\begin{enumerate}
\item Nous avons l'arbre suivant :
\begin{center}
\begin{tikzpicture}
\node (T) at (0,2) {T};
\node (B) at (0,0) {B};
\node (P) at (0,-2) {P};
%
\node (TM1) at (2,2.5) {M$_1$};
\node (TM2) at (2,1.5) {M$_2$};
\node (BM1) at (2,.5) {M$_1$};
\node (BM2) at (2,-.5) {M$_2$};
\node (PM1) at (2,-1.5) {M$_1$};
\node (PM2) at (2,-2.5) {M$_2$};
%
\node (TM1W1) at (4,2.75) {W$_1$};
\node (TM1W2) at (4,2.25) {W$_2$};
\node (TM2W1) at (4,1.75) {W$_1$};
\node (TM2W2) at (4,1.25) {W$_2$};
\node (BM1W1) at (4,.75) {W$_1$};
\node (BM1W2) at (4,.25) {W$_2$};
\node (BM2W1) at (4,-.25) {W$_1$};
\node (BM2W2) at (4,-.75) {W$_2$};
\node (PM1W1) at (4,-1.25) {W$_1$};
\node (PM1W2) at (4,-1.75) {W$_2$};
\node (PM2W1) at (4,-2.25) {W$_1$};
\node (PM2W2) at (4,-2.75) {W$_2$};
%
\draw (-2,0) -- (T.west);
\draw (-2,0) -- (B.west);
\draw (-2,0) -- (P.west);
%
\draw (T.east) -- (TM1.west);
\draw (T.east) -- (TM2.west);
\draw (B.east) -- (BM1.west);
\draw (B.east) -- (BM2.west);
\draw (P.east) -- (PM1.west);
\draw (P.east) -- (PM2.west);
%
\draw (TM1.east) -- (TM1W1.west);
\draw (TM1.east) -- (TM1W2.west);
\draw (TM2.east) -- (TM2W1.west);
\draw (TM2.east) -- (TM2W2.west);
\draw (BM1.east) -- (BM1W1.west);
\draw (BM1.east) -- (BM1W2.west);
\draw (BM2.east) -- (BM2W1.west);
\draw (BM2.east) -- (BM2W2.west);
\draw (PM1.east) -- (PM1W1.west);
\draw (PM1.east) -- (PM1W2.west);
\draw (PM2.east) -- (PM2W1.west);
\draw (PM2.east) -- (PM2W2.west);
\end{tikzpicture}
\end{center}

\uline{Il y a donc 12 configurations possibles.}

\item
\begin{enumerate}
\item La probabilité de choisir une tablette est $\dfrac{1}{3}$.
\item La probabilité de choisir un portable d'une puissance W$_1$ est $\dfrac{2}{12}=\dfrac{1}{6}$.

\item La probabilité de choisir un ordinateur de bureau de marque M$2$ est $\dfrac{2}{12}=\dfrac{1}{6}$.
\end{enumerate}
\end{enumerate}