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Commit c5a9fdac authored by Claudio's avatar Claudio
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probs evénement complémentaire

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......@@ -3482,7 +3482,7 @@ p(A)+p(\bar A)=1.
\end{equation}
On en déduit que
\begin{equation}
p(\bar A)=1-p(\bar A)=1-\frac{1}{3}=\frac{2}{3}.
p(A)=1-p(\bar A)=1-\frac{2}{3}=\frac{1}{3}.
\end{equation}
Dans ce cas on peut également calculer à priori $p(B)$
\begin{equation}
......
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