Discrete Mathematics and Probability Theory Summer 2015 Chung-Wei Lin Midterm 2

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1 CS 70 Discrete Mathematics and Probability Theory Summer 201 Chung-Wei Lin Midterm 2 PRINT Your Name:, (last (first SIGN Your Name: PRINT Your Student ID: CIRCLE Your Exam Room: 200 VLSB 10 EVANS OTHER Name of the person sitting to your left: Name of the person sitting to your right: The discussion section you usually attend: # Time TA # Time TA # Time TA am Caroline 4 4 pm Jerry 1 6pm Lexing pm Favian 4 pm Kunal 6 6pm Akshay pm Nathan Z. 7 4 pm Nathan M. 8 6pm Anna pm Diana After the exam starts, please write your student ID (or name on every odd page (we will remove the staple when scanning your exam. You should not write anything, including your SID or name, after time is up. We will not grade anything on the extra pages (Page 12 and Pages or outside of the space provided for a problem unless we are clearly told in the space provided for the question to look there. Except special notices, you must show your work to get credits. You may consult two single-sided sheets (or one double-sided sheet of notes. Apart from that, you may not look at books, notes, etc. Calculators, phones, and computers are not permitted. You have 110 minutes. There are 10 questions for a total of 100 points plus 1 bonus points. Use the number of points as a rough guide for the amount of time to allocate to that question. There are 16 pages (sides on the exam. Notify a proctor immediately if a page is missing. Do not turn this page until your instructor tells you to do so. CS 70, Summer 201, Midterm 2 1

2 1. Probability + Stable Marriage (4 points Men Preference Lists Women Preference Lists 1 C > A > B A 2 > 1 > 3 2 B > C > A B 3 > 1 > 2 3 B > C > A C 3 > 2 > 1 Given the preference lists above, if we randomly and uniformly decide a pairing, what is the probability that the pairing is stable and optimal for men or women? Answer: 1 6. There are totally 3! = 6 possible pairings. Run the men-proposing Stable Marriage Algorithm (SMA and get the male optimal pairing {(1,A,(2,C,(3,B}; run the women-proposing SMA and get the female optimal pairing {(1,A,(2,C,(3,B}. The two pairings are the same, so there is only one stable pairing and it is male optimal or female optimal. Since we randomly and uniformly decide a pairing, the probability is Probability + Polynomial (4 points P(x is a polynomial of degree at most 1, and each coefficient is decided randomly and uniformly from {0,1,2,3}. If we consider mod 4, observe P(0 = 2, and let random variable X be P(2. What is the distribution of X? Answer: Pr[X = 0] = 1 2 ; Pr[X = 2] = 1 2 or Pr[X = 0] = 1 2 ; Pr[X = 1] = 0; Pr[X = 2] = 1 2 ; Pr[X = 3] = 0. Since P(0 = 2, P(x must be ax + 2 where a {0,1,2,3} with probability 1 4 for each possible value. If a = 0,1,2,3, P(x = 2,x + 2,2x + 2,3x + 2 and P(2 = 2,0,2,0, respectively. Therefore, Pr[X = 0] = 1 2 ; Pr[X = 2] =

3 3. Apples (8 points, 4 points for each part (a You want to distribute 10 indistinguishable apples to Margo, Edith, Agnes, and Lucy. How many different ways are there? (You do not need to calculate the exact value of ( a b or a!. ( ( Answer: or It is exactly the number of non-negative integer solutions to x 1 + x 2 + x 3 + x 4 = 10. Therefore, it is ( 13 3 or ( (b Following Part (a, you also want to make sure Agnes gets at least 3 apples and Lucy gets at least 1 apple. How many different ways are there? (You do not need to calculate the exact value of ( a b or a!. ( ( 9 9 Answer: or. 3 6 Replace x 3,x 4 by y 3 + 3,y 4 + 1, respectively. It is exactly the number of non-negative integer solutions to x 1 + x 2 + (y (y = 10, which is equivalent to x 1 + x 2 + y 3 + y 4 = 6. Therefore, it is ( 9 3 or. ( 9 6 3

4 4. Superheros and Gifts (10 points, 4/6 points for each part (a Nick has gifts (we do not know they are distinguishable or indistinguishable. For each gift, he randomly and uniformly gives it to one of the three superheroes: Hulk, Iron Man, and Captain America (each superhero has a 1/3 probability to get the gift. However, Hulk will become angry if he does not get any gift. What is the probability that Hulk stays calm (not angry? (Show your work and write down your answer as an integer divided by another integer. (They may be useful: 2 2 = 4,2 3 = 8, 2 4 = 16, 2 = 32, 3 2 = 9, 3 3 = 27, 3 4 = 81, 3 = 243. Answer: 211 The probability that Hulk does not get any gift it ( 2 3 = Subtract it from 1: = = 211 (b Following Part (a, Hulk will also become angry if he finds one or both of Iron Man and Captain America get more gifts than him. What is the probability that Hulk stays calm? (Show your work and write down your answer as an integer divided by another integer. (They may be useful: ( ( 2 1 = 2, 3 ( 1 = 3, 4 ( 1 = 4, 4 ( 2 = 6, ( 1 =, 2 = Answer: There are four cases: 1. Hulk gets gifts: the probability is ( 1 ( 2 0 ( 3 3 = 1 2. Hulk gets 4 gifts: the probability is ( 1 4 ( 2 1 4( 3 3 = Hulk gets 3 gifts: the probability is ( 1 3 ( 2 2 3( 3 3 = Hulk gets 2 gifts and one of the other two gets 2 gifts (so the last one gets 1 gift: the probability is ( ( 2 (2 3 ( 1 2 ( 1 2 ( = Note that the (2 is to select one of Iron Man and Captain American to get 2 gifts. Take the summation and get 111 Note: you can also consider Case 4 and get the probability of distribution (do not care who gets 2 gifts or 1 gift is ( 3 2 ( = most gifts, so the probability is 1 3 ( The remaining cases have exactly one superhero getting the = 111 4

5 . Traffic Dilemma (10 points, 6/4 points for each part (a Your favorite team plays 0% games at Memorial Stadium, 40% games at AT&T Park, and 10% games at Coliseum. Its winning percentages are 80%, 0%, and 100% at Memorial Stadium, AT&T Park, and Coliseum, respectively. When driving, you hear news that the team just won a game. However, you forget where the game was, so you are not sure if you should detour to avoid traffic near Memorial Stadium. What is the probability that the game was at Memorial Stadium? Answer: 4 7. Let M,A,C be the events that the team plays at Memorial Stadium, AT&T Part, and Coliseum, respectively, and W be the event that the team wins a game. Pr[M W] = Pr[M W] Pr[W] = Pr[M W] Pr[M W] + Pr[A W] + Pr[C W] = Pr[W M] Pr[M] Pr[W M]Pr[M] + Pr[W A]Pr[A] + Pr[W C]Pr[C] = = = 4 7. (b Following Part (a, your friend reminds you that, although the driving time (from your current location to your destination without detouring will be increased from 3 minutes to 70 minutes if there is a game at Memorial Stadium, the driving time (also from your current location to your destination with detouring still needs 0 minutes. Given this information, should you detour? Justify your answer. Answer: Yes. Let random variable X be the driving time (from your current location to your destination without detouring. E(X = =, which is larger than 0. Therefore, detouring is better.

6 6. TRUE or FALSE First Round (18 points, 8/10 points for each part For any of the following statements, claim TRUE or FALSE first. If you claim TRUE, prove it. If you claim FALSE, disprove it. (a You roll a fair dice twice. Let A be the event that you get the same outcome on both rolls and B be the event that you get on your first roll. Statement: A and B are dependent events. TRUE FALSE Answer: FALSE. There are totally 36 sample points, and 6 sample points ((1,1,(2,2,(3,3,(4,4,(,,(6,6 have the same outcome on both rolls, so Pr[A] = There are also 6 sample points ((,1,(,2,(,3,(,4,(,,(,6 having on the first roll, so Pr[B] = There is only 1 sample point ((, having the same outcome on both rolls and having on the first roll, so Pr[A B] = Check Pr[A B] = 1 36 = Pr[A] Pr[B] and prove they are independent. 6

7 (b Assume each regrade request is submitted and processed independently and the probability that a request is determined as an invalid one is 4/. Every student can have 3 non-penalized invalid requests (an invalid request means the corresponding number of points not changed, but the 4th one will result in 4-point penalty. You want to submit 4 requests where 1 request is for points (the decision will be either more points or no change and each of the other 3 requests is for 1 point (the decision will be either 1 more point or no change. Statement: submitting those 4 requests is advantageous. TRUE FALSE (They may be useful: ( 4 2 = 0.64; ( 4 3 = 0.12; ( 4 4 = ; ( 4 = Answer: FALSE. Let random variable X be the total point gain, Y i be the point gain of request i, Z be the point deduction. E(Y 1 = = 1; E(Y 2 = = 0.2; E(Y 3 = = 0.2; E(Y 4 = E(Z = 4 ( 4 = 0.2; ( E(X = E(Y 1 +Y 2 +Y 3 +Y 4 Z ( 4 4 = ; = E(Y 1 + E(Y 2 + E(Y 3 + E(Y 4 E(Z = = , which is smaller than 0, meaning it is disadvantageous. 7

8 7. Combinatorial Proof (10 points Use a combinatorial proof to prove the following statement: ( 3n 1 2n = n 1 n i=1 ( i ( n i ( 2n i Answer: There are n males and 2n females applying for the CS major at UC Berkeley. The CS department has n seats available and 1 fellowship for a female student. Both of LHS and RHS are counting the number of ways giving admissions and the fellowship. LHS: First pick one female fellowship recipient from 2n female applicants (2n. Then, pick the other n 1 students from 3n 1 remaining applicants ( ( 3n 1 n 1. RHS: Pick i male applicants that are not admitted ( ( n i. The number is exactly the same as that of picking n i male applicants that are admitted. Then, pick i female applicants that are admitted ( ( 2n i. Note that there are n i+i students now. Next, pick a fellowship recipient from the i female students that are admitted (i. Lastly, take summation over all choices of i, where i cannot be 0 because at least one female must be picked ( n i=1. Since LHS and RHS count the same thing, they are equal.. 8

9 8. TRUE or FALSE Second Round (24 points, 12 points for each part For any of the following statements, claim TRUE or FALSE first. If you claim TRUE, prove it. If you claim FALSE, disprove it (e.g., provide a counterexample. (a There are 4 bins, and each bin has some black balls and some white balls. When you draw a ball from Bin i, the probability of getting a black ball is p i, where i = 1,2,3,4, and you are given that p 1 p 2 and p 3 p 4. Now, you put all balls in Bin 1 and Bin 3 to an empty bin, Bin, and you also put all balls in Bin 2 and Bin 4 to an empty bin, Bin 6. Define p, p 6 as the probabilities of getting a black ball from Bin and Bin 6, respectively. Statement: p p 6. TRUE FALSE Answer: FALSE. Let Bin 1 have 1 black ball and 2 white balls, so p 1 = 1 3. Let Bin 2 have 1000 black ball and 1000 white ball, so p 2 = 1 2. Let Bin 3 have 2000 black balls and 1000 white balls, so p 3 = 2 3. Let Bin 4 have 3 black balls and 1 white ball, so p 3 = 3 4. We can see p 1 p 2 and p 3 p 4. Bin has 2001 black balls and 1002 white balls, so p = Bin 6 has 1003 black balls and 1001 white balls, so p 6 = It is trivial that 2001, and this is a counterexample >

10 (b Statement: for any three events A 1,A 2,A 3 where 0 < Pr[A i ] < 1 for i = 1,2,3, if Pr[A 1 A 2 A 3 ] = Pr[A 1 ] Pr[A 2 ] Pr[A 3 ], then A 1,A 2,A 3 are pairwise independent. TRUE FALSE Answer: FALSE. Flip a fair coin; if H, flip another coin with H probability 1 4 ; if T, flip another coin with H probability 3 4. Let A 1 be the event getting H on the first time. Let A 2 be the event getting H on the first time (same as A 1. Let A 3 be the event getting H on the second time. Pr[A 1 ] = 1 2 ; Pr[A 2] = 1 2 ; Pr[A 3] = = 1 2 (HH or T H. Pr[A 1 A 2 A 3 ] = = 1 8 (HH. Pr[A 1 A 2 A 3 ] = 1 8 = Pr[A 1] Pr[A 2 ] Pr[A 3 ], but A 1 and A 2 dependent since Pr[A 1 A 2 ] = = Pr[A 1 ] Pr[A 2 ]. This is a counterexample. Note: here is another counterexample: Roll a fair 6-sided dice. Let A 1 be the event of rolling 1 3. Let A 2 be the event of rolling 3 6. Let A 3 be the event of rolling odds. Pr[A 1 ] = 1 2 ; Pr[A 2] = 2 3 ; Pr[A 3] = 1 2. The intersection of these events, A 1 A 2 A 3, is when we roll a 3. Pr[A 1 A 2 A 3 ] = 1 6 = Pr[A 1] Pr[A 2 ] Pr[A 3 ]. However, A 1 and A 2 are dependent, since Pr[A 1 A 2 ] = = Pr[A 1] Pr[A 2 ]. 10

11 9. Secret Sharing (12 points The United Nations consists of n countries (n > 3, and they are working with Thor and Loki. A vault can be opened with a secret s, and it should only be opened in any one of three situations: 1. All n countries in the UN help. 2. At least m countries (0 < m < n 2 get together with Thor. 3. At least m + 2 countries (0 < m < n 2 get together with Loki. Odin works over GF(p where p is sufficiently large and uses only one polynomial to implement a secret sharing scheme. In the scheme, P(x is with degree at most n 1, P(0 is the secret, the i-th country gets a point P(i, Thor gets (n m points P(n + 1,P(n + 2,...,P(n +(n m, and Loki gets (n m 2 points P(n + (n m + 1,P(n + (n m + 2,...,P(n + (n m + (n m 2. Does this scheme work? Hints: given n > 3 and 0 < m < n 2, we suggest you follow one of the following three paths: 1. Always Works! Claim the scheme works for all n,m and justify your claim. 2. Sometimes Works! Claim the scheme works for some (but not all n,m, provide the condition that it works, and justify your claim and condition. 3. Never Works! Claim the scheme does not work for all n,m and justify your claim. Answer: It never works. Thor and Loki can get together and have (n m+(n m 2 = 2n 2m 2 points (note that both of them have at least 1 point. There are two cases: 1. If 2n 2m 2 n, then they have enough points to reconstruct P(x, but this does not satisfy any of the three situations (note that there must be at least m countries, where m > If 2n 2m 2 < n, then they can work with n + 2m + 2 countries and get n points to reconstruct P(x, but this does not satisfy any of the three situations because n + 2m + 2 < m (note that this is equivalent to m < n 2, which is given in the question. 11

12 10. Probability + Error Correction (1 bonus points Figure 1: Minions using error correction codes [ Kevin wants to send a message of 10 ordered packets to Stuart. Each packet will first go through a hungry dragon. For each packet, there is a 1/10 probability to be eaten by the dragon. Each remaining packet will then go through a witch. For each remaining packet, there is a 8/9 probability to be changed by the witch. Given this scenario, how many packets should Kevin send so that the probability that Stuart can recover the message is maximized? Justify your answer (some approximation or estimation is encouraged for this question. The following table of ( n k may be useful. Answer: 14. n k The basic idea here is: having more packets going through the witch tends to be disadvantageous, but there should be enough packets so that Stuart receives fewer than 10 packets. Let n be the number of packets that Kevin sends. Following the idea above, n should be at least 10. Let A be the event that Stuart can recover the message, i be the number of packets eaten by the dragon, and j be the number of packets changed by the witch. From the perspective of Kevin, the probabilities that a packet is correct, eaten, and changed are 0.1, 0.1, and 0.8, respectively, so Pr[A] = n 10 i=0 n i 10 2 j=0 ( n i ( n i j 0.1 n i j 0.1 i 0.8 j. Following the idea again, we can start from n = 10. Pr[A] tends to first increase as n increases (when n is around 10, the probability that Stuart receives fewer than 10 packets is significant and then decrease as n increases (when n is larger, the probability that Stuart receives fewer than 10 packets is insignificant, and thus having more packets tends to be disadvantageous. 12

13 If n = 10, there must be 10 correct packets (10/0/0: Pr[A] = = If n = 11, there must be 11 correct packets or 10 correct packets and 1 eaten packet (11/0/0 or 10/1/1: ( 11 Pr[A] = = If n = 12, possible scenarios are 12/0/0, 11/1/0, 11/0/1 or 10/2/0: ( ( Pr[A] = ( = If n = 13, possible scenarios are 13/0/0, 12/1/0, 12/0/1, 11/2/0, 11/1/1 or 10/3/0: ( ( ( ( ( Pr[A] = ( = At this point, 12 is a reasonable answer. However, 14 is the best (most accurate answer because, if n = 14, Pr[A] = , and the probability drops significantly when n > 14. Note 1: you can approximate ( n k to help your calculation, e.g., ( or ( Note 2: the probabilities (10 10 from n = 1: 1.42, , , , , 0.10,... 13