\documentclass[12pt,a4paper]{article}
\usepackage[latin1]{inputenc}
\usepackage[francais]{babel}
\usepackage[T1]{fontenc}
\usepackage{kpfonts}
\usepackage[table]{xcolor}
\usepackage{tabularx}
\usepackage{cellspace}
\setlength{\cellspacetoplimit}{3pt}
\setlength{\cellspacebottomlimit}{3pt}
\usepackage{tkz-euclide}
\usetikzlibrary{fadings,calc,decorations.pathmorphing}
\usepackage[margin=2cm]{geometry}
\usepackage{titlesec}
\usepackage{fancyhdr}
\usepackage{enumitem}
\usepackage{eso-pic}
\usepackage[urlcolor=purple,colorlinks=true]{hyperref}
\usepackage{pas-math}
\usepackage{mathtools}
\definecolor{violet}{cmyk}{0.28,0.89,0,0.28}
\definecolor{vert}{cmyk}{0.17,0,0.17,0.07}
\newcommand{\chapHead}[1]{% #1 = Titre
\begin{tikzpicture}
\fill[blue!20] (0,0) rectangle+(\textwidth,-1.5cm);
\fill[orange!5] (0,-1) rectangle +(\textwidth,-5cm);
\fill[violet] (12cm,0) rectangle +(4cm,-4.5cm);
\fill[violet!80] (12cm,-4.5cm) rectangle+(0.5cm,0.5cm);
\fill[violet!70] (12.5cm,-4.5cm) rectangle+(0.5cm,0.5cm);
\fill[violet!60] (13cm,-4.5cm) rectangle+(0.5cm,0.5cm);
\fill[violet!60] (12cm,-4cm) rectangle+(0.5cm,0.5cm);
\fill[violet!50] (12.5cm,-4cm) rectangle+(0.5cm,0.5cm);
\fill[violet!40] (12cm,-3.5cm) rectangle+(0.5cm,0.5cm);
\fill[violet!40] (13.5cm,-4.5cm) rectangle+(0.5cm,0.5cm);
\fill[vert] (12cm,-4.5cm) rectangle +(4cm,-1.5cm);
% "Fiche"
\node[below] at (14cm,-0.5cm) {%
\begin{minipage}{4cm}
\begin{center}
\color{white}\fontfamily{lmss}\fontseries{bx}\selectfont{\Large Fiche}

\bigskip

\emph{par\\{\small Stéphane PASQUET}}
\end{center}
\end{minipage}
};
% Titre
\node[below right] (titre) at (0.25cm,-1.5cm) {%
\begin{minipage}{11cm}
\fontfamily{lmss}\fontseries{bx}\selectfont\color{purple}\Huge #1
\end{minipage}
};
\node[below right,text=purple] at ($(titre.south west)+(0,-1em)$) {\fontfamily{lmss}\fontshape{it}\selectfont Disponible sur \url{http://www.mathweb.fr}};
% Date
\node at (14cm,-5.5cm) {%
\begin{minipage}{4cm}
\begin{center}
\fontfamily{lmss}\fontseries{bx}\selectfont\color{black}\today
\end{center}
\end{minipage}
};
\end{tikzpicture}
}
% Sections
\titleformat{\section}[block]
{\fontfamily{lmss}\fontseries{bx}\fontsize{16}{14}\selectfont\color{violet}}
{\thesection}
{1em}
{\fontfamily{lmss}\fontseries{bx}\selectfont}
\setlength{\parindent}{0pt}

% Paragraph

\titleformat{\paragraph}[runin]
{\fontfamily{lmss}\fontseries{b}\selectfont\color{purple}}{\theparagraph}{1em}{}


% Itemize

\setlist[itemize,1]{label={\color{purple}\textbullet}}
\setlist[itemize,2]{label={\color{purple}$\star$}}

% enumerate

\setlist[enumerate,1]{label={\color{purple}\arabic*.}}

% footpage

\cfoot{\fontfamily{lmss}\fontseries{bx}\selectfont\color{white}\fcolorbox{violet}{violet}{\thepage}}
\renewcommand{\headrulewidth}{0pt}

% headpage
\lhead{}
\chead{}
\rhead{}

\begin{document}
\pagestyle{fancy}
\chapHead{Les probabilités en \\Terminale ES}

\section{Formules générales}

\begin{itemize}
\item $\proba{A\cap B}=\proba{A}+\proba{B}-\proba{A\cup B}$
\item $\proba{\evcont{A}}=1-\proba{A}$
\item Si A et B sont indépendants, $\proba{A\cap B}=\proba{A}\times\proba{B}$
\end{itemize}

\section{Probabilités conditionnelles}

\begin{itemize}
\item $\probacond{A}{B}=\dfrac{\proba{A\cap B}}{\proba{B}}$ (probabilité de A sachant que B est réalisé)
\item A et B sont incompatibles si $\probacond{A}{B}=\proba{A}$
\item Arbre schématisant l'ensemble des probabilités utiles :
\begin{center}
\begin{tikzpicture}
\node (omega) {$\Omega$};
\node (A) at ($(omega)+(2,1)$) {A};
\node (B) at ($(omega)+(2,-1)$) {B};
\draw (omega.east) -- (A.west) node[above left,midway,scale=0.7] {\proba{A}};
\draw (omega.east) -- (B.west) node[below left,midway,scale=0.7] {\proba{B}};
\node (EA) at ($(A)+(3,0.5)$) {E};
\node (EBA) at ($(A)+(3,-0.5)$) {\evcont{E}};
\draw (A.east) -- (EA.west) node[midway,scale=0.7,above] {\probacond{E}{A}};
\draw (A.east) -- (EBA.west) node[midway,scale=0.7,below] {\probacond{\evcont{E}}{A}};
%
\node (EB) at ($(B)+(3,0.5)$) {E};
\node (EBB) at ($(B)+(3,-0.5)$) {\evcont{E}};
\draw (B.east) -- (EB.west) node[midway,scale=0.7,above] {\probacond{E}{B}};
\draw (B.east) -- (EBB.west) node[midway,scale=0.7,below] {\probacond{\evcont{E}}{B}};
%
\node[right,scale=0.7] (PIAE) at ($(EA)+(1,0)$) {$\proba{A\cap E}=\proba{A}\times\probacond{E}{A}$};
\draw[->,>=stealth'] (EA.east) -- (PIAE.west);
%
\node[right,scale=0.7] (PIBE) at ($(EB)+(1,0)$) {$\proba{B\cap E}=\proba{B}\times\probacond{E}{B}$};
\draw[->,>=stealth'] (EB.east) -- (PIBE.west);
%
\draw[<-,>=stealth'] (PIAE.east) -- ($(PIAE.east)+(0.5,0)$);
\draw[<-,>=stealth'] (PIBE.east) -- ($(PIBE.east)+(0.6,0)$);
\draw ($(PIAE.east)+(0.5,0)$) -- ($(PIBE.east)+(0.6,0)$);
\draw[->,>=stealth'] ($0.5*(PIAE.east)+(0.25,0)+0.5*(PIBE.east)+(0.3,0)$) -- ($0.5*(PIAE.east)+(0.25,0)+0.5*(PIBE.east)+(0.3,0)+(0.5,0)$) node[right,scale=0.7] {$\proba{E}=\proba{A\cap E}+\proba{B\cap E}$};
\end{tikzpicture}
\end{center}
\end{itemize}

\section{Loi binomiale}

\begin{itemize}
\item Si une expérience à deux issues seulement (\og succès \fg{} et \og échec \fg) est répétée de façon indépendante $n$ fois, alors le nombre de succès est une variable aléatoire X suivant la loi binomiale $\mathscr{B}\coord{n}{p}$, où $p$ est la probabilité d'obtenir un succès lors d'une expérience.

\item Si X suit la loi $\mathscr{B}\coord{n}{p}$ alors, pour $0\leqslant k\leqslant n$ :
\begin{itemize}
\item $\probavar{X}{0}=(1-p)^n$
\item $\probavar{X}{1}=p(1-p)^{n-1}$
\newpage
\AddToShipoutPictureBG
{%
	\put(0,\LenToUnit{\dimexpr\paperheight-2cm})
	{%
		\begin{tikzpicture}
		\begin{scope}[rotate=-90]
		\fill[violet!80] (12cm,-4.5cm) rectangle+(0.5cm,0.5cm);
\fill[violet!70] (12.5cm,-4.5cm) rectangle+(0.5cm,0.5cm);
\fill[violet!60] (13cm,-4.5cm) rectangle+(0.5cm,0.5cm);
\fill[violet!60] (12cm,-4cm) rectangle+(0.5cm,0.5cm);
\fill[violet!50] (12.5cm,-4cm) rectangle+(0.5cm,0.5cm);
\fill[violet!40] (12cm,-3.5cm) rectangle+(0.5cm,0.5cm);
\fill[violet!40] (13.5cm,-4.5cm) rectangle+(0.5cm,0.5cm);
		\end{scope}
		\end{tikzpicture}
	}
	\put(\LenToUnit{\dimexpr\paperwidth-1.5cm},0)
	{%
		\begin{tikzpicture}
		\begin{scope}[rotate=90]
		\fill[violet!80] (12cm,-4.5cm) rectangle+(0.5cm,0.5cm);
\fill[violet!70] (12.5cm,-4.5cm) rectangle+(0.5cm,0.5cm);
\fill[violet!60] (13cm,-4.5cm) rectangle+(0.5cm,0.5cm);
\fill[violet!60] (12cm,-4cm) rectangle+(0.5cm,0.5cm);
\fill[violet!50] (12.5cm,-4cm) rectangle+(0.5cm,0.5cm);
\fill[violet!40] (12cm,-3.5cm) rectangle+(0.5cm,0.5cm);
\fill[violet!40] (13.5cm,-4.5cm) rectangle+(0.5cm,0.5cm);
		\end{scope}
		\end{tikzpicture}
	}
}
\item Pour calculer $\probavar{X}{k}$ à la calculatrice :

\begin{center}
\arrayrulecolor{violet}

\begin{tabular}{|*2{Sc|}}
\hline\rowcolor{violet!10}
\bfseries\textcolor{violet}{TI} & \bfseries\textcolor{violet}{CASIO}\\
\texttt{Distrib}~$\blacktriangleright$~\texttt{BinomFdp(n,p,k)} & \texttt{Menu STAT}~$\blacktriangleright$~\texttt{DIST}~$\blacktriangleright$~\texttt{BINM}~$\blacktriangleright$~Bpd
\\
\multicolumn{1}{|r|}{\raisebox{3em}{ou $\blacktriangleright$~\texttt{Binompdf(n,p,k)}}}
& 
\begin{tabular}{|Sl@{:~}Sl}
\multicolumn{2}{|Sl}{Binomial P.D}\\
Data & Variable\\
$x$ & $k$\\
Numtrial & $n$\\
$p$ & $p$
\end{tabular}
\\
& \multicolumn{1}{l|}{puis [F1]}\\\hline
\end{tabular}
\end{center}

\item Pour calculer $\probavar[\leqslant]{X}{k}$ à la calculatrice :

\begin{center}
\arrayrulecolor{violet}

\begin{tabular}{|*2{Sc|}}
\hline\rowcolor{violet!10}
\bfseries\textcolor{violet}{TI} & \bfseries\textcolor{violet}{CASIO}\\
\texttt{Distrib}~$\blacktriangleright$~\texttt{BinomFrep(n,p,k)} & \texttt{Menu STAT}~$\blacktriangleright$~\texttt{DIST}~$\blacktriangleright$~\texttt{BINM}~$\blacktriangleright$~Bcd
\\
\multicolumn{1}{|r|}{\raisebox{3em}{ou $\blacktriangleright$~\texttt{Binomcdf(n,p,k)}}}
& 
\begin{tabular}{|Sl@{:~}Sl}
\multicolumn{2}{|Sl}{Binomial C.D}\\
Data & Variable\\
$x$ & $k$\\
Numtrial & $n$\\
$p$ & $p$
\end{tabular}
\\
& \multicolumn{1}{l|}{puis [F1]}\\\hline
\end{tabular}
\end{center}
\end{itemize}


\item Si X suit la loi $\mathscr{B}\coord{n}{p}$ alors, son espérance mathématique est $\esp{X}=np$ et son écart-type est $\ectype{X}=\sqrt{np(1-p)}$.
\end{itemize}



\section{Loi uniforme}

Si X suit la loi uniforme sur \intervFF{a}{b} alors,
\begin{itemize}
\item $\text{P}(c\leqslant X\leqslant d)=\dfrac{d-c}{b-a}$.
\item $\esp{X}=\dfrac{a+b}{2}$.
\end{itemize}

\section{Loi normale}

\begin{itemize}
\item Si X suit la loi normale centrée réduite $\mathscr{N}\coord{0}{1}$ alors,
\begin{itemize}
\item $\text{P}(-1,96\leqslant X\leqslant 1,96)=0,95$.
\item $\text{P}(-2,58\leqslant X\leqslant 2,58)=0,99$.
\end{itemize}

\item Si X suit la loi $\mathscr{N}\coord{\mu}{\sigma^2}$ alors,
\begin{itemize}
\item $\text{P}(\mu-\sigma\leqslant X\leqslant \mu+\sigma)\approx0,68$.
\item $\text{P}(\mu-2\sigma\leqslant X\leqslant \mu+2\sigma)\approx0,95$.
\item $\text{P}(\mu-3\sigma\leqslant X\leqslant \mu+3\sigma)\approx0,997$.
\item La variable $\text{Z}=\dfrac{\text{X}-\mu}{\sigma}$ suit la loi $\mathscr{N}\coord{0}{1}$.
\newpage
\item Pour calculer $\probavar[\leqslant]{X}{k}$ à la calculatrice :

\begin{center}
\arrayrulecolor{violet}

\begin{tabular}{|*2{Sc|}}
\hline\rowcolor{violet!10}
\bfseries\textcolor{violet}{TI} & \bfseries\textcolor{violet}{CASIO}\\
\texttt{Distrib}~$\blacktriangleright$~\texttt{NormalFrep($-10^{99}$,k,$\mu$,$\sigma$)} & \texttt{Menu STAT}~$\blacktriangleright$~\texttt{DIST}~$\blacktriangleright$~\texttt{NORM}~$\blacktriangleright$~Ncd
\\
\multicolumn{1}{|r|}{\raisebox{3em}{ou $\blacktriangleright$~\texttt{Normalcdf($-10^{99}$,k,$\mu$,$\sigma$)}}}
& 
\begin{tabular}{|Sl@{:~}Sl}
\multicolumn{2}{|Sl}{Normal C.D}\\
Lower & -1\scalebox{0.7}{E}+99\\
Upper & $k$\\
$\sigma$ & $\sigma$\\
$\mu$ & $\mu$
\end{tabular}
\\
& \multicolumn{1}{l|}{puis [F1]}\\\hline
\end{tabular}
\end{center}

\item Pour calculer $\text{P}(a\leqslant \text{X}\leqslant b)$ à la calculatrice :

\begin{center}
\arrayrulecolor{violet}

\begin{tabular}{|*2{Sc|}}
\hline\rowcolor{violet!10}
\bfseries\textcolor{violet}{TI} & \bfseries\textcolor{violet}{CASIO}\\
\texttt{Distrib}~$\blacktriangleright$~\texttt{NormalFrep($a$,$b$,$\mu$,$\sigma$)} & \texttt{Menu STAT}~$\blacktriangleright$~\texttt{DIST}~$\blacktriangleright$~\texttt{NORM}~$\blacktriangleright$~Ncd
\\
\multicolumn{1}{|r|}{\raisebox{3em}{ou $\blacktriangleright$~\texttt{Normalcdf($a$,$b$,$\mu$,$\sigma$)}}}
& 
\begin{tabular}{|Sl@{:~}Sl}
\multicolumn{2}{|Sl}{Normal C.D}\\
Lower & $a$\\
Upper & $b$\\
$\sigma$ & $\sigma$\\
$\mu$ & $\mu$
\end{tabular}
\\
& \multicolumn{1}{l|}{puis [F1]}\\\hline
\end{tabular}
\end{center}

\item Pour calculer la valeur de $t$ telle que $\probavar[\leqslant]{X}{t}=p$ à la calculatrice :

\begin{center}
\arrayrulecolor{violet}

\begin{tabular}{|*2{Sc|}}
\hline\rowcolor{violet!10}
\bfseries\textcolor{violet}{TI} & \bfseries\textcolor{violet}{CASIO}\\
\texttt{Distrib}~$\blacktriangleright$~\texttt{FracNormale($\mu$,$\sigma$,$p$)} & \texttt{Menu STAT}~$\blacktriangleright$~\texttt{DIST}~$\blacktriangleright$~\texttt{NORM}~$\blacktriangleright$~InvN
\\
& 
\begin{tabular}{|Sl@{:~}Sl}
\multicolumn{2}{|Sl}{Inverse Normal}\\
Area & $p$\\
Upper & $t$\\
$\sigma$ & $\sigma$\\
$\mu$ & $\mu$
\end{tabular}
\\
& \multicolumn{1}{l|}{puis [F1]}\\\hline
\end{tabular}
\end{center}
\end{itemize}
\end{itemize}

\section{Intervalle de fluctuation asymptotique \& de confiance}

Si X$_n$ suit la loi $\mathscr{B}\coord{n}{p}$, avec $n\geqslant30$, $np\geqslant5$, $n(1-p)\geqslant5$ alors,
\begin{itemize}
\item l'intervalle de fluctuation asymptotique au seuil de \pourcent{95} de $\dfrac{\text{X}_n}{n}$ est :\[ \intervFF{p-1,96\sqrt{\dfrac{p(1-p)}{n}}}{p+1,96\sqrt{\dfrac{p(1-p)}{n}}}.\]

\item Un intervalle de confiance (ou fourchette de sondage) au seuil de \pourcent{95} est : \[\intervFF{f-\dfrac{1}{\sqrt{n}}}{f+\dfrac{1}{\sqrt{n}}}.\]
\end{itemize}
\end{document}