If $X$ and $Y$ are i.i.d. exponential as $f_X(x)=\lambda e^{-\lambda x}$ and $f_Y(y)=\lambda e^{-\lambda y}$, and $Z=\min(X,Y)$, then we may say that probability of event $Z=X$ or $Z=Y$ is $\frac 1 2$. This may be valid for any kind of distribution.   If $X$ and $Y$ are independent but not identical exponential as $f_X(x)=\lambda_x e^{-\lambda_x x}$ and $f_Y(y)=\lambda_y e^{-\lambda_y y}$, and $Z=\min(X,Y)$, then how I can find the probability of event $Z=X$ or $Z=Y$ ?.  Most importantly, I need how to find this probability when $X$ and $Y$ have any kind of continuous distributions which are independent.
Let $X, Y$ be an exponentially distributed random variables with parameters $a, b$. Then $X$ has pdf:    $$f_X(x) =\begin{cases} a e^{-a x},&amp; x\geq 0\\ 0,&amp; \text{otherwise}.\end{cases}$$ Suppose $X$ and $Y$ independent. Show that $$\mathbb P(X&gt;Y) = \frac{b}{a+b}.$$   Now I thought the following: $$f(x,y) = f_X(x)\ f_Y(y) = abe^{-ax -by},\qquad\text{for } x,y &gt; 0.$$ And then $$\mathbb P(X&gt;Y) = \int_0^\infty \int_0^x a b e^{-ax -by}\,dydx$$ However, if I solve this (manually or using Wolframalpha), I can't seem to end up with $\frac{b}{a+b}$. Any ideas?
Calculate the uncertainty in velocity of electron I got this question in one of my periodic test in coaching, I (seemingly) solved the question but later recognised what mess I did.  So, the question is :      Calculate the uncertainty in velocity of an electron. If the uncertainty in its position is ${10^ {-15}} \; \mathrm m$.   By simple calculations one can see that the answer would be ${5.79\times{10^{10}}} \; \mathrm{m/s}$. However, as one can see, the uncertainty in velocity is greater than the velocity of light then how is that possible or there is some other way to handle such problems.  P.S. I know the concept of dynamic mass but I am not very familiar with it. (I think that would work here, but don't know how?) Alternative activity monitors for OS X? I’m a bit intrigued by the new detailed activity monitors and progress indicators in Windows 8 (copying files, the task monitor, network activity, etc.) It got me thinking: surely there must be some better alternatives to the built-in Activity Monitor.app?  “Better” as in more detailed and nicer graphs. Specifically, it would be interested and see percentage disk I/O (out of disk capacity), memory usage, and CPU usage in the same nice way   The Building windows blog has an article  in Windows 8.  I just think that OS X’s task manager — which I have always found superior — just doesn’t cut it anymore compared to what is coming in Windows 8. I'm looking to find records in a table that match a specific number that the user enters.  So, the user may enter 12345, but this could be 123zz4-5 in the database. I imagine something like this would work, if PHP functions worked in MySQL. SELECT * FROM foo WHERE preg_replace(&quot;/[^0-9]/&quot;,&quot;&quot;,bar) = '12345'  What's the equivalent function or way to do this with just MySQL? Speed is not important. Which comes first: cat or dog? Which comes first in a sentence?  I know some word pairs such as bacon and eggs, where bacon always comes first. E.g:     Make me bacon and eggs for dinner, honey.   Cats and dogs are the two most common pets I know of. What is the preferred order?   Dogs are superior.     Don't put dogs and cats together, they might fight!  Cats are superior.     Don't put cats and dogs together, they might fight! Let $r$ be a relation on $A \times A$ such that $(a,b) r (c,d) \iff ad = bc.$ How can I show that this relation is transitive, ie. $(a, b)r(c,d)$ and $(c,d)r(e, f) \implies (a,b)r(e,f)$?  I tried to say that $(a,b) r (c,d)$ means that $c=ka$ and $d=kb,$ $k$ a coefficient, such that $ad = bc \iff akb = bka,$ and going on from there, but I'm not sure this is valid for all values of $a, b, c,$ and $d$. How can I show the transitivity of the relation? Asking days badges A couple of months ago, I suggested a set of badges for . The responses were... mixed. The  can be summarized:  In practice, all this would do is encourage people to ask yet more stupid, worthless questions and we have quite enough of that as it is.  We believe that one of the main things holding back many of our beta sites is that when people have questions they aren't in the habit of asking on the site. Even our flagship site, Stack Overflow, doesn't get many questions from its most knowledgeable users. Perhaps they don't have questions, but far more likely they find answers without needing to ask. Our subtle bias against asking (questions =&gt; clueless) means we lose out on what might be our best source of excellent questions. So we went back to the drawing board and designed a badge set that should not encourage worthless questions:  Asked a well-received question on X separate days, and maintained a positive question record.  In order to award badges, we need to define our terms:  positive question record   (total questions - negative questions - closed - deleted)/total questions &gt;= 0.5  It triple counts a question that is downvoted, closed, and deleted. If you ask one that hits that trifecta, you'll need to ask three better questions to dig yourself out. My own record of asking does not meet the criteria:   (44 - 0 - 12 - 12)/44 = 0.4545...  My  looks a lot better; I had a bunch of questions from the beta period that have been subsequently (and rightfully) deleted. With care, I could bring my record into positive territory with 4 questions. The takeaway is that people can't sweep past indiscretions under the rug by deleting them. And they can recover from a bad start as long as they don't ask more bad questions.  well-received question on X separate days A well-received question is one that's open, not deleted, and has a score &gt; 0. If you ask at least one well-received question in a  and none of your questions that day are deleted, downvoted, or closed, you get one more day of credit toward the badge. Whenever the badge criteria is checked, previous asking days are re-evaluated so that there's an incentive to fix downvoted and/or closed questions. The no-deleted-questions criteria is to prevent people from asking several questions at once and deleting any that aren't upvoted or answered to ask another day. (Though this is a losing strategy in the long run.)   My proposed badge names and levels:  Asked a well-received question on 5 separate days, and maintained a positive question record.  Asked a well-received question on 30 separate days, and maintained a positive question record.  Asked a well-received question on 100 separate days, and maintained a positive question record. In addition, since some outstanding individuals ask on more than 100 days, the gold badge may be awarded multiple times. Over on English, Oishi-sama (now a ) would have a remarkable . Bronze and silver badges are only awarded once. We've decided to hold off on multiple awards for now. See the . Neither the names nor the levels nor even the descriptions are set in stone. Please play around with  to get an idea of how the badge works using our public data. Since the query does not account for deleted questions, it's a bit more generous than the actual badges would be. This version of the badge set should encourage a habit of well-received questions over time and encourage people to fix poorly received questions.  I'd like to express my gratitude to the Parenting community for participating in our . I have a cartoon character (a lion) that is part of my business logo that I want to put into Blender and animate it somehow. Is there any way I can take a flat picture (JPEG) and transform it into a 3D image and then animate it in Blender? Suppose a gambler starts with one dollar and plays a game in which he or she wins one dollar with probability $p$ and loses one dollar with probability $1 - p$. Let $f_{n}$ be the probability that he or she first becomes broke at time $n$ for $n = 0, 1, 2\ldots$. Find a generating function for these probabilities.  Is it binomial distribution $?$ and then use the $\operatorname{mx}\left(s\right) = rx\left(\mathrm{e}^{s}\right)$ of the binomial distribution $?$. Help$\ldots!$. I am correcting my quiz. How loud/obvious is a wizard casting a spell? Some spells have somatic and vocal components, that's well known.  How loud must a wizard talk? How obvious/wide should his gestures be?  Some spells that would be awesome would be fairly limited if you need to shout them (ex: mage hand, range 30ft). On the meta home page I noticed the  is listed as doing something, but cannot figure out what it is.  There is nothing on the  or its , but the recent activity is listed with the same timestamp. I often cook fries in a skillet. I put about 1/4 cup vegetable oil into a large skillet and heat on high, then I toss in some frozen fries.   As soon as I toss them in there is an explosion of grease that makes a huge mess. After it simmers down though, I can cook the fries without spilling anymore grease.  Is there anyway to avoid the initial uproar? What is "Best Practice" For Comparing Two Instances of a Reference Type? I came across this recently, up until now I have been happily overriding the equality operator (==) and/or Equals method in order to see if two references types actually contained the same data (i.e. two different instances that look the same). I have been using this even more since I have been getting more in to automated testing (comparing reference/expected data against that returned). While looking over some of the  I came across an  that advises against it. Now I understand why the article is saying this (because they are not the same instance) but it does not answer the question:  What is the best way to compare two reference types? Should we implement ? (I have also seen mention that this should be reserved for value types only). Is there some interface I don't know about? Should we just roll our own?!  Many Thanks ^_^ Update Looks like I had mis-read some of the documentation (it's been a long day) and overriding  may be the way to go..  If you are implementing reference types, you should consider overriding the Equals method on a reference type if your type looks like a base type such as a Point, String, BigNumber, and so on. Most reference types should not overload the equality operator, even if they override Equals. However, if you are implementing a reference type that is intended to have value semantics, such as a complex number type, you should override the equality operator. I need a little hint in a proof on convergence of a sequence. The problem I have is: Suppose $(x_n)$ is a monotone decreasing sequence of real numbers such that $\sum x_n$ converges, prove that:  $$\lim_{n \to \infty} nx_n=0$$  My idea is: proving this fact aims to prove that given $\epsilon &gt;0$ there's some natural $n_0 \in \mathbb{N}$ such that if $n &gt;n_0$ we have $|nx_n|&lt;\epsilon$.  Now, I know that $\sum x_n$ converges, so that given $\epsilon'&gt;0$ there's some $k_0 \in \mathbb{N}$ such that if $k &gt; k_0$ we have:  $$\left|\sum_{i=1}^{k}x_i - S\right|&lt;\epsilon'$$  Where $S = \sum x_n$. Now, since $(x_n)$ is monotone decreasing we know that we must have $x_1 &gt; \cdots &gt; x_k$ so that the sum of all $x_i$ should be less or equal to $k x_k$. So I know that I have:  $$\left|\sum_{i=1}^{k}x_i - S\right|\leq\left|\sum_{i=1}^{k}x_i\right|+|S|\leq|kx_k|+|S|$$  I feel that the proof will come from this, however I'm stuck at this point. Can someone give just a little hint on how to proceed from here?  Thanks very much in advance for your help. Find all entire functions $f(z)$ such that $|f(z)|=1$ for $|z|=1$  Hint: First show that $f(z)$ is a polynomial.  Clearly one can not use Cauchy Estimates to prove that $f(z)$ is a polynomial, the other way is to prove that $f(z)$ has a pole at infinity which I am not sure how to prove that, are there any other ways to prove that a given entire function is a polynomial ?  Thank you ! I have been struggling to install VirtualBox Guest Additions in a Debian Virtual Machine (Debian 7, Debian 8 and Debian 9). I wish to print a Stack&lt;Integer&gt; object as nicely as the Eclipse debugger does (i.e. [1,2,3...]) but printing it with out = "output:" + stack doesn't return this nice result.  Just to clarify, I'm talking about Java's built-in collection so I can't override its toString().  How can I get a nice printable version of the stack? Add captcha threshold information to privileges page Related:   Related:   The captcha often gets to me when I'm editing a lot of posts in rapid succession. I was wondering if at some point you can get enough reputation to not have it interrupt you as much, so I looked on the SO . However, I didn't see anything there about captcha privileges. Then I dug up the related threads above, which indicate that indeed there are reputation levels at which you can reduce captcha interruption. Why not include this information in the list of benefits on the ? This is a  about Active Directory domain naming.   After experimenting with Windows domains and domain controllers in a virtual environment, I've realized that having an active directory domain named identically to a DNS domain is bad idea (Meaning that having example.com as an Active Directory name is no good when we have the example.com domain name registered for use as our website).   , but I'm still not sure about what other rules there are around naming Active Directory domains.  Are there any best practices on what an Active Directory name should or shouldn't be? LINQ to read XML I got this XML file:  &lt;root&gt;     &lt;level1 name="A"&gt;         &lt;level2 name="A1" /&gt;         &lt;level2 name="A2" /&gt;     &lt;/level1&gt;     &lt;level1 name="B"&gt;         &lt;level2 name="B1" /&gt;         &lt;level2 name="B2" /&gt;     &lt;/level1&gt;     &lt;level1 name="C" /&gt; &lt;/root&gt;   Could someone give me a C# code using LINQ, the simplest way to print this result: (Note the extra space if it is a level2 node)  A   A1   A2 B   B1   B2 C   Currently I got this code:  XDocument xdoc = XDocument.Load("data.xml")); var lv1s = from lv1 in xdoc.Descendants("level1")            select lv1.Attribute("name").Value;  foreach (var lv1 in lv1s) {     result.AppendLine(lv1);      var lv2s = from lv2 in xdoc...??? } Does a finally block always get executed in Java?