CS 237 Fall 2018, Homework 06 Solution

Transcription

1 0/9/20 hw06.solution CS 237 Fall 20, Homework 06 Solution Due date: Thursday October th at :59 pm (0% off if up to 24 hours late) via Gradescope General Instructions Please complete this notebook by filling in solutions where indicated. Be sure to "Run All" from the Cell menu before submitting. You may use ordinary ASCII text to write your solutions, or (preferably) Latex. A nice introduction to Latex in Jupyter notebooks may be found here: ( As with previous homeworks, just upload a PDF file of this notebook. Instructions for converting to PDF may be found on the class web page right under the link for homework. /9

2 0/9/20 hw06.solution In []: # Useful imports and definitions for CS 237 import numpy as np # arrays and functions which operate on array from numpy import linspace, arange import matplotlib.pyplot as plt # normal plotting #import seaborn as sns # Fancy plotting #import pandas as pd # Data input and manipulation from random import random, randint, uniform, choice, sample, shuffle, seed from collections import Counter %matplotlib inline # Calculating permutations and combinations efficiently def P(N,K): res = for i in range(k): res *= N N = N - return res def C(N,K): if(k < N/2): K = N-K X = []*(K+) for row in range(,n-k+): X[row] *= 2 for col in range(row+,k+): X[col] = X[col]+X[col-] return X[K] # Round to 4 decimal places for printing numeric answers. def round4(x): return round(x ,4) def round4_list(l): return [ round4(x) for x in L] # This function takes a list of outcomes and a list of probabilities and # draws a chart of the probability distribution. def draw_distribution(rx, fx, title='probability Distribution for X'): plt.bar(rx,fx,width=.0,edgecolor='black') plt.ylabel("probability") plt.xlabel("outcomes") if (Rx[-] - Rx[0] < 30): ticks = range(rx[0],rx[-]+) plt.xticks(ticks, ticks) plt.title(title) plt.show() Lab 06 Instructions Only problems this time, since some of them have many parts, and one or two are a little challenging... Problem concerns material that will be covered on Tuesday 0/6, but is covered also in the textbook. 2/9

3 0/9/20 hw06.solution Problem One (Geometric) Suppose a busy professor forgets to add the last homework problem to Gradescope twice in 5 homeworks. He decides to make up a silly problem about it on a homework sometime and keeps writing homeworks 6, 7, and so on. The question is, however, based on what happened in the first 5 homeworks, when does it next happen? Let us ignore the fact that there are only 0 homeworks planned in his course and consider the infinite number of homeworks he will write in the future, so that we can use the geometric distribution. We will only apply it, however, to the range,..., 5, so it this were to happen on homework 6, that would mean "X=" and so on. (a) What is the probability that he next does this on homework? (b) Given that he has done this on 2 of the first 5 homeworks, what is the probability that he will do it again for the first time in homeworks 7 through 9? (c) Supposing that he does not do it on homeworks 6 and 7, what is the probability that he does not do it again for the remaining 3 homeworks? Hint: Use p = 0.4. This is X G(0.4). (a) P(X = 3) = (0.6) 2 (0.4) = 0.44 (b) Past history has no effect, so this is just P(2 X 4) = (0.6)(0.4) + (0.6) 2 (0.4) + (0.6) 3 (0.4) = Alternately, you could calculate P( < X < 5) = P(X > ) P(X > 4) = (0.6) (0.6) 4 = (c) Again, past history (from the point of view of homeworks, 9, and 0) does not matter, so this is just P(X > 3) = = 0.26 Problem Two Wayne and Lenka are throwing darts at a target, and Wayne's probability of hitting the bullseye is p and Lenka's probability of hitting it is q (independently of Wayne). A round of the game is for Lenka to throw and then Wayne to throw. The game is to keep throwing until both of them hit the bulleye on the same round and then stop. (a) What is the probability that the game stops on the round? (b) What is the probability that Lenka first hits the target in the 4th round, but Wayne has not yet hit the target? (c) Suppose after 0 rounds (and no round where both have hit the target) they decide to change the rules and continue to play until at least one of them hits the target. How many more rounds would they expect to play on average? You must express your answers in terms of the parameters p and q (and N for (a)). N th 3/9

4 0/9/20 hw06.solution This is X G(pq). (a) (b) ( pq) N pq (( p)( q) ) 3 q( p) (c) As usual, past history has no relevance, so we are now in a geometric distribution with probability ( p)( q). Therefore the expected number of rounds is ( p)( q) = pq p q Problem Three Consider an Ethernet with N nodes. During lull periods of time, no nodes are broadcasting a message; during such lull periods, a particular node will send a message with probabiliy p, independently of the other nodes. However, if 2 or more nodes send a message simultaneously, we have a collision and the messages are corrupted and transmission stops. A node whose message has been cancelled in this way will try again until the message is successfully transmitted. How many times on average will a given node attempt to transmit a message until it is successfully transmitted? There is a subtlety here which I did not intend (DWPs!). Does "times" in the question mean (a) "how many lull periods" (in which case you have to take account of the probability that the particular node attempted to send a message) or (b) "attempts" (in which case the particular node has already tried, and p is.0 that the attempt was made). Either assumption would be appropriate. For (a): The probability in any particular lull period that the station will attempt and successfully transmit or retransmit a message is p( p) N. This is because for the station to successfully transmit or retransmit its message, none of the other stations should transmit messages in the same period. The number of transmissions and retransmissions of a message until the success is geometric with parameter p( p) N. Therefore, on average, the number of transmissions and retransmissions is For (b) you would simply use p =, so. p( p) N. ( p) N Problem Four (Poisson) Assume that arrival of earthquakes is a Poisson Process, and suppose that in a certain region of California, earthquakes occur at the average rate of 7 per year. </p> (a) What is the probability of no earthquakes in a particular year? (b) Suppose a year ago, there were earthquakes in this region, and this year again there were ; what is the probability that next year there will be at least earthquakes? (c) What is the probability that in exactly three of the next eight years, no earthquakes occur? 4/9

5 0/9/20 hw06.solution This is X Poi(7). (a) e P(X = 0) = = ! (b) Trick question! The previous years are irrelevant, since each unit interval is independent of the others. This is NOT the memoryless property, just independence. P(X ) =.0 = (c) Use the answer from (a) for P(X = 0) ; then we have Y B(, ), so P(Y = 3) = ( ) 3 7 k=0 e 7 7 k k! (0.0009) 3 (0.999) 5 = Problem Five Passengers are making reservations for the Jetblue flight at 9pm the night before Thanksgiving, at a Poisson rate of 3 reservations per hours. If 24 seats are made available for the flight two days (4 hours) before the flight, and passengers make reservations at the same rate for the whole 4 hours, what is the probability that by the time of the flight, all 24 seats have been reserved? Putting the rate in terms of 2 days, we have λ = = per 2 day period. The Poisson will tell us how many people attempted to make reservations, and so we have to use the inverse method: what is the probability that less than 24 people tried to make reservations, and then subtract from : 23 i=0 P(X < 24) = P(X = i) = = 0.99 = i=0 e i i! Problem Six In this problem we will compare the Binomial and the Poisson as an approximation to the Binomial. The probability that a patient will have a bad reaction to a new drug being tested is The new drug is administered to 2000 people. (a) What is the probability that exactly 3 individuals will develop a bad reaction (use the Binomial). (b) What is the probability that least 2 individuals will develop a bad reaction (use the Binomial) and what is the absolute difference between your answers for (a) and (b)? (c) Redo (a) but using the Poisson, letting λ = the expected value of the Binomial. (d) Redo (b) using the Poisson, and give the absolute difference between your answers for (c) and (d). (a) X ~ B(2000,0.00) P(X=3) = C(2000,3) 0.00^ ^997 = 0.05 (b) P(X>=2) =.0 - P(X=0) - P(X=) = (c) X ~ Poi( ) P(X=3) = e^(-2) 2^3 / 3! = 0.04 (d) P(X>=2) =.0 - P(X=0) - P(X=) = /9

6 0/9/20 hw06.solution Problem Seven This problem again compares the Binomial and the Poisson. Suppose in the previous problem, the probability that the patient will develop a bad reaction is 0.3 and the number of patients is 200. (a) What is the probability that precisely 60 individuals will develop a bad reaction? Use the Binomial. (b) What is the probability that at most individuals 0 will develop a bad reaction? Use the Binomial. (c) Repeat (a) using the Poisson. (d) Repeat (b) using the Poisson. (e) How does this compare with the previous problem? (a) X ~ B(200,0.3) P(X=60) = C(200,60) 0.3^60 0.7^40 = (b) X ~ B(200,0.3) P(X<=0) = sum x = 0 to 0: C(200,x) 0.3^x 0.7^(200-x) = 5.52 x 0^-9 (c) X ~ Poi(2000.3) P(X=60) = e^(-60) 60^60 / 60! = (d) e x X Poi( ) P(X 0) = = x=0 x! 0 5 (e) For (a) and (c), the Poisson is about 6% too low; for (b) and (d) the estimate is again too low, this time by a factor of / 324. Not a good approximation at all! Problem Eight Wayne rolls a fair die until he gets a 2. Lenka rolls the same die until she gets an odd number. What is the probability that Lenka rolls the die more times than Wayne does? Hint: Find the probability that Wayne rolls k times and Lenka rolls more than k times (remembering that these are independent), and then sum over all possible k. 6/9

7 0/9/20 hw06.solution Let X be the number of rolls until Wayne gets a 2 and Y be the number of rolls of the die until Lenka rolls an odd number. We seek P(Y > X). The probability that Lenka has to roll more than k times is: P(Y > k) = (0.5) k since for Y > k, she must have an even number on each of the first k rolls. If Wayne has to roll the die k times and Lenka has to roll more than k, then the probability is P(X = k) P(Y > k), and we must sum over all possible k, so k= P(X = k) P(Y > k) = k= k 5 k ( ( 2 ) 6 ) 6 6 = 5 6 k= ( 5 k 2 ) = = 7 Problem Nine (Continuous Distributions) Let X be a continuous random variable with a frequency distribution (PMF) of the form f (x) = { x 4 0 if x 3 otherwise which can be graphed as follows: In [2]: plt.figure(figsize=(, 5)) plt.title("pmf") plt.plot(np.arange(0,4,0.00),[x/4 if <= x <= 3 else 0 for x in np.arange(0,4,0.00)]) plt.show() 7/9

8 0/9/20 hw06.solution (a) Determine the formula for the CDF using geometrical techniques (i.e., not using integrals, but considering what happens to the area to the left of a point a by considering the area of geometrical shapes) (b) Determine the formula for the CDF F X (x) using an integral. (c) Plot the CDF F X (x) (using the code above as a model). (d) Find P(X 2) (e) Find E(X) F X For (d) -- (e) you must use mathematical techniques and not just calculate it using iterative techniques in Python code. You may use Python to calculate results of mathematical formulae. (a) The figure is composed of a rectangle of height 0.25 and width 2.0, plus a triangle of height 0.5 and width 2.0. We must find the area to the left of a given point a, as shown here: In [3]: a = 2.37 plt.figure(figsize=(, 5)) plt.plot([0,4],[0,0],color="k") plt.plot(np.arange(0,4,0.00),[x/4 if <= x <= 3 else 0 for x in np.arange(0,4,0.00)],color= "k") plt.plot([a,a],[0,a/4], color="r") plt.text(a-0.04,a/4+0.03,"a",color="r") plt.plot(np.arange(,3,0.00),[0.25 if <= x <= 3 else 0 for x in np.arange(,3,0.00)],color ="k") plt.show() /9

9 0/9/20 hw06.solution Doing these separately, at a point a 3, we have a rectangle of width (a ) and height 0.25, so its area is ; we also have a a triangle of base (a ) and height (since the slope of the diagonal line is /4), whose area is. So we have 4 a (a ) 2 2(a ) + (a ) 2 a 2 + = =. 4 (b) The indefinite integral of f (x) is: (a ) 2 a 4 x dx = 4 x 2 + C So then, breaking it into 3 pieces, we have: (c) The plot is as follows: F(x) = 0 x 2 if x if x 3 otherwise In [4]: plt.figure(figsize=(, 5)) plt.title("cdf") plt.plot(np.arange(0,4,0.00),[0 if x <= else (x**2-)/ if <= x <= 3 else for x in np.a range(0,4,0.00)]) plt.show() (d) 2 2 P(X 2) = P(X < 2) = F(2) = = = = (e) 3 x 2 x E(X) = dx = = = = /9