Fall 2017 March 13, Written Homework 4

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1 CS1800 Discrete Structures Profs. Aslam, Gold, & Pavlu Fall 017 March 13, 017 Assigned: Fri Oct Due: Wed Nov Instructions: Written Homework 4 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. We require that all homework submissions be neat and easily readable. We recommend using a word processor like Microsoft Word or LaTeX for your submissions. If you scan your homework, you may lose points if the scan is not legible. To get full credit, show INTERMEDIATE steps leading to your answers, throughout. You can give answers that are probabilities as either fully simplified fractions or decimal answers to at least two nonzero digits. Problem 1 [ pts (4,4,6,4,4]: Black and White An urn contains 10 balls: 6 white balls numbered 1 through 6, and 4 black balls numbered 1 through 4. We are simultaneously and randomly drawing balls out of the Find the probability of event A: the two balls are white. (1a Sample space has S = ( 10 = 45. Event A has A = ( 6 = 6 5/ = 15. Thus Pr[A] = 15/45 = 1/3.. Find the probability of event B : the two balls are odd. (1b There are 3 + = 5 odd balls. B = ( 5 = 5 4/ = 10. Thus Pr[B] = 10/45 = /9. 3. Are events A and B independent? Prove your answer. No: A B = ( 3 = 3 thus Pr[A B] = 3/45 = 1/15 but Pr[A] Pr[B] = 1/3 /9 = /7. 4. Let X be the random variable whose value is the number of white balls in the drawing. 1

2 (a Write the probability distribution of X in form of a table: Pr[X=0] Pr[X=1] Pr[X=] Pr[X = 0] = ( 4 / ( 10 = 6/45 = /15 Pr[X = 1] = (6 4/ ( 10 = 4/45 = 8/15 Pr[X = ] = 1/3 by the earlier problem Pr[X=0] Pr[X=1] Pr[X=] /15 8/15 1/3 (b Find the expected value E[X]. 0 (/ (8/15 + (1/3 = 8/ /15 = 18/15 = 6/5 = 1.. Problem [4 pts (4,4,4,5,7] Random Stocks The company EquiCola s stock price fluctuates in the following way from one day to the next. With probability 0.5, the price remains the same the next day. With probability 0.3, the price increases by one dollar. With probability 0., the price decreases by one dollar. 1. What is the expected change in value for the stock from one day to the next? = 0.1 or ten cents.. What is the expected change in value for the stock over the course of a week (7 days? What principle allows you to make this conclusion? By linearity of expectation, we expect a gain of 0.7. The overall change is the sum of the daily changes. 3. What is the probability that the stock does not have a single decline over the course of the whole week? (To two decimal places = What is the probability that the stock declines no more than once over the course of the week? (To two decimal places = What is the probability that the price changes every day for the full week, and ends a dollar higher than it started? (To two significant digits. (0.3 4 (0. 3( 7 4 = 0.00

3 Problem 3 [18 pts (,4,4,4,4]: Texas Hold Em Note: For this problem, be careful about taking into account cards that you know aren t in the deck or in the opponents hands. You have just sat down to a game of Texas Hold Em, a poker variant played with a normal 5 card deck (-10, Jack, Queen, King, Ace, each in 4 suits: Clubs, Spades, Hearts, Diamonds. The game has the following steps. (1 Each player is dealt two cards face down. ( There is a round of betting. Players who don t want to bet can fold and bow out without revealing cards. (3 Three cards are revealed simultaneously in the center (the flop. (4 There is another round of betting. (5 Another card is revealed in the center (the turn, and there is another round of betting where players may fold. (6 A final card is revealed in the center, and there is a final round of betting. Players reveal their hands, and whoever can make the best 5-card poker hand, using some combination of their cards and the 5 shared cards, wins the pot of money. (You can assume in the problems that follow that we will tell you the relevant facts about values of poker hands. i. Calculate the number of possible two-card poker hands you could be dealt. 5 choose = 5 * 51 / = 136 ii. It turns out you are dealt two Jacks the black ones (Clubs and Spades. Calculate (a the number of possible two-card poker hands you could be dealt that are pairs of Jacks, Queens, Kings, or Aces, (b the probability of being dealt such a pair, to two significant figures. (a 4 values * 4 choose = 4 * 4 * 3 / = 4 (b 4/136 = iii. Two other players fold, leaving just one other player, and the flop is revealed: it s Ten of Hearts, Ten of Diamonds, Ten of Clubs. You now have a full house two of one card, three of another which is a pretty good hand. But given that the other player is holding neither your two cards nor any of these three, what is the likelihood the other player is holding the last ten, and therefore beating you with a four-of-a-kind? (To three significant digits. 1 * (46 choose 1 / (47 choose possible deals to other player = 46/1081 = iv. What is the probability now that the turn and the river will both be Jacks? (To three significant digits. 1 / (47 choose = 1/1081 = v. The turn comes, and it s the Jack of Hearts. That turned out well for you. But what is the probability now that your opponent is holding the final ten? (To three significant digits. 1* (45 choose 1 / (46 choose = 45 /1035 =

4 In the final showdown, you each show your cards. Your opponent wins with the Ace of Diamonds and the Ten of Spades. You resist the urge to blurt, What are the chances? because you know. Problem 4 [18 pts (5,4,4,5]: At the Airport 1. Suppose a bomb detector at the airport has a 95% chance of detecting a bomb if there is one in a piece of luggage, but it has a 1% chance of falsely detecting a bomb within an arbitrary innocent piece of luggage. Suppose only 1 in 100,000 pieces of luggage actually contains a bomb. Calculate the conditional probability of a piece of luggage containing a bomb, given that the detector is claiming there is such a bomb inside. P r(bomb detection P r(detection bombp r(bomb = / = P r(nobomb detection P r(detection nobombp r(nobomb = / = 0.01 P r(bomb detection = /( = What is the probability that three bomb detections are all false alarms? ( = How many detection events must occur until there is one real bomb among them in expectation? By linearity of expectation, each detection event contributes of a bomb. 1/ = Now suppose there is a liquid detector with a 95% chance of detecting liquid if luggage contains some, and a 1% chance of detecting liquid if there is none. 1 in 5 pieces of luggage actually contains a liquid. What is the probability that a piece of luggage contains a liquid, given that the detector claims there is liquid? P r(liquid detection P r(detection liquidp r(liquid = /5 = 0.19 P r(noliquid detection P r(detection noliquidp r(noliquid = /5 = P r(liquid detection = 0.19/( = Problem 5 [18 pts (6, 1]: Bags of Words 1. Make an intuitive prediction as to which of the words entropy and bookkeeper has the higher entropy (treating the letters as symbols, and explain your prediction. Then calculate the entropies of the two words and determine whether you were right. (For each word, treat its own symbols as the only symbols that exist. We d expect entropy to have more entropy because it has less repetition; a random letter would be a little less predictable than one from bookkeeper. Entropy has 7 letters, all different, so the expected entropy is exactly the information of one of these letters, log 7 =.8. Bookkeeper has 10 symbols: 1 b, o s, k s, 3 e s, 1 p, 1 r. The entropy is 0.1 log log log log log log 0.1 = =.43. 4

5 . A Markov babbler is a program trained to generate random sequences of words that match the transition probabilities of some target text it was trained on. These transition probabilities are conditional probabilities of each possible next word given the previous word, and they can be calculated by counting how often each word follows each other one. For example, if the training text were TO BE OR NOT TO BE THAT BE THE QUESTION, P( BE TO = 1 since BE is the only word that ever follows TO in the training text, but P( OR BE = 1/3 since three different words can follow BE. The babbler might then end up saying something like TO BE THAT BE OR NOT TO BE THAT BE THE QUESTION. (It would probably be more interesting with more training text. It s called a Markov babbler because it effectively samples a Markov chain and babbles the word corresponding to each new state as it enters it. Consider the babbler trained on the short text LOGIC IS THE PLAN THE PLAN IS MATH. Assume it wraps around to the beginning so that LOGIC is considered to follow MATH. (a Construct the transition matrix that corresponds to the Markov chain that describes this babbler s behavior. Please assume the order of the columns and rows is LOGIC, IS, THE, PLAN, MATH. LOGIC IS THE PLAN MATH LOGIC IS THE PLAN MATH (b Use the iterative method described in lecture and the textbook to find the frequency of each word in the babbler s output as it runs forever. If you use code, please include it in your submission. LOGIC = 0.15, IS = 0.5, THE = 0.5, PLAN = 0.5, MATH = 0.15 (c Guess the simple fractions your decimal numbers are approximating, and verify that this is the correct stationary distribution by substituting into appropriate equations derived from the transition matrix. LOGIC = 1/8, IS = 1/4, THE = 1/4, PLAN = 1/4, MATH = 1/8 LOGIC = MATH: 1/8 = 1/8, check IS = LOGIC + 1/ PLAN: 1/4 = 1/8 + 1/*1/4 = 1/4, check THE = 1/ IS + 1/ PLAN: 1/ * 1/4 + 1/ * 1/4 = 1/4, check PLAN = THE: 1/4 = 1/4, check MATH = 1/ IS: 1/8 = 1/4*1/, check 5