Lets say I toss a coin $n$ times where $n \geq 1$. Is it always the case that the probability of getting an even number of heads is $\frac{1}{2}$. If so can someone explain mathematically why? edit: as an additional question, If I introduce some subset of the n coins as unfair coins(i.e. probability of getting heads may not be 0.5) how does this effect the probability of getting an even number of heads?
A coin is tossed $n$ times. What is the probability of getting odd number of heads?   I started this chapter sometimes ago and faced in front of a tough problem. At first I started considering cases. Case-I : The probability of getting 1 head.Case-II : The probability of getting 3 head and so on. But there are many cases. So how can I solved this . Please help me. Thank you!
What's the easiest way to install a missing Perl module? The Data Interoperability Tools .tbx appears in my system toolboxes, however when I double click quick export or import, a red X appears.  I have gone into the administrator settings and the tool doesn't appear to be installed. How do I install it?  I am using ArcGIS 10.2 for desktop trial.  After I attempt to evaluate extensions, I receive a message which states all requested features have already been authorized on this machine. Is the universe fundamentally deterministic? I saw a code change at work, where the mode values were changed from 777 to 0777 to make nfs setattr work. What is the difference in the 2 values? I'm going to university in October and thought I'd have a go at a few questions from one of their past papers. I have completed the majority of this question but I'm stuck on the very last part. In honesty I've been working on this paper a while now and I'm a bit tired so I'm probably giving up earlier than I usually would.  I won't write out the full question, only the last part:     Let $$S_r(n) = \sum_{m=0}^n\ (-1)^m m^r {n \choose m}$$ where r is a   non-negative integer . Show that $S_r(n)=0$ for $r&lt;n$. Evaluate   $S_n(n)$.   I have shown that $S_r(n)=0$ for $r&lt;n$ by taking $(1+z)^n= \sum_{m=0}^n\ z^m {n \choose m}$, letting $D_r(f(z))=z\frac d{dz}(z\frac d{dz}...(\frac d{dz}(f(z)))...)$ where $z\frac d{dz}$ is applied $r$ times and applied it to both sides.  The left hand side gives a polynomial, degree n which has factor $(1+z)$ for all $r&lt;n$ and the right hand side gives $\sum_{m=0}^n\ z^m (m)^r {n \choose m}$. Setting $z=-1$ yields the required result.   There is some build up to this, so I'm fairly certain that this was the intended method.  I'm stuck however on the very last part. I have tried finding a form for $D_r((1+z)^n)$, but I'm fairly sure that this isn't the correct approach, as the wording of the question implies that $S_n(n)$ needs to be considered separately.  I'm surprised that I didn't find that this question had already been asked, so apologies if it has been.  Thank you. Where are the audio files of "Voice Memos" stored on macOS Mojave? There are no options for Voice Memos application to specify path to save.  Searching for the recording name also did not reveal any file in Finder.  I wish to edit the audio file in Audacity but I need to find the saved file first. Projecting a point onto a convex set decreases its distance from each point of that set For a convex set $K$, $x\in \mathbb{R}^d$ and $y=\Pi_K(x) $ (the projection of $x$ onto $K$), we want to show that for any $z\in K$ it holds that: $\|x-z\| \ge \|y-z\|$.  $$ $$  I'm not sure if the following proof is valid:  $$ y=\Pi_K(x) = \arg \min_{y\in K} ||y-x||  \Rightarrow \forall z\in K \langle y-x,y-z\rangle =0 $$ $$ ||y-x||^2_2+||y-z||^2_2=||x-z||^2_2  \quad (Pythagoras)$$  $$ \Rightarrow ||y-z||^2_2 \le ||x-z||^2_2 $$ $$ \Rightarrow||y-z|| \le ||x-z|| $$ Will taking a gap year after my Master's degree improve or degrade my potential of doing a PhD? Taking notes, I was going to write this:     therefore p′ is a shortest path.   But the topic is APSP which stands for "all pairs shortest paths" so I decided to just write SP instead and...  Is it written with a like this     therefore p′ is a SP   or with an like this (because in my head I would "read" the S)     therefore p′ is an SP Failed review audit - for me quite clearly off-topic, can it be removed from the audit set? I've seen multiple servers where you can connect from any (recent) major version. (Example, 1.9, 1.8 OR 1.7) And they all seem to be able to do it. Alas, a simple google search did not fix the problem. Any ideas? Thanks guys.  (PS: The Minecraft server is running on the latest Spigot for 1.10 as of 6/27/16) I have a Category entity which has a Nullable ParentId field. When the method below is executing and the categoryId is null, the result seems null however there are categories which has null ParentId value.  What is the problem in here, what am I missing?  public IEnumerable&lt;ICategory&gt; GetSubCategories(long? categoryId) {     var subCategories = this.Repository.Categories.Where(c =&gt; c.ParentId == categoryId)         .ToList().Cast&lt;ICategory&gt;();      return subCategories; }   By the way, when I change the condition to (c.ParentId == null),  result seems normal. I've seen documentation whereby an \alph command is put around the \begin{enumerate} somewhere, but I'm not entirely sure how that operates... biblatex and autonum don't work together Accidently deleted boot option on Dell laptop Prove $| \sum_{i \leq n} \frac{\mu(i)}{i} | \leq 1$ So I am trying to prove the above stated formula. I am completely new to number theory so I thank you in advance for baring with me. I start off by using the definition:  $\frac{\phi(n)}{n} = \sum_{d|n} \frac{\mu(d)}{d}$  Where $\phi$ is Eulers function, and $\mu$ is the Mobius mu function. From here, I use the Mobius inversion formula to give:  $\frac{\mu(n)}{n} = \sum_{d|n}d\frac{\phi(\frac{n}{d})}{n}\mu(d)$  Now summing $n$ up to $x$:  $\sum_{n \leq x}\frac{\mu(n)}{n} = \sum_{n \leq x}\sum_{d|n}d\frac{\phi(\frac{n}{d})}{n}\mu(d)$  Now I'm not fully sure on how I can rearrange things inside the two summations on the right hand side, so leaving things as they are; using that $\sum_{d|n} \mu(d) = 1$, if and only if $n =1$. Hence:  $\sum_{n \leq x}\frac{\mu(n)}{n} = \sum_{d|1}d\phi(\frac{1}{d})\mu(d)$  Thus $d=1$. Subbing this in:  $\sum_{n \leq x}\frac{\mu(n)}{n} = \phi(1) = 1$  I am not sure what I have missed (or done wrong) as to have an equality and not a triangle relation. Any suggestions or ideas would be greatly appreciated Polynomial $P(a)=b,P(b)=c,P(c)=a$ Let $a,b,c$ be $3$ distinct integers, and let $P$ be a polynomial with integer coefficients.Show that in this case the conditions $$P(a)=b,P(b)=c,P(c)=a$$ cannot be satisfied simultaneously.  Any hint would be appreciated. Why do we use $x - y$ rather than $x + y$ in the definition of the convolution? Is it just convention? (If we are thinking of convolutions as weighted averages, for instance against "good kernels," it should make no difference.)  Why $(f * g) (x) = \int f(y) g(x - y) dy$ rather than $(f * g) (x) = \int f(y) g(x + y) dy$?  Edit: I'm finding it really hard to choose a best answer. There are at least three very good ones here. What temperature can you attain with a solar furnace? Point reflection over a line I'm having trouble understanding the solution presented below (It's from a textbook). I tried to get something similar, but to no avail. Help me find the way he derived those a,b,x2 and y2 expressions.   Point p is a point which we want to reflect (It's a simple structure with two integers x and y representing coordinates on a 2d plane); (x0,y0) and (x1,y1) are coordinates of endpoints of a line.  Expected result is a new point with reflected coordinates.   Point mirror(Point p, int x0, int y0, int x1, int y1) {     double dx,dy,a,b;    long x2,y2;    Point p1; //reflected point to be returned      dx  = (double) (x1 - x0);    dy  = (double (y1 - y0);     a   = (dx * dx - dy * dy) / (dx * dx + dy*dy);    b   = 2 * dx * dy / (dx*dx + dy*dy);     x2  = Math.round(a * (p.x - x0) + b*(p.y - y0) + x0);     y2  = Math.round(b * (p.x - x0) - a*(p.y - y0) + y0);     p1 = Point((int)x2,(int)y2);      return p1;  }