
\begin{tikzpicture}
\def \h{4}
\node () at (0,1) {$X=0$};
\node[rectangle,draw=bonus] (G0) at (0,0) {$\begin{matrix}\bigcirc&\bigcirc\\\bigcirc&\bigcirc\end{matrix}$};
\node () at (\h,1) {$X=1$};
\node[rectangle,draw=bonus] (G1) at (\h,0){$\begin{matrix}\bigcirc&\bigcirc&\bigtriangleup\\&\bigcirc&\bigtriangleup\end{matrix}$};
\node () at (2*\h,1) {$X=2$};
\node[rectangle,draw=bonus] (G2) at (2*\h,0){$\begin{matrix}\bigcirc&\bigcirc&\times\\&\bigcirc&\times\end{matrix}$};
    \pause

        \node (L0) at (0,-2) {$
        \begin{matrix}
            P(Y=\bigcirc|X=\emph{0}) = 1\\[6pt]
            P(Y=\bigtriangleup|X=\emph{0}) = 0\\[6pt]
            P(Y=\times|X=\emph{0}) = 0
        \end{matrix}
        $};
        \node (L1) at (\h,-2) {$
        \begin{matrix}
            P(Y=\bigcirc|X=\emph{1}) = \frac{3}{5}\\[6pt]
            P(Y=\bigtriangleup|X=\emph{1}) = \frac{2}{5}\\[6pt]
            P(Y=\times|X=\emph{1}) = 0
        \end{matrix}
        $};
        \node (L2) at (2*\h,-2) {$
        \begin{matrix}
            P(Y=\bigcirc|X=\emph{2}) = \frac{3}{5}\\[6pt]
            P(Y=\bigtriangleup|X=\emph{2}) =0 \\[6pt]
            P(Y=\times|X=\emph{2}) = \frac{2}{5}
        \end{matrix}
        $};

\draw[->] (G0) to (L0);
\draw[->] (G1) to (L1);
\draw[->] (G2) to (L2);

\draw (-2,-1) rectangle (1.9,-3);
    \node (x0) at (0,-3) {};
\draw (-2+\h,-1) rectangle (1.9+\h,-3);
    \node (x1) at (\h,-3) {};
\draw (-2+2*\h,-1) rectangle (1.9+2*\h,-3);
\pause
    \draw[blue] (-2,-1) rectangle (1.9,-1.6);
    \draw[blue] (-2+\h,-1) rectangle (1.9+\h,-1.6);
    \draw[blue] (-2+2*\h,-1) rectangle (1.9+2*\h,-1.6);
\pause
    \node (x2) at (2*\h,-3) {};

    \node (f0) at (0,-4) {$f(\emph{0}) = \bigcirc$};
    \node (f1) at (\h,-4) {$f(\emph{1}) = \bigcirc$};
    \node (f2) at (2*\h,-4) {$f(\emph{2}) = \bigcirc$};

    \draw[->] (x0) to (f0);
    \draw[->] (x1) to (f1);
    \draw[->] (x2) to (f2);
    \pause
\node[anchor=west] () at (-2,-4.5) {Exactitude $=\frac{10}{14}(\approx 0.7)$};
\node[anchor=west] () at (-2,-5) {$A(\bigcirc)=1$~$A(\bigtriangleup)=0$~$A(\times)=0$ Exactitude équilibrée $=\frac{1}{3}$};
\end{tikzpicture}
