Let a family have two children. It is known that one of the children is a boy. What is the probability that both the children are boys. So for this we build the sample space $S=\{(b,b)(b,g)(g,g)\}$ Let our event E be the case where both children are boys $E=\{ (b,b) \}$ Let the conditional be F $F=\{(b,g),(b,b)\}$ Hence $P(E|F)=\frac{P(E\cap F)}{P(F)}=\frac{1/3}{2/3}=\frac{1}{2}$ But the answer in my book is given as $\frac{1}{3}$ and I can't seem to understand why.
In a family with two children, what are the chances, if one of the children is a girl, that both children are girls?  I just dipped into a book, The Drunkard's Walk - How Randomness Rules Our Lives, by Leonard Mlodinow, Vintage Books, 2008. On p.107 Mlodinow says the chances are 1 in 3.   It seems obvious to me that the chances are 1 in 2. Am I correct? Is this not exactly analogous to having a bowl with an infinite number of marbles, half black and half red? Without looking I draw out a black marble. The probability of the second marble I draw being black is 1/2.
I really read many tutorials, but crontab doesn't start notify send.  What I've tried:   etc/crontab  added at the end:  15 * * * * notify-send 'crontab' 'test'  with crontab interface:  crontab -e   added at the end:  15 * * * * notify-send 'crontab' 'test' Migrate to *any* site? Wouldn't it be a good idea to be able to vote for questions to be moved to any Stack Exchange site? For example, I just voted  as off topic, but that was all I could do. I wish I could've voted it be moved to .  It's OK to have the predefined list of sites that is now when voting to move, but it would be nice to have a more link, where we could pick any Stack Exchange site.  Here's a quick mockup of how I see this implemented:    When the more link is clicked, a list of all the Stack Exchange sites should appear. A site URL changed and now its icon in SEDE is broken Orthogonality of Hermite functions I would like to prove to myself that Hermite functions, defined by $\varphi_n(x)=(-1)^n e^{x^2/2}\frac{d^n e^{-x^2}}{dx^n}$, $n\in\mathbb{N}$ are an orthogonal system in $L^2(\mathbb{R})$, i.e. that, for any $m\ne n$, $$\int_{-\infty}^\infty e^{x^2} \frac{d^m e^{-x^2}}{dx^m} \frac{d^n e^{-x^2}}{dx^n}dx=0 $$ I have tried by integrating by parts as the book where I find such a statement, Kolmogorov-Fomin's Элементы теории функций и функционального анализа, says that it can be proved, but I am landing nowhere since I have no experience in manipulating such integrals... Thank you so much for any answer! Is Java "pass-by-reference" or "pass-by-value"? I always thought Java uses pass-by-reference. However, I've seen a couple of blog posts (for example, ) that claim that it isn't (the blog post says that Java uses pass-by-value). I don't think I understand the distinction they're making. What is the explanation? I never had this problem until today. When I watch a youtube video everything is fine, until I make the video fullscreen. It actually continues to play perfectly when fullscreen, but when I then exit fullscreen my sidebar and topbar are gone, and the screen is just frozen. Also my cursor is gone. It does seem to respond to ALT-F4, but it takes about 15 seconds to even close the window, and after that everything is unusable (basically frozen). What's strange though, when I re-enter fullscreen after exiting (and unity being frozen) the video plays perfectly still.   For now my workaround is opening tty1 and executing pkill -u username which logs off the user. I have looked at /var/log/syslog but it seems to be empty.  Is this freezing a common problem, if not, how can I make steps to fix this?    EDIT with specs:  I'm on a simple Samsung laptop with:  Ubuntu 14.10 64-bit,  Intel® Core™ i5-3210M,  Intel® Ivybridge Mobile (So integrated intel hd graphics 4000),  8GB ram. I don't think i've ever changed my video drivers, but it's reported as i915.  Ubuntu is up to date, and my browser is Chrome 40.0.2214.91 (64-bit). (Will test out now with firefox, but already posting this because my pc might freeze again)     EDIT 2: It doesn't freeze on Firefox. Will try reinstalling chrome now... EDIT 3: Reinstalling chrome didn't help. Will try chrome 32-bit too now... Do I need a transit visa in Zurich on the way to India? I am travelling from US to India, with an Indian passport but expired F1 visa and going there to get my H1B visa stamping. I have a stop at Zurich for 2 hours while flying to India. Do I need a transit visa for my trip? Let $p\in[1,\infty]$, $a\in\ell^\infty$ and $$T:\ell^p\to\ell^p\;,\;\;\;x\mapsto ax.$$ Note that $\left\|T\right\|=\left\|a\right\|_{\ell^\infty}$. It's easy to see that if $a\in\ell^1$, then $T$ is nuclear and hence compact.  Can we more generally conclude that $T$ is compact?  Assume $p&lt;\infty$. The intuitive approach would be to define $$T_n:\ell^p\to\ell^p\;,\;\;\;x\mapsto(a_1x_1,\ldots,a_nx_n,0,\ldots)$$ for $n\in\mathbb N$ and to show that $\left\|T_n-T\right\|\to0$. Now, $$\left\|(T_n-T)x\right\|^p=\sum_{i=n+1}^\infty|a_ix_i|^p\tag1$$ for all $x\in\ell^p$ and $n\in\mathbb N$. The right-hand side of $(1)$ clearly converges to $0$ as $n\to\infty$, but this only yields convergence in the strong operator topology. On the other hand, if $(a_n)_{n\in\mathbb N}$ would be nonincreasing and $a_n\to0$ (e.g. $a_n=\frac1n$), then $$\left\|(T_n-T)x\right\|^p\le|a_{n+1}|^p\left\|x\right\|_{\ell^p}^p\tag2$$ for all $x\in\ell^p$ and $n\in\mathbb N$ and we could immediately conclude $\left\|T_n-T\right\|\to0$. EDIT: Maybe the general result can be obtained from the convergence in strong operator topology (which holds by $(1)$) by noting that $\ell^p$ has the approximation property; but I'm not sure whether this is true. A good approximation of $(1+x)^n$ is $1+xn$ when $|x|n &lt;&lt; 1$.  Does this approximation have a name?  Any leads on estimating the error of the approximation? Where is the line drawn with domain names which include a trademark? A search on google for "iphone developer" turns up loads of websites which have "iphone" in them, a trademarked name by Apple.  So I'm led to believe that having a domain such as iphonedeveloper.com is ok? Well, you're still using Apple's trademark, but it would be hard to brand yourself otherwise. You're an IPHONE DEVELOPER... right?  Well, what if I want to provide a website where users pay to get a list of the best offers from Ebay? I might have a domain like ebaydeals.com (I don't... i'm just speculating!).  Now I've heard that places like Ebay are really hot on the trigger and fire out emails to people who register domains like that straight away. But whats the difference?  In both cases I'm making money from the trademark, effectively, so is it just down to how lenient the company who owns the trademark is? Or are there rules? Is there a specific "line" you don't cross?  Thanks This is the first part of the first half of the a proof in the Axiom of Choice. The proof that the Axiom of Choice is equivalent to the choice function. REsources: Keith Devlin's Joy of Sets: pg. 58. F.R. Drake and Singh Intermediate Set Theory: pg. 110).   I am confused on proving disjointness and will ask further questions of proof once I get this first part v. clear for myself.   AC1: Let $\mathcal F$ be a family of pairwise disjoint, non-empty sets. Then there is a set $\mathcal M$ that consists of precised one element from each member of $\mathcal F$  AC2: For every family $\mathcal F$ of non-empty sets, there is a choice function $\mathcal f$. (I.e. $\mathcal f(X)$ is a member of X, for each X in $\mathcal F$.   Disjointness: Two sets are disjoin if Q $\cap$ R = $\emptyset$. Meaning that they have zero elements in common. These sets do not overlap.   Show: AC1 $\leftrightarrow$ AC2   (1) ($\rightarrow$) Let $\mathcal F$ be a set of nonempty sets.   (2)For each X $\in$ $\mathcal F$, let $X^*$ = (X $\times$ {X}) = {(x, X) | x $\in$ X})  (3) Show Disjointness.  An outline of how book proves  disjointness:  (a) if u = {(X $\times$ {X}) | x $\in$ X} then u is of the form $\lt  z, t \gt$ where z = X and t=x, i.e. x $\in$ X.   (b) then if u $\in$ ({X} $\times$ x) $\cap$ ({Y} $\times$ Y) $\rightarrow$ u = $\lt z, t \gt$ with z = X and z=Y, i.e. X=Y. Which means ({X} $\times$ x) = ({Y} $\times$ y).  Questions: (1) Apparently we cannot prove disjointness in $\mathcal F$ but must prove in a different set. If so, why is that?  (2) why is the case that if u $\in$ ({X} $\times$ X) $\cap$ ({Y} $\times$ Y) then z =X and z=Y? I am not following that step at all.   (3) why does the axiom of choice have to do with ordered pairs? Why are we setting X $\in$ $\mathcal F$, let $X^*$ = (X $\times$ {X})?    Thank you! Upgrading my Wubi 11.04 to 11.10 I want to upgrade my current Wubi installation (Ubuntu 11.04) to Ubuntu 11.10. Is this possible to do inside Wubi? Or do I have delete Wubi from Windows, therefore deleting my current Ubuntu installation; and then install 11.10 via the Wubi installer?  Thanks Using StartCoroutine() and WaitForSeconds() I am trying to perform two operations. First I want to increase the size of a GameObject and then delay for 0.1s before returning it back to its original size. However, when I try using StartCoroutine(), it keeps giving me the error :  An object reference is required to access not-static members   even though I made my Delay() function static. Can somebody please tell me what I'm doing wrong?  public static void MoveSelected(int choice) {     GameObject rock1 = GameObject.Find("Rock1");     GameObject paper1 = GameObject.Find("Paper1");     GameObject scissors1 = GameObject.Find("Scissors1");      Vector3 scl;      if (choice == 1)     {         scl = rock1.transform.localScale;         scl.x *= 1.1f;         scl.y *= 1.1f;         rock1.transform.localScale = scl;          StartCoroutine(Delay());          scl.x /= 1.1f;         scl.y /= 1.1f;         rock1.transform.localScale = scl;     }     else if (choice == 2)     {         scl = paper1.transform.localScale;         scl.x *= 1.1f;         scl.y *= 1.1f;         paper1.transform.localScale = scl;          StartCoroutine(Delay());          scl.x /= 1.1f;         scl.y /= 1.1f;         rock1.transform.localScale = scl;     }     else if (choice == 3)     {         scl = scissors1.transform.localScale;         scl.x *= 1.1f;         scl.y *= 1.1f;         scissors1.transform.localScale = scl;          StartCoroutine(Delay());          scl.x /= 1.1f;         scl.y /= 1.1f;         rock1.transform.localScale = scl;     }  }  static IEnumerator Delay() {     yield return new WaitForSeconds(0.1f); } Remove Background in Inkscape How to avoid perceived circularity when defining a formal language? Suppose we want to define a first-order language to do set theory (so we can formalize mathematics). One such construction can be found . What makes me uneasy about this definition is that words such as "set", "countable", "function", and "number" are used in somewhat non-trivial manners. For instance, behind the word "countable" rests an immense amount of mathematical knowledge: one needs the notion of a bijection, which requires functions and sets. One also needs the set of natural numbers (or something with equal cardinality), in order to say that countable sets have a bijection with the set of natural numbers.  Also, in set theory one uses the relation of belonging "$\in$". But relation seems to require the notion an ordered pair, which requires sets, whose properties are described using belonging...  I found the following in Kevin Klement's,  (pages 2-3).  "You have to use logic to study logic. There’s no getting away from it. However, I’m not going to bother stating all the logical rules that are valid in the metalanguage, since I’d need to do that in the metametalanguage, and that would just get me started on an infinite regress. The rule of thumb is: if it’s OK in the object language, it’s OK in the metalanguage too."  So it seems that, if one proves a fact about the object language, then one can also use it in the metalanguage. In the case of set theory, one may not start out knowing what sets really are, but after one proves some fact about them (e.g., that there are uncountable sets) then one implicitly "adds" this fact also to the metalanguage.  This seems like cheating: one is using the object language to conduct proofs regarding the metalanguage, when it should strictly be the other way round.  To give an example of avoiding circularity, consider the definition of the integers. We can define a binary relation $R\subseteq(\mathbf{N}\times\mathbf{N})\times(\mathbf{N}\times\mathbf{N})$, where for any $a,b,c,d\in\mathbf{N}$, $((a,b),(c,d))\in R$ iff $a+d=b+c$, and then defining $\mathbf{Z}:= \{[(a,b)]:a,b\in\mathbf{N}\}$, where $[a,b]=\{x\in \mathbf{N}\times\mathbf{N}: xR(a,b)\}$, as in  question or  on Wikipedia. In this definition if set theory and natural numbers are assumed, then there is no circularity because one did not depend on the notion of "subtraction" in defining the integers.  So my question is:     Question Is the definition of first-order logic circular?   If not, please explain why.   If the definitions are circular, is there an alternative definition which avoids the circularity?   Some thoughts:   Perhaps there is the distinction between what sets are (anything that obeys the axioms) and how sets are expressed (using a formal language). In other words, the notion of a set may not be circular, but to talk of sets using a formal language requires the notion of a set in a metalanguage. In foundational mathematics there also seems to be the idea of first defining something, and then coming back with better machinery to analyse that thing. For instance, one can define the natural numbers using the Peano axioms, then later come back to say that all structures satisfying the axioms are isomorphic. (I don't know any algebra, but that seems right.) Maybe sets, functions, etc., are too basic? Is it possible to avoid these terms when defining a formal language? What invisible flicker do different types of light bulbs have? I have a shapefile and I want to change the order of the features. Is that possible? I want to sort the features in the shapefile because I'm embedding this shapefile in a mobile app. The app will display a list of features, and although I could do the sorting at load time, I'd prefer to have them sorted. Can't install Oracle Java 8 in Ubuntu 16.04 When I press the power button a dialog opens, but I want to shutdown the system without receiving "Do you really really want to shutdown?". How can I choose a direct shutdown action on a pressed power button?   I had the same problem with the 11.04 version, but I don't remember the solution and yesterday I upgraded to Ubuntu 11.10. Microsoft Edge keeps taking over PDF file open/association