I need your support to draw an arrow like this in Tikz. White inside, black border. I’d like to insert it between two nodes and change the length, width and head extend. Accordingly with the distance. I got this once \draw[-{Triangle[width=8pt,length=5pt]}, line width=3pt] (bloco2.east) -- (bloco3.west); but I can’t fill it in white. I would be grateful for easy solutions. Thanks!
I would like to draw an arrow like this using TikZ:    This is taken from . I didn't succeed in reproducing the arrow; is somewhat beyond my (extremely limited) TikZ abilities. I don't need all the workflow stuff. I simply want to draw such an arrow at a certain place in my text, e.g. -->  \documentclass{article} \usepackage{&lt;???&gt;} \usepackage{tikz} \usetikzlibrary{&lt;???&gt;} \begin{document} Donec vehicula augue eu neque. Pellentesque habitant morbi \tikz &lt;???code for the arrow???&gt; ; tristique senectus et netus \end{document}   PS: Some explanations in your solution would be much appreciated, otherwise it will be difficult to customize it.
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I am wondering how to find instead the expected number of people sharing a birthday in a group of $n$ people. I remember that expectation means the weighted sum of the probabilities of each outcome:  $$E[X]=\sum_{i=0}^{n-1}x_ip_i$$  And here $x$ must mean the number of collisions involving $i+1$ people, which is $n\choose i$. All $n$ people born on different days means no collisions, $i=0$; two people born on the same day means $n$ collisions, $i=1$; all $n$ people born on the same day means $n$ collisions, $i=n-1$.  Since the probabilities of three or more people with the same birthday are vanishingly small compared to two people with the same birthday, and decreases faster than $x$ increases, is it correct to say that this expectation can be approximated by  $$E[X]\approx {n\choose 0}p_{no\ collisions}+{n\choose 1}p_{one\ collision}$$  This doesn't look right to me and I'd appreciate some guidance.    Sorry - edited to change ${n\choose 1}$ to ${n\choose 0}$ in second equation. 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