This is a question from Sheldon M Ross' Intro to Probability Models.      Suppose that each coupon obtained is, independent of what has been previously   obtained, equally likely to be any of $m$ different types. Find the expected number of coupons one needs to obtain in order to have at least one of each type.       Hint: Let $X$ be the number needed. It is useful to represent $X$ by   $X = \sum_{i=1}{X_i}$ where each $X_i$ is a geometric random variable.   I did it using this hint and I now get $m^2$ as $E[X]$. I'm not sure if this is right and more importantly, I'm not able to understand why/how it can be modeled as the sum of many geometric distributions.  My initial approach was to define the indicator random variable   $$X_i = \begin{cases}        1 &amp; \textrm{if ith coupon is new}  \\          0 &amp; \textrm{otherwise}     \end{cases} $$    and then take $X = \sum_{i=1}{X_i}$ but I'm not sure how to proceed with that.
What is the average number of times it would it take to roll a fair 6-sided die and get all numbers on the die?  The order in which the numbers appear does not matter.  I had this questions explained to me by a professor (not math professor), but it was not clear in the explanation.  We were given the answer $(1-(\frac56)^n)^6 = .5$ or $n = 12.152$  Can someone please explain this to me, possibly with a link to a general topic?
We currently have two rewards checking accounts.  We are nearing the maximums permitted on each account, after which we receive .5% and .25% APY.  It's within our means to perhaps open another one, but at that point we are looking at using our debit cards 50 to 55 times a month.  I feel that running that number of debit transactions is getting out of hands as it's just the two of us.  I realize there are easy ways to game the system, but I want these things to stay around and another person doing just such that could lead to the ends of these accounts.  We are saving towards a new house down payment.  As such we don't require these funds to be fully liquid, but would still like to remain free of as much risk as possible -- however I realize we are unlikely to find anything near our rewards checking accounts (4% APY interest for meeting various rules: direct deposit, a certain number of debit card transactions, and not receiving a paper statement), but would hope we can get at least half of that.  We aren't fully sure when we will be buying, but we are probably a couple of years out.  For that reason I don't think CDs would work.  Even if we ladder them, when we need the money we will need it pretty much instantly.  Do I bite the bullet and just find the best available money market account?  What other options do I have that let me deposit monthly, and not have a timebomb associated to accessing that money that isn't in the initial lump sum? I have this function $$f(x) = x\cos(1/x)$$ defined for all $x \geq 1$.      How do I prove that    $$f(x+2)-f(x)&gt;2$$    for all $x\geq 1$?   I found $$f'(x)= \cos(1/x) + \frac{\sin(1/x)}{x}$$ and $$f''(x)=-\frac{\cos(1/x)}{x^3}.$$ How do i use this information? i m using ubuntu 10.10. please tell me how to type in hindi fonts in open office word. please help Is $\sqrt{x^2}=|x|$ or $=x$? Isn't $(x^2)^\frac12=x?$ $|x|=\sqrt{x^2}$ as  shows. But, as $(x^2)^\frac12=x$, I can't understand where am I wrong interpreting Square-root. Why won't my follower, Serana, follow me? I told my follower (Serana) to wait. But this is different. I told her to go somewhere and wait so it will activate the enemies. Well it did and I killed them. But Serana seems to be not following me anymore. I can't tell her to follow me. And she is constantly in a squatting position. The only thing I can do is tell her to wait somewhere else to move her. And I can't part ways because I'm doing a main quest and she has to be there. Can anybody give me a tip on what to do? Random spawn location despite sleeping in a bed I am trying to make Minecraft harder, and one of the ways I wish to do this is to have a "pseudo-hardcore" mode where dying spawns you in a random location rather than completely locking you out of the world.  I intend to do this by setting the spawn radius to be very high.  However, I also want to make this "bed-proof" - that is, if someone places a bed and sleeps in it, I want the player to still be spawned in a random location, regardless of their personal spawn point.  How do I do this? Command blocks are fine, but data packs are not preferred. How do I set up SSH so I don't have to type my password when connecting to a host? Proving $\,a\mid bc\!\iff\! \frac{a}{(a,b)}\!\mid c\ $ [general Euclid's Lemma] I tried to start by showing that $ \frac{a}{\gcd(a,b)} $ is always an integer, let's call it $d$, because $a$ is always a multiple of $(a,b)$ based on the definition of a g.c.d. I then tried to show that if $a | b$ then $a|bc$ from the theorem that states that $a|b$ implies that $a|bc$ for any integer $c$, but I cant seem to prove that $a | b$ or that $a|bc$ only if $d \mid c$ (where $d = \frac{a}{\gcd(a,b)}$ ). Any help would be much appreciated. Calculate posterior distribution (gamma-prior, poisson-likelihood) I want to calculate the posterior distribution given a gamma-prior and a poisson likelihood. The task is from a textbook and I just have the solutions (without a walkthrough). Please find all given information in the following:  Given information:  $m$ countries each with a population of $n_i$; $y_i$ is the number of people having a disease; $y_i$ is a poisson random variable with mean $\frac{n_i*\lambda}{100}$  The prior distribution for $\lambda$: $\lambda \sim Gamma(a_0,b_0)$  Task:  Show that the posterior distribution is also a gamma distribution with $a_m = a_0 + \sum_{i = 1}^{m}y_i$ and $b_m = b_0 + \sum_{i = 1}^{m} \frac{n_i}{100}$  My solution:  $Posterior \propto Likelihood \times Prior$  $p(\theta|y) \propto \prod_{i=1}^{m}(\theta^{y_i}e^{-\theta})\times \theta^{a_0-1}e^{-b_0\theta}$  $p(\theta|y) \propto \theta^{\sum_{i = 1}^{m}y_i}e^{-m\theta}\times \theta^{a_0-1}e^{-b_0\theta}$  $p(\theta|y) \propto \theta^{\sum_{i = 1}^{m}y_i+a_0-1}e^{-\theta(m+b_0)}$  $a_m = a_0 + \sum_{i = 1}^{m}y_i$  $b_m = b_0 + m \neq b_0 + \sum_{i = 1}^{m} \frac{n_i}{100}$  I think I am doing a mistake, because I don't know how to incorporate the information that the mean of $y_i$ is given as $\frac{n_i*\lambda}{100}$. Or do I just have to transform $b_m$ somehow? Can someone may give me a hint?   Thank you already in advance!  Edit 1: Considering first advices  $Likelihood: p(y | \theta) \propto \prod_{i=1}^{m}(\frac{n_i\lambda}{100})^{y_i}e^{-\frac{n_i\lambda}{100}}$  $\propto \prod_{i = 1}^{m} ((\frac{n_i}{100})^{y_i})\lambda^{\sum_{i=1}^{m}y_i}e^{-\lambda\sum_{i=1}^{m}\frac{n_i}{100}}$  $\propto \lambda^{\sum_{i=1}^{m}y_i}e^{-\lambda\sum_{i=1}^{m}\frac{n_i}{100}}$  Is the transformation of $\prod_{i=1}^{m}$ I did from the first to the second step correct? Generally accepted root mean square (RMS) error for rectifying topographic maps This is a generic inquiry on the common practice/standard for determining the acceptable values of RMS when georeferencing topographic maps.  Is there an absolute value?  Some literature suggests that it should be "less than or equal to 1/2 of the side of a cell which make up the total resolution of the image." Show that $(\mathbb{Q}^*,\cdot)$ and $(\mathbb{R}^*,\cdot)$ aren't cyclic I'm reading a book about abstract algebra, but I'm having trouble solving this excercise: "Show that $(\mathbb{Q}^*,\cdot)$ and $(\mathbb{R}^*,\cdot)$ aren't cyclic"  Where $(\mathbb{Q}^*,\cdot)$ is the group of nonzero rational numbers under multiplication and $(\mathbb{R}^*,\cdot)$ is the group of nonzero real numbers under multiplication.    Here is my attempt for the first.  Suppose $(\mathbb{Q}^*,\cdot)$ is cyclic, then $\mathbb{Q}^*=\langle\frac{p}{q}\rangle=\{(\frac{p}{q})^n,n\in\mathbb{Z}\}$, where $p$ and $q$ are coprime.  $\frac{2p}{q}$ is also in $\mathbb{Q}^*$ so it must be equal to $(\frac{p}{q})^n$ for some $n\in\mathbb{Z}$.  To solve $\frac{2p}{q}=(\frac{p}{q})^n$, I take a logarithm of both sides and end up with $1+\log_\frac{p}{q}(2)=n$, since $n$ is an integer $\log_\frac{p}{q}(2)$ must be an integer too, but it is possible only when $\frac{p}{q}=2^{\frac{1}{k}}, k\in\mathbb{N}$, (i.e. $\frac{p}{q}$ is a k-th root of $2$), but $k$ must be $1$ for $2^\frac{1}{k}$ to be rational so $\frac{p}{q}=2$ contradicting the hypothesis of $p$ and $q$ being coprime.    However I don't know whether this is a proper proof and the same reasoning cannot be applied to $\mathbb{R}^*$, I'd like you to just give me an hint towards a proof, without telling me the whole proof, if possible. How can I make subsections lose the section number within the section and gain the section number outside it? For a paper I am typesetting, the author wants subsections to be written with single numerals.  So far, OK,   \renewcommand{\thesubsection}{\arabic{subsection}}   does the job.  Yet frequently, he wants to cite the subsection in a later section, and wants both section and subsection numbers in references; i.e., he wants to quote Section 1.1 from Section 2.      My current workaround is to give both the section and subsection numbers in the reference.   MWE  \documentclass[11pt]{amsart} \usepackage{hyperref} \usepackage[backend=bibtex,citestyle=alphabetic]{biblatex} \renewcommand{\thesubsection}{\arabic{subsection}}  \begin{document} \section{A section} \label{sec:first} \subsection{A subsection} \label{subsect:first} Something cool. \section{New section} Now I want to cite the coolness in Section~\ref{sec:first}.\ref{subsect:first}. \end{document}     This works fine, except that the hyperref is set up so that if you click on the first 1, you go to Section 1, not Subsection 1.  Yet I would prefer that both numerals go to the same place in the hyperref links.       I could just wrap each such citation in \renewcommand{\thesubsection}'s, but is there  an easier way to add section numbers when citing subsections from a different section?   Note: Since I'm using biblatex, titlesec is not applicable. Helpless after exhausting daily limit for reviewing Suggested Edits I have faced this problem earlier. But today I got to face this problem twice in a span of couple of minutes.. I have unfortunately finished my daily quota of reviewing Suggested Edits. But I want to further edit a post for which an edit has already been suggested by another user having less than 2000 Reputation.  I come across a post and it says that an edit has already been suggested which is pending for Review:   I click on edit (1) and a pop-up of Suggested Edit appears:    However, the problem is this Suggested Edit doesn't thoroughly deal with the post. It just fixes some capitalization and nothing else — no grammar fixes, no sentence restructuring to make it more clear, no code formatting, no dividing it into lists, etc. — in short, perfect candidate to be rejected. (The image for the suggested-edit I have uploaded didn't have those problems - however, there were those problems in the edits that I encountered earlier.) The problem before me is that I am helpless. I can't approve, reject or improve the edit since I have already gone past my daily quota of Reviewing them. Neither can I make the edits to the post since this suggested edit is pending review. I am completely helpless!! Whats the opinion of the community in this matter?  Should we (those with more than 2000 Rep) be able to edit them? Don't count it in our Review Progress obviously - not sure what should be done with the suggested edit already proposed.  Or, Let it be, someone else would come and edit them and other Reviewers would deal with the Suggested Edit appropriately.  Or, Something else.    Note: This might not be much of a problem for Stack Overflow since the Review Queue is very busy here. However, this is a bit of a problem for smaller sites. So, treat it as a question posted at Meta.SE and not so much at Meta.SO.  Dirty Hack: Ah... found a dirty hack to work around this situation. However, this shouldn't be used unless absolutely necessary. Credit goes to  in . To edit a post enter this link in your browser: http://${site_url}/posts/${post_id}/edit. So, to edit this particular post one would enter: http://meta.stackoverflow.com/posts/171129/edit. I repeat, the hack shouldn't be used unless necessary. I keep going around in circles with this problem:  Given the functions $f:X\to Y$ and $g:Y\to Z$, if $g\circ f:X\to Z$ is bijective, then prove $f$ must be one-to-one and $g$ must be onto. I was able to prove this without difficulty. However the second part of the problem has me stumped:  a) Find functions $f,g$ where $g\circ f$ is bijective but $f$ fails to be onto.  b) Find functions $f,g$ where $g\circ f$ is bijective but $g$ fails to be one-to-one.  My attempt for (a): Let $f:\mathbb{R}\to \mathbb{R}^+$ be defined by $f(x)=e^x$ and $g:\mathbb{R}^+\to \mathbb{R}$ be defined by $g(x)=\ln(x)$. Then $g\circ f(x)=g(f(x))=g(e^x)=\ln(e^x)=x$ which is bijective, $g$ is onto and $f$ fails to be onto... or so I thought. Upon closer inspection $f$ $\underline{is}$ onto for the given codomain $\mathbb{R}^+$.  And now I'm stuck in a rut. Anyone out there have any hints? call function declared below Is it possible to call a function which is declared below in bash?  Example  if [ "$input" = "yes" ]; then     YES_FUNCTION elif [ "$input" = "no" ]; then     NO_FUNCTION else     exit 0; fi  YES_FUNCTION() {   .....   ..... }  NO_FUNCTION() {   .....   ..... } List of functions not integrable in elementary terms When teaching integration to beginning calculus students I always tell them that some integrals are "impossible" (with a bit of expansion on what that actually means).  However I must admit that the examples I give mostly come from "folklore" or guesswork.  Can anyone point me to a list (not a complete list of course!) of fairly simple elementary functions whose antiderivatives are not elementary?  I'm thinking of things like $\exp(x^2)$ which is the standard example, $\sin(\exp(-x))$ perhaps, things like this, not hugely complicated formulae. How to log out of a Stack Exchange site? I have continuous problems of network dropping connections on wifi with ubuntu 16.04, I have an ASUS R510J, this is my network card:  ##### lspci #############################  03:00.0 Network controller [0280]: Realtek Semiconductor Co., Ltd. RTL8821AE 802.11ac PCIe Wireless Network Adapter [10ec:8821]         Subsystem: AzureWave RTL8821AE 802.11ac PCIe Wireless Network Adapter [1a3b:2161]         Kernel driver in use: rtl8821ae  04:00.1 Ethernet controller [0200]: Realtek Semiconductor Co., Ltd. RTL8111/8168/8411 PCI Express Gigabit Ethernet Controller [10ec:8168] (rev 12)         Subsystem: ASUSTeK Computer Inc. RTL8111/8168/8411 PCI Express Gigabit Ethernet Controller [1043:200f]         Kernel driver in use: r8169   sudo service network-manager restart doesnt's help, as I never regain internet connection after it drops, and I have to restart the PC. On Ethernet cable I have no problem.  What do you think is the issue? drivers?  thank you Why do the less than symbol () appear wrong as upside down exclamation (¡) or question mark (¿)? When typing &lt; or &gt; in LaTeX and compiling with pdflatex, the less than and greater than symbols appear at upside down exclamation points. I'm not in math mode. How to delete Windows.old in Windows 8 It seems to be a common thing that a Windows Update creates a Windows.old folder under C:\Windows.old My question is now, can it be deleted with Windows 8 and how can I do that?  If I try to manually delete it in Windows Explorer it does not let me, a permissions issue.