\documentclass[12pt]{article} \usepackage[margin=1.15in]{geometry} \geometry{letterpaper} \usepackage{graphicx} \usepackage{amsmath, amssymb, amsthm} \usepackage{hyperref, multicol} \hypersetup{colorlinks=false, allcolors=blue} \newcommand{\noin}{\noindent} \begin{document} \noindent {\large \textbf{Stat 110 Homework 2, Fall 2017}} \bigskip \noindent \textbf{Due}: Friday 9/22 at 5:00 pm, submitted as a PDF via the \href{https://canvas.harvard.edu/courses/27764}{{course webpage}}. Please check carefully to make sure you upload the correct file. Your submission must be a single PDF file, no more than $20$ MB in size. It can be typeset or scanned, but must be clear and easily legible (not blurry or faint) and correctly rotated. No submissions on paper or by email will be accepted. A useful site for compressing and performing various conversions for PDFs is \href{https://smallpdf.com}{smallpdf.com}. \medskip \noin The following problems are from Chapter 2 of the book. Please show your work and give clear, careful, convincing justifications (using \emph{words and sentences} to explain your logic, not just formulas). See the syllabus for the collaboration policy. \bigskip \noin 1. (BH 2.8) The screens used for a certain type of cell phone are manufactured by 3 companies, A, B, and C. The proportions of screens supplied by A, B, and C are 0.5, 0.3, and 0.2, respectively, and their screens are defective with probabilities 0.01, 0.02, and 0.03, respectively. Given that the screen on such a phone is defective, what is the probability that Company A manufactured it? \bigskip \noin 2. (BH 2.10) Fred is working on a major project. In planning the project, two milestones are set up, with dates by which they should be accomplished. This serves as a way to track Fred's progress. Let $A_1$ be the event that Fred completes the first milestone on time, $A_2$ be the event that he completes the second milestone on time, and $A_3$ be the event that he completes the project on time. Suppose that $P(A_{j+1}|A_j)=0.8$ but $P(A_{j+1}|A_j^c)=0.3$ for $j=1,2$, since if Fred falls behind on his schedule it will be hard for him to get caught up. Also, assume that the second milestone supersedes the first, in the sense that once we know whether he is on time in completing the second milestone, it no longer matters what happened with the first milestone. We can express this by saying that $A_1$ and $A_3$ are conditionally independent given $A_2$ and they're also conditionally independent given $A_2^c$. \medskip \noin (a) Find the probability that Fred will finish the project on time, given that he completes the first milestone on time. Also find the probability that Fred will finish the project on time, given that he is late for the first milestone. \medskip \noin (b) Suppose that $P(A_1) = 0.75$. Find the probability that Fred will finish the project on time. \bigskip \noin 3. (BH 2.11) An \emph{exit poll} in an election is a survey taken of voters just after they have voted. One major use of exit polls has been so that news organizations can try to figure out as soon as possible who won the election, before the votes are officially counted. This has been notoriously inaccurate in various elections, sometimes because of \emph{selection bias}: the sample of people who are invited to and agree to participate in the survey may not be similar enough to the overall population of voters. Consider an election with two candidates, Candidate A and Candidate B. Every voter is invited to participate in an exit poll, where they are asked whom they voted for; some accept and some refuse. For a randomly selected voter, let $A$ be the event that they voted for A, and $W$ be the event that they are willing to participate in the exit poll. Suppose that $P(W|A)=0.7$ but $P(W|A^c)=0.3$. In the exit poll, $60\%$ of the respondents say they voted for A (assume that they are all honest), suggesting a comfortable victory for A. Find $P(A)$, the true proportion of people who voted for A. \bigskip \noin 4. (BH 2.13) Company A has just developed a diagnostic test for a certain disease. The disease afflicts $1\%$ of the population. As defined in Example 2.3.9, the \emph{sensitivity} of the test is the probability of someone testing positive, given that they have the disease, and the \emph{specificity} of the test is the probability that of someone testing negative, given that they don't have the disease. Assume that, as in Example 2.3.9, the sensitivity and specificity are both $0.95$. Company B, which is a rival of Company A, offers a competing test for the disease. Company B claims that their test is faster and less expensive to perform than Company A's test, is less painful (Company A's test requires an incision), and yet has a higher overall success rate, where overall success rate is defined as the probability that a random person gets diagnosed correctly. \medskip \noin (a) It turns out that Company B's test can be described and performed very simply: no matter who the patient is, diagnose that they do not have the disease. Check whether Company B's claim about overall success rates is true. \medskip \noin (b) Explain why Company A's test may still be useful. \medskip \noin (c) Company A wants to develop a new test such that the overall success rate is higher than that of Company B's test. If the sensitivity and specificity are equal, how high does the sensitivity have to be to achieve their goal? If (amazingly) they can get the sensitivity equal to $1$, how high does the specificity have to be to achieve their goal? If (amazingly) they can get the specificity equal to $1$, how high does the sensitivity have to be to achieve their goal? \bigskip \noin 5. (BH 2.14) Consider the following scenario, from Tversky and Kahneman [this is from work for which Kahneman, a psychologist, won the Nobel Prize in Economics; \emph{The Undoing Project: A Friendship That Changed Our Minds} is a good book by Michael Lewis about their friendship and collaboration]: \begin{quote} Let $A$ be the event that before the end of next year, Peter will have installed a burglar alarm system in his home. Let $B$ denote the event that Peter's home will be burglarized before the end of next year. \end{quote} \medskip \noin (a) Intuitively, which do you think is bigger, $P(A|B)$ or $P(A|B^c)$? Explain your intuition. \medskip \noin (b) Intuitively, which do you think is bigger, $P(B|A)$ or $P(B|A^c)$? Explain your intuition. \medskip \noin (c) Show that for \emph{any} events $A$ and $B$ (with probabilities not equal to $0$ or $1$), $P(A|B) > P(A|B^c)$ is equivalent to $P(B|A) > P(B|A^c)$. \medskip \noin (d) Tversky and Kahneman report that $131$ out of $162$ people whom they posed (a) and (b) to said that $P(A|B) > P(A|B^c)$ and $P(B|A) < P(B|A^c)$. What is a plausible explanation for why this was such a popular opinion despite (c) showing that it is impossible for these inequalities both to hold? \bigskip \noin 6. (BH 2.20) The Jack of Spades (with cider), Jack of Hearts (with tarts), Queen of Spades (with a wink), and Queen of Hearts (without tarts) are taken from a deck of cards. These four cards are shuffled, and then two are dealt. \medskip \noin (a) Find the probability that both of these two cards are queens, given that the first card dealt is a queen. \medskip \noin (b) Find the probability that both are queens, given that at least one is a queen. \medskip \noin (c) Find the probability that both are queens, given that one is the Queen of Hearts. \bigskip \noin 7. (BH 2.41) You are the contestant on the Monty Hall show. Monty is trying out a new version of his game, with rules as follows. You get to choose one of three doors. One door has a car behind it, another has a computer, and the other door has a goat (with all permutations equally likely). Monty, who knows which prize is behind each door, will open a door (but not the one you chose) and then let you choose whether to switch from your current choice to the other unopened door. \medskip \noin Assume that you prefer the car to the computer, the computer to the goat, and (by transitivity) the car to the goat. \medskip \noin (a) Suppose for this part only that Monty always opens the door that reveals your less preferred prize out of the two alternatives, e.g., if he is faced with the choice between revealing the goat or the computer, he will reveal the goat. Monty opens a door, revealing a goat (this is again for this part only). Given this information, should you switch? If you do switch, what is your probability of success in getting the car? \medskip \noin (b) Now suppose that Monty reveals your less preferred prize with probability $p$, and your more preferred prize with probability $q=1-p$. Monty opens a door, revealing a computer. Given this information, should you switch (your answer can depend on $p$)? If you do switch, what is your probability of success in getting the car (in terms of $p$)? \end{document}