Math 55: Discrete Mathematics UC Berkeley, Fall 2011 Homework # 7, due Wedneday, March 14 Happy Pi Day! (If any errors are spotted, please email them to morrison at math dot berkeley dot edu..5.10 A croissant shop has plain croissants, cherry croissants, chocolate croissants, almond croissants, apple croissants, and broccoli croissants. How many ways are there to choose (a a dozen croissants? Solution. This is equivalent to putting 12 indistinguishable objects into distinguishable boxes. (The boxes are the types of croissants, and putting objects into them corresponds to choosing that many of that type. The number of ways of doing this is + 12 1 = 188. 12 (b three dozen croissants? Solution. Similar to part (a, this is equivalent to putting 3 indistinguishable objects into distinguishable boxes. The number of ways of doing this is + 3 1 = 749, 398. 3 (c two dozen croissants with at least two of each kind? 1
Solution. We may consider the first dozen croissants already picked out (two of each of the six kinds. Thus the number is the number of ways of choosing the remaining dozen, which by part (a is 188. (d two dozen croissants with no more than two broccoli croissants? Solution. We will add up three cases: no broccoli croissants, exactly one broccoli croissant, and exactly two broccoli croissants. These numbers are 5 + 24 1 = 20, 475 24 5 + 23 1 = 17, 550 23 5 + 22 1 = 14, 950. 22 (The first is choosing 24 croissants from the 5 non-brocolli types; the second is choosing just 23, since we already have one broccoli; the third is choosing just 22, since we already have two broccolis. Adding up, we have ways total. 20, 475 + 17, 550 + 14, 950 = 52, 975 (e two dozen croissants with at least five chocolate croissants and at least three almond croissants? Solution. We have already chosen the first eight, so we need only choose the remaining sixteen. There are + 1 1 = 20, 349 1 ways to do this. (f two dozen croissants with at least one plain croissant, at least two cherry croissants, at least three chocolate croissants, at least one almond croissants, at least two apple croissants, and no more than three broccoli croissants? 2
Solution. We have already chosen the first nine croissants, so we need merely count how many ways there are to choose the last fifteen, without choosing more than three broccoli croissants. Similar to part (c, this is the sum of 5 + 15 1 = 3, 87 15 5 + 14 1 = 3, 00 14 5 + 13 1 = 2, 380 13 5 + 12 1 = 1, 820. 12 These add to 3, 87 + 3, 00 + 2, 380 + 1, 820 = 11, 13 so there are 11, 13 ways of choosing..5.22 How many ways are there to distribute 12 indistinguishable balls into distinguishable boxes? Solution. There are + 12 1 =, 188 12 ways..5.34 How many strings with five or more characters can be formed from the letters in SEERESS? Solution. We ll count up how many can be formed with seven characters, with six characters, and with five characters, then add them up. For length seven, we are putting {1, 2,..., 7} into the three boxes S, E, and R, with S receiving three numbers, E receiving three numbers, and R receiving one number. (For instance, if S gets 1, 2, 7, E gets 3
4, 5, and R gets 3, this corresponds to the anagram SSREEES. There are 7! 3! 3! 1! = 140 ways to do this. For length six, we have several cases. If we left out an R, we are putting {1,..., } into two distinguishable boxes, each of which gets three numbers; there are! 3! 3! = 20 ways to do this. If we left out an S, we are putting {1,..., } into three boxes which get three, two, and one! objects, respectively; there are 3! 2! 1! = 0 ways to do this. Leaving out an E symmetrically gives 0. Hence there are 20 + 0 + 0 = 140 strings of length six. For length five, we once again have several cases: 5! Leave out an S and an R: we get an R: also 10. 3! 2! = 10. Leave out an E and Leave out two S s: we get 5! 3! 1! 1! = 20. Leave out two E s: also 20. Leave out an S and an E: we get 5! 2! 2! 1! = 30. This gives a total of 10 + 10 + 20 + 20 + 30 = 90 strings of length five. All told, we get that 140 + 140 + 30 = 310 strings with five or more characters can be formed from the letters in SEERESS..5.42 In bridge, the 52 cards of a standard deck are dealt to four players. How many different ways are there to deal bridge hands to four players? Solution. This is equivalent to putting 52 distinguishable objects into 4 distinguishable boxes, with each box getting 13 objects. There are ways to do this. 52! 5.3 1028 13! 13! 13! 13! 4
.5.50 How many ways are there to distribute five distinguishable objects into three indistinguishable boxes? Solution. Using Stirling numbers, the answer is 3 S(5, j = S(5, 1 + S(5, 2 + S(5, 3. j=1 Let s compute these out. S(5, 1 = 1 1 1 1 ( 1 i (1 i 5 1! i i=0 1 = 1 5 = 1. 0 S(5, 2 = 1 2 1 ( 2 ( 1 i 2! i i=0 = 1 ( 2 2 5 2 0 = 1 ( 2 5 2 = 15. 2 (2 i 5 2 1 5 1 S(5, 3 = 1 3 1 ( 3 ( 1 i 3! i i=0 = 1 ( 3 3 5 0 (3 i 5 3 2 5 + 1 = 1 (243 9 + 3 = 25. 3 1 5 2 Adding these up gives a total of 41..5. How many terms are there in the expansion of (x + y + z 100? Solution. The number of terms in the expansion is the number of ways of writing x a y b z c, where a + b + c = 100 and a, b, c are nonnegative 5
integers. This is the number of ways of putting 100 indistinguishable balls into three distinguishable boxes. There are 3 + 100 1 = 5151 100 ways to do this. 7.1.12 What is the probability that a five-card poker hand contains exactly one ace? Solution. There are four possibilities for the ace, and then we must choose 4 cards from the 48 non-ace cards. This gives us 48 4 4 five-card hands with exactly one ace. Dividing by the total number of hands, which is 52 5, we find 4 (48 4 ( 52.299, 5 so the probability that a five-card poker hand contains exactly one ace is about.299. 7.1.21 What is the probability that a fair die never comes up an even number when it is rolled six times? Solution. The probability that any given die roll comes up odd is 1/2. Since the rolls are independent, the probability it comes up odd all six times is 1/2 = 1/4. This is the probability that a fair die never comes up an even number when it is rolled six times. 7.1.2 Find the probability of selecting none of the correct six integers in a lottery, where the order in which these integers are selected does not matter, from the positive numbers not exceeding (a 40. (b 48. (c 5.
(d 4. Solution. For 40, the number of ways of selecting all wrong numbers is the number of ways of selecting six numbers from the 34 incorrect numbers. There are ( 34 ways to do this. Since there are 40 ways to choose numbers in total, the probability of selecting none of the correct six integers is ( 34 ( 40.35 The same argument gives the answers in the other cases: ( 42 ( 48.43 ( 50 ( 5.49 ( 58 ( 4.54 7.1.3 Which is more likely: rolling a total of 8 when two dice are rolled or rolling a total of 8 when three dice are rolled? Solution. There are five ways to roll an 8 when two dice are rolled: (2,, (, 2, (3, 5, (5, 3, (4, 4; since there are 3 rolls possible, the probability of rolling an 8 is 5/3.139. The ways to roll an 8 when three dice are rolled, before we consider reordering, are: (1, 1,, (1, 2, 5, (1, 3, 4, (2, 2, 4, (2, 3, 3 Two have three distinct numbers, and so can each happen in six ways. Three have two distinct numbers, and so can happen in three ways. This gives 2 + 3 3 = 21 ways to roll an 8. Since there are 3 = 21 rolls possible, the probability of rolling an 8 is 21/21.097. Thus it is more likely to roll a total of 8 when two dice are rolled than when three dice are rolled. 7
7.2.2 Find the probability of each outcome when a loaded die is rolled, if a 3 is twice as likely to appear as each of the other five numbers on one die. Solution. Let p be the probability of rolling a 1. The probability of 3 is 2p, and the probability of each other number is p. Since probabilities must add to 1, we have p + p + 2p + p + p + p = 1, so p = 1/7. Hence the probability of a 3 being rolled is 2/7, and the probability of any given other number being rolled is 1/7. 7.2.8 What is the probability of these events when we randomly select a permutation of {1, 2,..., n} where n 4? (a 1 precedes 2. Solution. The number of strings with 1 preceding 2 can be calculated ( as follows. Choose the two spots for 1 and 2; there are n 2 ways to do this. Then fill in the other other n 2 spaces with the remaining n 2 numbers; there are (n 2! ways to do this. This gives ( n 2 (n 2! ways to do this. Dividing by the n! permutations possible, this gives a probability of ( n 2 (n 2! n! = n(n 1/2 n(n 1 = 1/2. Alternative method: let A be the set of strings with 1 preceding 2, and B be the set of strings with 2 preceding 1. A and B are disjoint, and A B has all the strings. Moreover, A and B are in bijection: send a string to another string of A to a string of B by switching 1 and 2. Hence A and B are of the same size, so much each contain half of all strings. Thus the probability of choosing an element of A is 1/2. (b 2 precedes 1. Solution. A nigh-identical argument to part (a gives 1/2. (c 1 immediately precedes 2. 8
Solution. We may treat 1 and 2 as a single entity that moves as one, meaning we re permuting n 1 elements. There are (n 1! ways to do this, and so there are (n 1! permutations with 1 immediately preceding 2. Dividing by the n! permutations total gives a probability of 1/n that 1 immediately precedes 2. Alternative method: First we calculate the probability of 1 being in a position that has a number after it; all but the n th position do, so there s a probability of (n 1/n of this happening. Once we have this, we ask what the probability of 2 being in the correct spot is. Only one of the remaining n 1 spots is directly after 1, so there s a 1/(n 1 chance of this. Multiplying these gives us our probability of 1/n. (d n precedes 1 and n 1 precedes 2. Solution. Let s count up the number of strings satisfying this property. First we ll choose the four spots for 1, 2, n 1, and n. There are ( n 4 ways to do this. Then there are six orders we may place these four numbers into these four spots, giving us a factor of. Then we must fill in the remaining n 4 spots with the remaining n 4 numbers; there are (n 4! ways to do this. Multiplying together, there are ( n 4 (n 4! permutations with the desired property. Dividing by the n! total permutations gives a probability of ( n 4 (n 4! n(n 1(n 2(n 3/4! = = 1 n! n(n 1(n 2(n 3 4. Alternative method: the event n precedes 1 and n 1 precedes 2 are independent, and the probability of each is 1/2 (by part (a, so the probability of both is 1/2 1/2 = 1/4. (e n precedes 1 and n precedes 2. Solution. First we ll count up the number of strings where n precedes 1 and 1 precedes 2. First we ll choose the three spots for these numbers; there are ( n 3 ways to do this. Then we ll fill in the n 3 remaining spots; there are (n 3! ways to do this. This gives a total of ( n 3 (n 3! = n!/3! ways to do this. Now we ll count up the number of strings where n precedes 2 and 2 precedes 1. By symmetry, there are n!/3! strings like this. 9
Hence there are 2n!/3! = n!/3 strings with n preceding 1 and n preceding 2 (since either 1 precedes 2 or 2 precedes 1. This means the probability of a random permutation having n preceding 1 and n preceding 2 is n!/3 = 1 n! 3. Alternative method: there are six possible orderings for 1, 2, n. Let A i be the set of permutations containing 1, 2, n in the i th order, so that there are six sets, A 1, A 2,..., A. These sets are disjoint and A 1 A 2... A contains all the permutations. Moreover, A i is in bijection with A j for all i and j: send a string in A i to a string in A j with 1, 2, n switched to the appropriate ordering. Thus all the sets are the same size, and each must contain 1/ of all permutations. Since we re allowing two possible orderings, n 1 2 and n 2 1, the permutations we are interested in make up 1/ + 1/ = 1/3 of all permutations. 7.2.13 Show that if E and F are events, then p(e F p(e + p(f 1. Proof. Note that p(e F is at most 1, since it is a probability. Using the law of the excluded middle, this means that 1 p(e F = p(e + p(f p(e F. Adding p(e F to both sides gives as desired. p(e F p(e + p(f 1, ( 7.2.1 Show that if E and F are independent events, then E and F are also independent events. Solution. We know that p(e F = p(ep(f, and we want to show that p ( E F = p ( E p ( F. 10
Using de Morgan s law, the rule of the excluded middle, and the independence of E and F, we have p ( E F =p ( E F as desired. =1 p (E F =1 p(e p(f + p(e F =1 p(e p(f + p(ep(f =(1 p(e(1 p(f =p ( E p ( F, 7.2.24 What is the conditional probability that exactly four heads appear when a fair coin is flipped five times, given that the first flip came up tails? Solution. Let A be the event that exactly four heads appear, and let B be the event that the first flip comes up tails. We have p(a B = p(a B p(b = 1/25 1/2 = 1 2 4 = 1 1. 7.2.34 Find each of the following probabilities when n independent Bernoulli trials are carried out with probability of success p. (a the probability of no successes. Solution. The probability of no successes is the probability of each outcome being a failure. Since each failure has probability (1 p, and the trials are independent, there is a probability of (1 p n of each outcome being a failure. (b the probability of at least one success. Solution. The complement of at least one success is no successes. The probabilities of these two outcomes must add to 1. We already calculated the probability of no successes as (1 p n, so the probability of at least one success is 1 (1 p n. (c the probability of at most one success. 11
Solution. The probability of exactly one success is np(1 p n 1 : there are n trials where the success could happen, and the probability of each such outcome is p(1 p n 1. We ve already calculated the probability of exactly zero successes as (1 p n. The probability of either zero or one is the probabilities added minus the probability of both happening; since both can t happen, we have a probability of np(1 p n 1 + (1 p n that there will be at most one success. (d the probability of at least two successes. Solution. This is the complement of at most one success. Hence its probability is 1 minus the answer from part (c, namely 1 np(1 p n 1 (1 p n. 12