
Avant tout, faisant un diagramme de Venn du tableau :
\begin{center}
\begin{tikzpicture}
\node[fill = white, inner sep = 0pt, outer sep = 0pt, draw = numpagecorrige, very thick] (img) {
\begin{tikzpicture}
\begin{scope}[rotate=20]
\filldraw[draw=purple,decorate,fill=purple!50,opacity=0.5] (4,1) circle[x radius=3cm,y radius=1.5cm];
\end{scope}
\begin{scope}[rotate=-30,shift={(5,5.5)}]
\filldraw[draw=violet,fill=violet!50,opacity=0.5] (0,0) circle[x radius=2cm,y radius=1cm];
\end{scope}
\node[purple] at (1.5,1) {\bfseries 10};
\node[purple] at (3,2.5) {\bfseries B};
\node[violet] at (8,2) {\bfseries 6};
\node[violet] at (6.5,2) {\bfseries A};
\node[blue!50!black] at (5.8,2.75) {\bfseries 7};
\node[green!50!black,inner sep = 1ex] at (9.5,4.5) {\bfseries 3};
\node[green!50!black,inner sep = 1ex] at (0.5,4.5) {$\Omega$};
\end{tikzpicture}};
\drawshadow{img}{black}
\end{tikzpicture}
\end{center}
On voit alors qu'il y a en tout $10+7+6+3=26$ éléments dans l'univers.
\begin{enumerate}
\item $\proba{A\cap B}=\dfrac{7}{26}$.
\item $\proba{A\cup B}=\dfrac{10+7+6}{26}=\dfrac{23}{26}$.
\item $\proba{\evcont{A\cup B}}=1-\proba{A\cup B}=1-\dfrac{23}{26}=\dfrac{3}{26}$.
\item $\proba{\evcont{A\cap B}}=1-\proba{A\cap B}=1-\dfrac{7}{26}=\dfrac{19}{26}$.
\end{enumerate}
