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\begin{document}%%%%%%%%%%b{gedeonInfo}%%%%%%%%%
%%--Document N� 13866550439488--
%%--Corrig� du document N� 1386342287785--
%%%%%%%%%%e{gedeonInfo}%%%%%%%%%



\begin{claim}
\item
\begin{enumerate}
\item Il n'y a qu'une fa\c{c}on de tirer \textbf{la} boule de num\'{e}ro $k$
puis $\dbinom{k-1}{2}$\ fa\c{c}ons de tirer les $2$ boules qui portent un
num\'{e}ro inf\'{e}rieur \`{a} $k$ (donc parmi $k-1$ boules).%

\[
\fbox{Il y a donc $\dbinom{k-1}{2}$ tirages de 3 boules dont le plus grand
num\'{e}ro est $k$.}%
\]


\item Si l'on note $\Omega_{k}$ l'ensemble des tirages dont le plus grand
num\'{e}ro est $k,$ on a:%
\begin{align*}
\Omega_{3}\cup\Omega_{4}\cup\Omega_{5}\cup\Omega_{6}\cup\Omega_{7}  &
=\Omega\text{ (ensemble de tous les tirages possibles)}\\
\Omega_{k}\cap\Omega_{k^{\prime}}  & =\emptyset\text{ pour }k\neq k^{\prime}%
\end{align*}


D'o\`{u} $Card(\Omega_{3})+Card(\Omega_{4})+Card(\Omega_{5})+Card(\Omega
_{6})+Card(\Omega_{7})=Card(\Omega)$

et comme il y a $\dbinom{7}{3}$ tirages possibles, ie $Card(\Omega)=\dbinom
{7}{3},$ on a:%
\[
\fbox{$%
%TCIMACRO{\dsum \limits_{k=3}^{7}}%
%BeginExpansion
{\displaystyle\sum\limits_{k=3}^{7}}
%EndExpansion
\dbinom{k-1}{2}=\dbinom{7}{3}$}%
\]

\end{enumerate}
\end{claim}

\begin{claim}
On peut sch\'{e}matiser les donn\'{e}es par l'arbre suivant:%
\[%
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%cropbottom "0";  filename 'Include/arbre.eps';file-properties "XNPEU";}}}%
%BeginExpansion
\raisebox{-0cm}{\includegraphics[
height=2.8227in,
width=2.4425in
]%
{Include/arbre.eps}%
}%
%EndExpansion
\]


\begin{enumerate}
\item
\begin{enumerate}
\item
\begin{align*}
P(D)  & =P(D\cap F_{1})+P(D\cap F_{2})+P(D\cap F_{3})\\
& =0,2\times0,03+0,5\times0,02+0,3\times0,05\\
& =0,031
\end{align*}


\item
\[
P_{D}\left(  F_{1}\right)  =\dfrac{P\left(  D\cap F_{1}\right)  }{P(D)}%
=\dfrac{0,2\times0.03}{0.031}\simeq0,194\text{ \`{a} }10^{-3}%
\]

\end{enumerate}

\item On ach\`{e}te 12 ampoules et l'on note $X$ le nombre d'ampoules
\textbf{non} d\'{e}fectueuses.

La variable al\'{e}atoire $X$ suit alors une loi binomiale de param\`{e}tres
$p=P(\overline{D})=0,969$ et $n=12,$ ie:%

\[
\text{pour }0\leq k\leq12,\quad P(X=k)=\dbinom{n}{k}p^{k}(1-p)^{n-k}%
\quad\text{avec }p=0,969\text{ et }n=12
\]


Or $P(R)=P(X=12)+P(X=11)$, d'o\`{u}:%
\[
P(R)=\dbinom{12}{12}\times0,969^{12}\times0,031^{0}+\dbinom{12}{11}%
\times0,969^{11}\times0,031^{1}\simeq0,948397
\]%
\[
\fbox{$P(R)=0,948$ \`{a} $10^{-3}$ pr\`{e}s}%
\]


\item
\begin{enumerate}
\item
\[
P_{1}=1-P(X\leq25000)=1-\int_{0}^{25000}\lambda e^{-\lambda t}%
dt=1-(1-e^{25000\lambda})=e^{-25000\lambda}%
\]


D'o\`{u}:%
\[
\fbox{$P_{1}=e^{-\dfrac{1}{2}}=\dfrac{1}{\sqrt{e}}$}%
\]


\item D'apr\`{e}s le calcul pr\'{e}c\'{e}dent:%
\[
\fbox{$p_{2}=e^{-1}=\dfrac{1}{e}$}%
\]


\item
\[
P(X\geq50000\quad/\quad X\geq25000)=\dfrac{P(X\geq50000\text{ et }X\geq
25000)}{P(X\geq25000)}=\dfrac{P(X\geq50000)}{P(X\geq25000)}=\dfrac{e^{-1}%
}{e^{-\frac{1}{2}}}%
\]
D'o\`{u}%
\[
\fbox{$P(X\geq50000\quad/\quad X\geq25000)=e^{-\tfrac{1}{2}}$}%
\]

\end{enumerate}
\end{enumerate}
\end{claim}


\end{document}