
\begin{enumerate}
\item \youtube{https://www.youtube.com/watch?v=pliXtKFy4RA}
\vspace*{-1em}
\begin{center}
\begin{tikzpicture}[yscale=0.7]
\node (O) at (0,0) {$\Omega$};
%
\node (B1) at (2,2) {$B_1$};
\node (B2) at (2,0) {$B_2$};
\node (B3) at (2,-2) {$B_3$};
%
\node (B12) at (4,2.5) {$B_2$};
\node (B13) at (4,1.5) {$B_3$};
%
\node (B21) at (4,.5) {$B_1$};
\node (B23) at (4,-.5) {$B_3$};
%
\node (B31) at (4,-1.5) {$B_1$};
\node (B32) at (4,-2.5) {$B_2$};
%
\draw (O.east)--(B1.west) node[midway,fill=pagecorrige!30] {$\frac{1}{3}$};
\draw (O.east)--(B2.west) node[midway,fill=pagecorrige!30] {$\frac{1}{3}$};
\draw (O.east)--(B3.west) node[midway,fill=pagecorrige!30] {$\frac{1}{3}$};
%
\draw (B1.east)--(B12.west) node[midway,fill=pagecorrige!30,scale=0.7] {$\frac{1}{2}$};
\draw (B1.east)--(B13.west) node[midway,fill=pagecorrige!30,scale=0.7] {$\frac{1}{2}$};
\draw (B2.east)--(B21.west) node[midway,fill=pagecorrige!30,scale=0.7] {$\frac{1}{2}$};
\draw (B2.east)--(B23.west) node[midway,fill=pagecorrige!30,scale=0.7] {$\frac{1}{2}$};
\draw (B3.east)--(B31.west) node[midway,fill=pagecorrige!30,scale=0.7] {$\frac{1}{2}$};
\draw (B3.east)--(B32.west) node[midway,fill=pagecorrige!30,scale=0.7] {$\frac{1}{2}$};
%
\node (B123) at (6,2.5) {$B_3$};
\node (B132) at (6,1.5) {$B_2$};
\node (B213) at (6,.5) {$B_3$};
\node (B231) at (6,-.5) {$B_1$};
\node (B312) at (6,-1.5) {$B_2$};
\node (B321) at (6,-2.5) {$B_1$};
\draw (B12.east)--(B123.west) node[midway,fill=pagecorrige!30] {1};
\draw (B13.east)--(B132.west) node[midway,fill=pagecorrige!30] {1};
\draw (B21.east)--(B213.west) node[midway,fill=pagecorrige!30] {1};
\draw (B23.east)--(B231.west) node[midway,fill=pagecorrige!30] {1};
\draw (B31.east)--(B312.west) node[midway,fill=pagecorrige!30] {1};
\draw (B32.east)--(B321.west) node[midway,fill=pagecorrige!30] {1};
\end{tikzpicture}
\end{center}

Il y a donc 6 tirages possibles, et chaque possibilité a une probabilité de $\dfrac{1}{3}\times\dfrac{1}{2}\times1=\dfrac{1}{6}$ d'être obtenue. Ces tirages sont donc équiprobables.

\item
\begin{enumerate}
\item 
\begin{itemize}
\item \og La boule $B_1$ est extraite avant $B_2$ \fg{} est possible dans 3 cas : $B_1-B_2-B_3$, $B_1-B_3-B_2$ et $B_3-B_1-B_2$.

Ainsi, $P(A)=3\times\dfrac{1}{6}$, soit $\boxed{P(A)=\dfrac{1}{2}}$.

\item $B_1$ ets extraite au premier tirage dans 1 cas sur 3 car il y a 3 boules possibles au premier tirage.

Donc $\boxed{P(B)=\dfrac{1}{3}}$.

\item La boule $B_1$ est extraite au deuxième tirage dans 2 cas sur 6, donc $\boxed{P(C)=\dfrac{1}{3}}$.
\end{itemize}

\item $P(A \cap B) = \dfrac{1}{3}$ et $P(A)\times P(B)=\dfrac{1}{2}\times\dfrac{1}{3}=\dfrac{1}{6}\neq P(A\cap B)$.

\underline{Donc A et B ne sont pas indépendants.}

\item $P(A \cap C) = \dfrac{1}{6}$ et $P(A)\times P(C)=\dfrac{1}{2}\times\dfrac{1}{3}=\dfrac{1}{6}=P(A\cap C)$.

\underline{Donc A et C sont indépendants.}
\end{enumerate}

\end{enumerate}
