. the base case P ( is true, 2. for every integer k, if P ( is true, then also P ( is true. 1. the base case P ( 2 is true, 2. for every integer k 2, if P ( k is true, then also P ( k + 1 is true. ii. Consider the sequence a 0, a 1, a 2,... defined recursively by a 0 = 0 and a n+1 = a n + 2(n + 1 We wish to prove, by induction, that a n = n(n + 1 for all n N. Complete the following proof by filling in the blanks. Base case. The equality a n = n(n + 1 is satisfied when n = and left hand sides have the value., as then both the right Inductive step. Let k be any natural number and suppose that the inductive hypothesis a k = is true. [We must now show that a k+1 = is true.] By the recursive definition of the sequence, we have therefore by inductive hypothesis a k+1 = a k + 2(k + 1 a k+1 = + 2(k + 1 = (k + 1( This completes the inductive step. Base case. The equality a n = n(n + 1 is satisfied when n = 0, as then both the right and left hand sides have the value 0. 6

7 Inductive step. Let k be any natural number and suppose that the inductive hypothesis a k = k(k + 1 is true. [We must now show that a k+1 = (k + 1(k + 2 is true.] By the recursive definition of the sequence, we have a k+1 = a k + 2(k + 1 therefore by inductive hypothesis a k+1 = k(k (k + 1 = (k + 1( k + 2 This completes the inductive step. 7