
\begin{enumerate}
\item L'arbre est le suivant :
\begin{center}
\begin{tikzpicture}
\node (O) at (0,0) {$\Omega$};
\node (P1) at (2,2) {P (1 \euro)};
\node (F1) at (2,-2) {F (0 \euro)};
%
\node (P1P2) at (5,3) {P (2 \euro)};
\node (P1F2) at (5,1) {F (0 \euro)};
\node (F1P2) at (5,-1) {P (2 \euro)};
\node (F1F2) at (5,-3) {F (0 \euro)};
%
\node (P1P2P3) at (8,3.5) {P (2 \euro)};
\node (P1P2F3) at (8,2.5) {F (0 \euro)};
\node (P1F2P3) at (8,1.5) {P (2 \euro)};
\node (P1F2F3) at (8,0.5) {F (0 \euro)};
\node (F1P2P3) at (8,-.5) {P (2 \euro)};
\node (F1P2F3) at (8,-1.5) {F (0 \euro)};
\node (F1F2P3) at (8,-2.5) {P (2 \euro)};
\node (F1F2F3) at (8,-3.5) {F (0 \euro)};
%
\draw (O.east) -- (P1.west) node[midway,fill=pagecorrige!30] {$\frac{1}{2}$};
\draw (O.east) -- (F1.west) node[midway,fill=pagecorrige!30] {$\frac{1}{2}$};
\draw (P1.east) -- (P1P2.west) node[midway,fill=pagecorrige!30] {$\frac{1}{2}$};
\draw (P1.east) -- (P1F2.west) node[midway,fill=pagecorrige!30] {$\frac{1}{2}$};
\draw (F1.east) -- (F1P2.west) node[midway,fill=pagecorrige!30] {$\frac{1}{2}$};
\draw (F1.east) -- (F1F2.west) node[midway,fill=pagecorrige!30] {$\frac{1}{2}$};
\draw (P1P2.east) -- (P1P2P3.west) node[midway,fill=pagecorrige!30,scale=0.6] {$\frac{1}{2}$};
\draw (P1P2.east) -- (P1P2F3.west) node[midway,fill=pagecorrige!30,scale=0.6] {$\frac{1}{2}$};
\draw (P1F2.east) -- (P1F2P3.west) node[midway,fill=pagecorrige!30,scale=0.6] {$\frac{1}{2}$};
\draw (P1F2.east) -- (P1F2F3.west) node[midway,fill=pagecorrige!30,scale=0.6] {$\frac{1}{2}$};
\draw (F1P2.east) -- (F1P2P3.west) node[midway,fill=pagecorrige!30,scale=0.6] {$\frac{1}{2}$};
\draw (F1P2.east) -- (F1P2F3.west) node[midway,fill=pagecorrige!30,scale=0.6] {$\frac{1}{2}$};
\draw (F1F2.east) -- (F1F2P3.west) node[midway,fill=pagecorrige!30,scale=0.6] {$\frac{1}{2}$};
\draw (F1F2.east) -- (F1F2F3.west) node[midway,fill=pagecorrige!30,scale=0.6] {$\frac{1}{2}$};
%
\draw[->,>=latex] (P1P2P3.east) -- ($(P1P2P3.east)+(1,0)$) node[right] {$\text{X}=1+2+2=5$};
\draw[->,>=latex] (P1P2F3.east) -- ($(P1P2F3.east)+(1,0)$) node[right] {$\text{X}=1+2+0=3$};
\draw[->,>=latex] (P1F2P3.east) -- ($(P1F2P3.east)+(1,0)$) node[right] {$\text{X}=1+0+2=3$};
\draw[->,>=latex] (P1F2F3.east) -- ($(P1F2F3.east)+(1,0)$) node[right] {$\text{X}=1+0+0=1$};
\draw[->,>=latex] (F1P2P3.east) -- ($(F1P2P3.east)+(1,0)$) node[right] {$\text{X}=0+2+2=4$};
\draw[->,>=latex] (F1P2F3.east) -- ($(F1P2F3.east)+(1,0)$) node[right] {$\text{X}=0+2+0=2$};
\draw[->,>=latex] (F1F2P3.east) -- ($(F1F2P3.east)+(1,0)$) node[right] {$\text{X}=0+0+2=2$};
\draw[->,>=latex] (F1F2F3.east) -- ($(F1F2F3.east)+(1,0)$) node[right] {$\text{X}=0+0+0=0$};
\end{tikzpicture}
\end{center}

\item On a : $\text{X}=\{0\,;1\,;2\,;3\,;4\,;5\}$ d'après l'arbre précédent, d'où la loi de probabilité suivante :
\begin{center}
\begin{tabular}{|>{\columncolor{numpagecorrige!10}\bfseries\arraybackslash}Sr|*6{Sc|}}
\hline
$x_i$ & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline
P(X$=x_i$) & $\dfrac{1}{2^3}=\dfrac{1}{8}$ & $\dfrac{1}{8}$ & $2\times\dfrac{1}{8}=\dfrac{1}{4}$ & $\dfrac{1}{4}$ & $\dfrac{1}{8}$ & $\dfrac{1}{8}$\\
\hline
\end{tabular}
\end{center}

\item $\probavar[\geqslant]{X}{3}=\probavar{X}{3}+\probavar{X}{4}+\probavar{X}{5}=\dfrac{1}{2}$.
\end{enumerate}
