This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 6.042/18.062J Mathematics for Computer Science Tom Leighton and Eric Lehman December 14, 2004 Final Exam
YOUR NAME: : You may use two 8.511" sheets with notes in your own handwriting on both sides, but no other reference materials. Calculators are not allowed. You may assume all results presented in class. Write your solutions in the space provided. If you need more space, write on the back of the sheet containing the problem. Please keep your entire answer to a prob lem on that problem's page. Be neat and write legibly. You will be graded not only on the correctness of your answers, but also on the clarity with which you express them. GOOD LUCK! Problem Points Grade Grader 1 13 2 15 3 12 4 12 5 12 6 12 7 12 8 12 Total 100 Final Exam 2 Problem 1. [13 points] Give an inductive proof that the Fibonacci numbers Fn and Fn+1 are relatively prime for all n 0. The Fibonacci numbers are defined as follows: F0 = 0 F1 = 1 Fn = Fn1 + Fn2 (for n 2) Solution. We use induction on n. Let P (n) be the proposition that Fn and Fn+1 are relatively prime. Base case: P (0) is true because F0 = 0 and F1 = 1 are relatively prime. Inductive step: Assume that P (n) is true where n 0; that is, Fn and Fn+1 are relatively prime. We must show that Fn+1 and Fn+2 are relatively prime as well. If Fn+1 and Fn+2 had a common divisor d > 1, then d would also divide the linear combination Fn+2  Fn+1 = Fn , contradicting the assumption that Fn and Fn+1 are relatively prime. So Fn+1 and Fn+2 are relatively prime. The theorem follows by induction. Final Exam 3 Problem 2. [15 points] The double of a graph G consists of two copies of G with edges joining corresponding vertices. For example, a graph appears below on the left and its double appears on the right. Some edges in the graph on the right are dashed to clarify its structure. (a) Draw the double of the graph shown below. Solution. Final Exam 4 (b) Suppose that G1 is a bipartite graph, G2 is the double of G1 , G3 is the double of G2 , and so forth. Use induction on n to prove that Gn is bipartite for all n 1. Solution. We use induction. Let P (n) be the proposition that Gn is bipartite. Base case: P (1) is true because G1 is bipartite by assumption. Inductive step: For n 1, we assume P (n) in order to prove P (n + 1). The graph Gn+1 consists of two subgraphs isomorphic to Gn with edges joining corresponding vertices. Remove these extra edges. By the assumption P (n), we can color each vertex of one subgraph black or white so that adjacent vertices get different colors. If we color the corresponding vertices in the other subgraph oppositely, then adjacent vertices get different colors within that subgraph as well. And now if we add back the extra edges, each of these joins a white vertex and a black vertex. Therefore, Gn+1 is bipartite. The theorem follows from the principle of induction. Final Exam 5 Problem 3. [12 points] Finalphobia is a rare disease in which the victim has the delusion that he or she is being subjected to an intense mathematical examination. A person selected uniformly at random has finalphobia with probability 1/100. A person with finalphobia has shaky hands with probability 9/10. A person without finalphobia has shaky hands with probability 1/20. What is the probablility that a person selected uniformly at random has finalphobia, given that he or she has shaky hands? Solution. Let F be the event that the randomlyselected person has finalphobia, and let S be the event that he or she has shaky hands. A tree diagram is worked out below:
has shaky hands? yes has finalphobia? yes 1/100 9/10 1/10 no X 1/1000 F X S X Probability 9/1000 no 99/100 yes 1/20 19/20 no X 99/2000 1801/2000 The probability that a person has finalphobia, given that he or she has shaky hands is: Pr (F  S) = Pr (F S) Pr (S) 9/1000 = 9/1000 + 99/2000 18 = 18 + 99 18 = 117 Final Exam 6 Problem 4. [12 points] Suppose that you roll five 6sided dice that are fair and mutually independent. For the problems below, answers alone are sufficient, but we can award par tial credit only if you show your work. Also, you do not need to simplify your answers; you may leave factorials, binomial coefficients, and arithmetic expressions unevaluated. (a) What is the probability that all five dice show different values? Example: (1, 2, 3, 4, 5) is a roll of this type, but (1, 1, 2, 3, 4) is not. Solution. The probability space is the uniform distribution on the 65 possible num bers rolled on the five (distinguishable) dice. So the probability that all dice are different is the number of outcomes in which the dice have distinct values divided by 65 . There are (6)5 = 6 5 4 3 2 such outcomes so Pr (all rolls distinct) = 65432 5! = 4 5 6 6 An alternative approach uses the observation that the conditional probability that the i + 1st die value differs from the preceding rolls, given that the first i values differ, is (6  i)/6 for 1 i 4, and the probability that all five values are different is the product of these conditional probabilities, namely Pr (all rolls distinct) = 5 4 3 2 5! = 4 6 6 6 6 6 (b) What is the probability that two dice show the same value and the remaining three dice all show different values? Example: (6, 1, 6, 2, 3) is a roll of this type, but (1, 1, 2, 2, 3) and (4, 4, 4, 5, 6) are not. Solution. There are 5 possible pairs of rolls that might have the same value and 2 6 possibilities for what this value is. There 5 4 3 possible distinct values for the remaining three rolls. So 5 6543 100 2 Pr (exactly two values the same) = = 3 5 6 6 An alternative way to count is: there are 6 sets of four values among the five 4 dice, 5 4 choices for which of these values is repeated, and by the Bookkeeper rule, = 5!/2 permutations of a sequence consisting of five values, one of which 2,1,1,1 appears twice. So, 6 4 5!/2 100 4 Pr (exactly two values the same) = = 3 5 6 6 Final Exam 7 (c) What is the probability that two dice show one value, two different dice show a second value, and the remaining die shows a third value? Example: (6, 1, 2, 1, 2) is a roll of this type, but (4, 4, 4, 4, 5) and (5, 5, 5, 6, 6) are not. Solution. There are 6 sets of two values that might duplicated. There are 5 be 2 2 rolls where larger duplicated value may come up and 3 remaining rolls where the 2 smaller duplicated value may come up. There is only 1 remaining roll where the nonduplicated value may then come up, and 4 remaining values it could take. So, 6 5 3 4 50 2 2 2 Pr (exactly two pairs of same values) = = 3 5 6 6 Alternatively, there are 6 sets of three values among the five dice, 3 3 5 choices for which of these values is not repeated, and by the Bookkeeper rule, 2,2,1 permuta tions of a sequence consisting of five values, two of which appear twice. So, 6 5 3 50 3 2, 2, 1 Pr (exactly two pairs of same values) = = 3 5 6 6 Final Exam 8 Problem 5. [12 points] An electronic toy displays a 4 4 grid of colored squares. At all times, four are red, four are green, four are blue, and four are yellow. For example, here is one possible configuration: R Y B B B B R G Y G R Y R G G Y 2 3 4 5 1 For parts (a) and (b) below, you need not simplify your answers. (a) How many such configurations are possible? Solution. This is equal to the number of sequences containing 4 R's, 4 G's, 4 B's, and 4 Y's, which is: 16! (4!)4 (b) Below the display, there are five buttons numbered 1, 2, 3, 4, and 5. The player may press a sequence of buttons; however, the same button can not be pressed twice in a row. How many different sequences of n buttonpresses are possible? Solution. There are 5 choices for the first button press and 4 for each subsequence press. Therefore, the number of different sequences of n button presses is 5 4n1 . (c) Each button press scrambles the colored squares in a complicated, but nonran dom way. Prove that there exist two different sequences of 32 button presses that both produce the same configuration, if the puzzle is initially in the state shown above. (Hint: 432 = 1616 > 16!) Solution. We use the Pigeonhole Principle. Let A be the set of all sequences of 32 button presses, let B be the set of all configurations, and let f : A B map each sequence of button presses to the configuration that results. Now: A > 432 > 16! > B Thus, by the Pigeonhole Principle, f is not injective; that is, there exist distinct el ements a1 , a2 A such taht f (a1 ) = f (a2 ). In other words, there are two different sequences of button presses that produce the same configuration. Final Exam 9 Problem 6. [12 points] MIT students sometimes delay laundry for a few days. Assume all random values described below are mutually independent. (a) A busy student must complete 3 problem sets before doing laundry. Each prob lem set requires 1 day with probability 2/3 and 2 days with probability 1/3. Let B be the number of days a busy student delays laundry. What is Ex (B)? Example: If the first problem set requires 1 day and the second and third problem sets each require 2 days, then the student delays for B = 5 days. Solution. The expected time to complete a problem set is: 1 2 1 4 +2 = 3 3 3 Therefore, the expected time to complete all three problem sets is: Ex (B) = Ex (pset1) + Ex (pset2) + Ex (pset3) 4 4 4 = + + 3 3 3 =4 (b) A relaxed student rolls a fair, 6sided die in the morning. If he rolls a 1, then he does his laundry immediately (with zero days of delay). Otherwise, he delays for one day and repeats the experiment the following morning. Let R be the number of days a relaxed student delays laundry. What is Ex (R)? Example: If the student rolls a 2 the first morning, a 5 the second morning, and a 1 the third morning, then he delays for R = 2 days. Solution. If we regard doing laundry as a failure, then the mean time to failure is 1/(1/6) = 6. However, this counts the day laundry is done, so the number of days delay is 6  1 = 5. Alternatively, we could derive the answer as follows: Ex (R) = k=0 Pr (R > k) 2 3 5 5 5 = + + + ... 6 6 6 2 5 5 5 = 1+ + + ... 6 6 6 5 1 6 1  5/6 =5 = Final Exam 10 (c) Before doing laundry, an unlucky student must recover from illness for a number of days equal to the product of the numbers rolled on two fair, 6sided dice. Let U be the expected number of days an unlucky student delays laundry. What is Ex (U )? Example: If the rolls are 5 and 3, then the student delays for U = 15 days. Solution. Let D1 and D2 be the two die rolls. Recall that a die roll has expectation 7/2. Thus: Ex (U ) = Ex (D1 D2 ) = Ex (D1 ) Ex (D2 ) 7 7 = 2 2 49 = 4 (d) A student is busy with probability 1/2, relaxed with probability 1/3, and unlucky with probability 1/6. Let D be the number of days the student delays laundry. What is Ex (D)? Leave your answer in terms of Ex (B), Ex (R), and Ex (U ). Solution. Ex (D) = 1 1 1 Ex (B) + Ex (R) + Ex (U ) 2 3 6 Final Exam Problem 7. [12 points] I have twelve cards: 1 1 2 2 3 3 4 4 5 5 6 6 11 I shuffle them and deal them in a row. For example, I might get: 1 2 3 3 4 6 1 4 5 5 2 6 What is the expected number of adjacent pairs with the same value? In the example, there are two adjacent pairs with the same value, the 3's and the 5's. We can award partial credit only if you show your work. Solution. Consider an adjacent pair. The left card matches only one of the other 11 cards, which is equally likely to be in any of the 11 other positions. Therefore, the proba bility that an adjacent pair matches is 1/11. Since there are 11 adjacent pairs, the expected number of matches is 11 1/11 = 1 by linearity of expectation. Final Exam 12 Problem 8. [12 points] Each time a baseball player bats, he hits the ball with some proba bility. The table below gives the hit probability and number of chances to bat next season for five players. player prob. of hit Player A 1/3 Player B 1/4 Player C 1/4 Player D 1/5 Player E 2/5 # chances to bat 300 200 400 250 500 (a) Let X be the total number times these five players hit the ball next season. What is Ex (X)? Solution. The expected number of events that happen is equal to the sum of the event probabilities. So we have: Ex (X) = 300 1 1 1 1 2 + 200 + 400 + 250 + 500 3 4 4 5 5 = 100 + 50 + 100 + 50 + 200 = 500 (b) Give a nontrivial upper bound on Pr (X 1500) and justify your answer. Do not assume that hits happen mutually independently. Solution. We use the Markov Inequality: Pr (X c) This gives: Pr (X 1500) Ex (X) 1500 1 = 3 Ex (X) c (c) Using a Chernoff inequality, give a nontrivial upper bound on Pr (X 400). For this part, you may assume that all hits happen mutually independently. Solution. We use the Chernoff Inequality:
2 Pr (X (1  ) Ex (X)) e Ex (X) /2 This gives: Pr (X 400) = Pr (X (1  1/5) Ex (X)) e(1/5) = e10
2 500/2 MIT OpenCourseWare http://ocw.mit.edu 6.042J / 18.062J Mathematics for Computer Science
Fall 2010 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. ...
View
Full
Document
This note was uploaded on 01/19/2012 for the course CS 6.042J / 1 taught by Professor Tomleighton,dr.martenvandijk during the Fall '10 term at MIT.
 Fall '10
 TomLeighton,Dr.MartenvanDijk
 Computer Science

Click to edit the document details