\begin{tikzpicture} \def \h{4} \node () at (-1.5,-1) {$S=\bigcirc$}; \node[rectangle,draw=,anchor=north] (G0) at (0,0) {$\begin{matrix}0&0\\0&0\\1&1\\&1\\2&2\\&2\end{matrix}$}; \node () at (\h-1.5,-1) {$S=\bigtriangleup$}; \node[rectangle,draw=,anchor=north] (G1) at (\h,0){$\begin{matrix}1&1\end{matrix}$}; \node () at (2*\h-1.5,-1) {$S=\times$}; \node[rectangle,draw=,anchor=north] (G2) at (2*\h,0){$\begin{matrix}2&2\end{matrix}$}; \pause \node (L0) at (0,-4.5) {$ \begin{matrix} P(Y=\emph{0}|S=\bigcirc) = \frac{4}{10}\\[6pt] P(Y=\emph{1}|S=\bigcirc) = \frac{3}{10}\\[6pt] P(Y=\emph{2}|S=\bigcirc) = \frac{3}{10} \end{matrix} $}; \node (L1) at (\h,-4.5) {$ \begin{matrix} P(Y=\emph{0}|S=\bigtriangleup) = 0\\[6pt] P(Y=\emph{1}|S=\bigtriangleup) = 1\\[6pt] P(Y=\emph{2}|S=\bigtriangleup) = 0 \end{matrix} $}; \node (L2) at (2*\h,-4.5) {$ \begin{matrix} P(Y=\emph{0}|S=\times) = 0\\[6pt] P(Y=\emph{1}|S=\times) = 0\\[6pt] P(Y=\emph{2}|S=\times) =1 \end{matrix} $}; \draw[->] (G0) to (L0); \draw[->] (G1) to (L1); \draw[->] (G2) to (L2); \def \s{0.6} \def \o{0} \draw (-2,-3.5) rectangle (1.9+2*\h,-4.1); \node (x0) at (-2-\o,-3.5-\o) {}; \draw (-2,-3.6-\s) rectangle (1.9+2*\h,-4.2-\s); \node (x1) at (-2-\o,-3.6-\s-\o) {}; \draw (-2,-3.7-2*\s) rectangle (1.9+2*\h,-4.3-2*\s); \node (x2) at (-2-\o,-3.7-2*\s-\o) {}; \pause \draw[blue] (-2,-3.5) rectangle (2,-4.1); \draw[blue] (-2+\h,-3.6-\s) rectangle (2+\h,-4.2-\s); \draw[blue] (-2+2*\h,-3.7-2*\s) rectangle (1.9+2*\h,-4.3-2*\s); \pause \node[anchor=west] () at (-2,-6) {$a(\emph{2}) = \times$}; \node[anchor=west] () at (-2,-6-\s) {$a(\emph{1}) = \bigtriangleup$}; \node[anchor=west] () at (-2,-6-2*\s) {$a(\emph{0}) = \bigcirc$}; \node[anchor=west] (f2) at (-2.1,-6) {}; \node[anchor=west] (f1) at (-2.1,-6-\s) {}; \node[anchor=west] (f0) at (-2.1,-6-2*\s) {}; \draw[->] (x0) to[bend right] (f0); \draw[->] (x1) to[bend right] (f1); \draw[->] (x2) to[bend right] (f2); \pause \node[anchor=west] () at (1,-6) {Exactitude $=\frac{8}{14}(\approx 0.57)$}; \node[anchor=west] () at (1,-6.5) {$A(\bigcirc)=\frac{4}{10}$~$A(\bigtriangleup)=1$~$A(\times)=1$}; \node[anchor=west]() at (1,-7) {Exactitude équilibré $=\frac{24}{30}=0.8$}; \end{tikzpicture}