HW Solution 5 Due: Sep 13

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1 ECS 315: Probability and Random Processes 01/1 HW Solution 5 Due: Sep 13 Lecturer: Prapun Suksompong, Ph.D. Instructions (a) ONE part of a question will be graded (5 pt). Of course, you do not know which part will be selected; so you should work on all of them. (b) It is important that you try to solve all problems. (5 pt) (c) Late submission will be heavily penalized. (d) Write down all the steps that you have done to obtain your answers. You may not get full credit even when your answer is correct without showing how you get your answer. Problem 1. When n is large, binomial distribution Binomial(n, p) becomes difficult to compute directly because of the need to calculate factorial terms. In this question, we will consider an approximation when p is close to 0. In such case, the binomial can be approximated 1 by the Poisson distribution with parameter α = np. (a) Let X Binomial(1, 1/36). (For example, roll two dice 1 times and let X be the number of times a double 6 appears.) Evaluate p X (x) for x = 0, 1,. (b) Compare your answers in the previous part with the Poisson approximation. (c) Compare MATLAB plots of p X (x) and the pmf of P(np). (d) In one of the New York state lottery games, a number is chosen at random between 0 and 999. Suppose you play this game 50 times. Use the Poisson approximation to estimate the probability that you will never win and compare this with the exact answer. Solution: 1 More specifically, suppose X n has a binomial distribution with parameters n and p n. If p n 0 and np n α as n, then α αk P [X n = k] e k!. 5-1

2 (a) 0.713, 0.445, (b) , 0.388, (c) See Figure?? Binomial pmf Poisson pmf x Figure 5.1: Poisson Approximation rrett, 009, Q.41 Let W be the number of wins. Then, W Binomial(50, p) where p = 1/1000. Hence, ( ) 50 P [W = 0] = p 0 (1 p) If we approximate W by Λ P(α). Then we need to set In which case, α = np = = 1 4. α α0 P [Λ = 0] = e 0! = e α which is very close to the answer from direct calculation. 5-

3 and they can be considered to be independent for mutation A computer system uses pass Determine the following probabilities. The binomial table in six characters and each character is one Appendix A can help. or 10 integers (0 9). Suppose there a (a) No samples are mutated. system with unique passwords. A ha ECS 315 (b) At most one sample is mutated. HW Solution 5 Due: Sep 13 (with replacement) one billion passwo 01/1 (c) More than half the samples are mutated. set, and a match to a user s password is An article in Information Security Technical Report (a) What is the distribution of the num [ Malicious Software Past, Present and Future (004, Vol. 9, (b) What is the probability of no hits? pp. 6 18)] provided the following data on the top ten malicious software instances for 00. The clear leader in the num A statistical process control ch (c) What are the mean and variance of Problem. An article in Information Security Technical Report [ Malicious Software Past, Present and Future ber of registered (004, Vol. incidences 9, pp. for 618)] the year provided 00 was the the Internet data (shown in Figure 5.1) of 0 parts from a metal punching pro on the top ten malicious worm software Klez, and instances it is still one for of the 00. most widespread The clear threats. leader inhour. thetypically, number 1% of of the parts require registered incidences for This the virus year was 00 first detected was the on 6 Internet October worm 001, and Klez. it has This the virus number was of parts firstin the sample of 0 detected on 6 October held 001, the top andspot it among has held malicious the top software spot among for the longest maliciousprocess software problem for the is suspected if X exc longest period in the history period in of the virology. history of virology. Place Name % Instances 1 I-Worm.Klez 61.% I-Worm.Lentin 0.5% 3 I-Worm.Tanatos.09% 4 I-Worm.BadtransII 1.31% 5 Macro.Word97.Thus 1.19% 6 I-Worm.Hybris 0.60% 7 I-Worm.Bridex 0.3% 8 I-Worm.Magistr 0.30% 9 Win95.CIH 0.7% 10 I-Worm.Sircam 0.4% than three standard deviations. (a) If the percentage of parts that req 1%, what is the probability that X more than three standard deviation (b) If the rework percentage increas probability that X exceeds 1? (c) If the rework percentage increase probability that X exceeds 1 in at le hours of samples? Because not all airline passen reserved seat, an airline sells 15 ticket only 10 passengers. The probability th show up is 0.10, and the passengers be (a) What is the probability that every up can take the flight? (b) What is the probability that the flig The 10 most widespread malicious programs for 00 seats? Figure 5.: The 10 most widespread (Source Kaspersky malicious Labs). programs for 00 (Source Kaspersky Labs) This exercise illustrates that p Suppose that 0 malicious software instances are reported. schedules and costs. A manufacturing pr Suppose that 0 malicious software instances are reported. Assume that the malicious Assume that the malicious sources can be assumed to be independent. to be independent. purchased from a supplier. However, ty orders to fill. Each order requires one sources can be assumed (a) What is the probability (a) What is that the probability at least one that at instance least one instance is Klez? is Klez? ponents are identified as defective, and (b) What is the probability that three or more instances are assumed to be independent. (b) What is the probability Klez? that three or more instances are Klez? (a) If the manufacturer stocks 100 co (c) What are the mean and standard deviation of the number probability that the 100 orders (c) What are the expected of Klez valueinstances and standard among the deviation 0 reported? of the number of Klez reordering instances components? among the 0 reported? Heart failure is due to either natural occurrences (b) If the manufacturer stocks 10 co (87%) or outside factors (13%). Outside factors are related to probability that the 100 orders Solution: Let N induced be thesubstances numberor oforeign instances objects. (among Natural occurrences the 0) that are are Klez. reordering Then, components? caused by arterial blockage, disease, and infection. Suppose (c) If the manufacturer stocks 105 co N binomial(n, p) where n = 0 and p = that 0 patients will visit an emergency room with heart failure. probability that the 100 orders (a) P [N 1] = 1 P Assume [N < that 1] = causes 1 P of [N heart = failure 0] = 1 p between N (0) individuals = 1 ( ) 0 0 are reordering components? independent Consider the lengths of stay at (a) What is the probability that three individuals have conditions caused by outside factors? pendently arrive for service. department in Exercise 3-9. Assume (b) P [N 3] (b) = What 1 is P the [N probability < 3] = 1that three (P [N or more = 0] individuals + P [N = have 1] + P (a) [N What = ]) is the probability that the le conditions caused ( ) by outside factors? one person is less than or equal to 4 (c) What are the mean 0 = 1 (0.61) and standard k (0.3878) deviation 0 k of the number (b) What is the probability that exactly of individuals kwith conditions caused by outside factors? than 4 hours? k=0 5-3

4 (c) EN = np = = σ N = Var N = np(1 p) = Problem 3. The random variable V has pmf { 1 + c, v {,, 3} p V (v) = v 0, otherwise. (a) Find the value of the constant c. (b) Find P [V > 3]. (c) Find P [V < 3]. (d) Find P [V > 1]. (e) Let W = V V + 1. Find the pmf of W. (f) Find EV (g) Find E [V ] (h) Find Var V (i) Find σ V (j) Find EW Solution: (a) The pmf must sum to 1. Hence, The value of c must be Note that this gives 1 ( ) + c + 1 () + c + 1 (3) + c = 1. c = 1 3 ( ) = p V ( ) = p V () = and p V (3) =

5 (b) P [V > 3] = 0 because all elements in the support of V are 3. (c) P [V < 3] = 1 p V (3) = (d) P [V > 1] = 1 because the square of any element in the support of V is > 1. (e) W = V V + 1. So, when V =,, 3, we have W = 7, 3, 7, respectively. Hence, W takes only two values, 7 and 3. the corresponding probabilities are P [W = 7] = p V ( ) + p V (3) = and P [W = 3] = p V () = Hence, the pmf of W is given by 41, w = 3, , w = 3, 67 p W (w) =, w = 7, 0.6, w = 7, 108 0, otherwise. 0, otherwise. (f) EV = (g) EV = (h) Var V = EV (EV ) = (i) σ V = Var V.1638 (j) EW = Problem 4. Suppose X is a uniform discrete random variable on { 3,, 1, 0, 1,, 3, 4}. Find (a) EX (b) E [X ] (c) Var X (d) σ X Solution: All of the calculations in this question are simply plugging in numbers into appropriate formula. See the corresponding MATLAB file. 5-5

6 (a) EX = 0.5 (b) E [X ] = 5.5 (c) Var X = 5.5 (d) σ X =.913 Alternatively, we can find a formula for the general case of uniform random variable X on the sets of integers from a to b. Note that there are n = b a + 1 values that the random variable can take. Hence, all of them has probability 1. n (a) EX = b i 1 n = 1 n (b) First, note that b i = 1 n n(a+b) = a+b. ( ) (i + 1) (i ) i (i 1) = i (i 1) 3 ( ) = 1 (i + 1) i (i 1) i (i 1) (i ) 3 = 1 ((b + 1) b (b 1) a (a 1) (a )) 3 where the last equality comes from the fact that there are many terms in the first sum that is repeated in the second sum and hence many cancellations. Now, Therefore, i = = 1 3 (i (i 1) + i) = i (i 1) + i ((b + 1) b (b 1) a (a 1) (a )) + n (a + b) i 1 n = 1 a + b ((b + 1) b (b 1) a (a 1) (a )) + 3n = 1 3 a 1 6 a ab b b (c) Var X = E [X ] (EX) = 1 1 (d) σ X = n Var X = 1. 1 (b a) (b a + ) = 1 1 (n 1)(n + 1) = n

7 Problem 5. (Expectation + pmf + Gambling + Effect of miscalculation of probability) In the eighteenth century, the famous French mathematician Jean Le Rond d Alembert, author of several works on probability, analyzed the toss of two coins. He reasoned that because this experiment has THREE outcomes, (the number of heads that turns up in those two tosses can be 0, 1, or ), the chances of each must be 1 in 3. In other words, if we let N be the number of heads that shows up, Alembert would say that p N (n) = 1/3 for N = 0, 1,. [Mlodinow, 008, p 50 51] We know that Alembert s conclusion was wrong. His three outcomes are not equally likely and hence classical probability formula can not be applied directly. The key is to realize that there are FOUR outcomes which are equally likely. We should not consider 0, 1, or heads as the possible outcomes but rather the sequences (heads, heads), (heads, tails), (tails, heads), and (tails, tails). These are the 4 possibilities that make up the sample space. The actual pmf for N is 1/4, n = 0,, p N (n) = 1/ n = 1, 0 otherwise. Suppose you travel back in time and meet Alembert. You could make the following bet with Alembert to gain some easy money. The bet is that if the result of a toss of two coins contains exactly one head, then he would pay you $150. Otherwise, you would pay him $100. (a) Let R be the amount of money that Alembert gets from this bet. Then, R = 150 if you win and R = +100 otherwise. Use Alembert s miscalculated probabilities to determine the pmf of R (from Alembert s belief). (b) Use Alembert s miscalculated probabilities (or the corresponding miscalculated pmf that you got from part (a)) to calculate ER, the expected amount of money that he expects to gain. Remark: You should find that ER > 0 and hence Alembert will be very happy to accept your bet. (c) Use the actual probabilities, to determine the pmf of R. (d) Use the actual pmf, to determine ER. Remark: You should find that ER < 0 and hence Alembert should not accept your bet if he calculates the probabilities correctly. (e) Let Y be the return of this bet for you. Then, Y = +150 if you win and Y = 100 otherwise. Use the actual probabilities to determine the pmf of Y. 5-7

8 (f) Use the actual probabilities, to determine EY. Remark: You should find that EY > 0. This is the amount of money that you expect to gain each time that you play with Alembert. Of course, Alembert, who still believes that his calculation is correct, will ask you to play this bet again and again believing that he will make profit in the long run. By miscalculating probabilities, one can make wrong decisions (and lose a lot of money)! Solution: (a) P [R = 150] = P [N = 1] and P [R = +100] = P [N 1] = P [N = 0] + P [N = ]. So, p N (1), r = 150, p R (r) = p N (0) + p N (), r = +100, 0, otherwise. Using Alembert s miscalculated pmf, 1/3, r = 150, p R (r) = /3, r = +100, 0, otherwise (b) From p R (r) in part (a), we have ER = r p R(r) = 1 ( 150) = (c) Again, Using the actual pmf, 1, r = 150, 1 p R (r) = + 1, r = +100, 4 4 0, otherwise p N (1), r = 150, p R (r) = p N (0) + p N (), r = +100, 0, otherwise = { 1, r = 150 or + 100, 0, otherwise. (d) From p R (r) in part (c), we have ER = r p R(r) = 1 ( 150) = 5. (e) Observe that Y = R. Hence, using the answer from part (d), we have p R (r) = { 1, r = +150 or 100, 0, otherwise. 5-8

9 (f) Observe that Y = R. Hence, EY = ER. Using the actual probabilities, ER = 5 from part (d). Hence, EY =