
\begin{enumerate}
\item Construisons un arbre pour dénombrer les jetons selon leur couleur et leur forme:
\begin{center}
\begin{tikzpicture}[yscale=0.8]
\node (R1) at (6,3) {$R$};
\node (C1) at (6,2) {$C$};
\node (R2) at (6,1) {$R$};
\node (C2) at (6,0) {$C$};
\node (R3) at (6,-1) {$R$};
\node (C3) at (6,-2) {$C$};
%
\node (V) at (3,2.5) {$V$};
\node (B) at (3,0.5) {$B$};
\node (N) at (3,-1.5) {$N$};
%
\draw (0,0.5) -- (V.west) node[midway,fill=pagecorrige!30] {10};
\draw (0,0.5) -- (B.west) node[midway,fill=pagecorrige!30] {12};
\draw (0,0.5) -- (N.west) node[midway,fill=pagecorrige!30] {18};
%
\draw (V.east) -- (R1.west) node[midway,fill=pagecorrige!30] {6};
\draw (V.east) -- (C1.west) node[midway,fill=pagecorrige!30] {4};
\draw (B.east) -- (R2.west) node[midway,fill=pagecorrige!30] {2};
\draw (B.east) -- (C2.west) node[midway,fill=pagecorrige!30] {10};
\draw (N.east) -- (R3.west) node[midway,fill=pagecorrige!30] {14};
\draw (N.east) -- (C3.west) node[midway,fill=pagecorrige!30] {4};
\end{tikzpicture}
\end{center}

\item 
\begin{enumerate}
\item D'après l'arbre construit précédemment, on a:
\begin{itemize}
\item $P(A) = \dfrac{10}{10+12+18}=\dfrac{1}{4}$.
\item $P(B) = \dfrac{4+10+4}{30}=\dfrac{18}{40}=\dfrac{9}{20}$.
\item $P(C) = \dfrac{4+4}{40}=\dfrac{1}{5}$.
\end{itemize}

\item 
\begin{itemize}
\item $P(\evcont{A}) = 1-P(A) = 1-\dfrac{1}{4}=\dfrac{3}{4}$.
\item $P(\evcont{B}) = 1-P(B) = 1-\dfrac{9}{20}=\dfrac{11}{20}$.
\item $P(\evcont{C}) = 1-P(C) = 1-\dfrac{1}{5}=\dfrac{4}{5}$.
\end{itemize}

\item L'événement contraire de $C$ est:
\begin{center}
$\evcont{C}$ : \og le jeton n'est pas carré ou il est bleu \fg.
\end{center}
\begin{itemize}
\item On compte les jetons qui ne sont pas carrés ou qui sont bleus pour calculer:
\[
P(\evcont{C}) = \dfrac{6+2+14+10}{40}=\dfrac{32}{40}=\dfrac{\cancel{8}\times4}{\cancel{8}\times5}=\dfrac{4}{5}.
\]
\item On utilise la formule du cours:
\[
P(\evcont{C}) = 1 - P(C) = 1 - \dfrac{1}{5} = \dfrac{4}{5}.
\]
\end{itemize}
\end{enumerate}
\end{enumerate}
