\partie{A}

\begin{enumerate}
\item D'apr\`es l'\'enonc\'e, on a :
\[
\proba{A_1}=0,9 \quad ; \quad \proba{\evcont{A_1}}=1-0,9=0,1 \quad ; \quad \probacond{A_2}{A_1}=0,8\quad ; \quad \probacond{\evcont{A_2}}{\evcont{A_1}}=0,3
\]
\newpage
\item L'arbre des probabilit\'es est le suivant :
\begin{center}
\begin{tikzpicture}[xscale=1.5]
\node[right] (a1) at (1.5,1) {A$_1$};
\node[right] (a1b) at (1.5,-1) {$\overline{\text{A}_1}$};
\draw (0,0) -- (1.5,1) node[midway,fill=white] {\small 0,9};
\draw (0,0) -- (1.5,-1) node[midway,fill=white] {\small 0,1};
\node[right] at (3.5,-0.5) {A$_2$};
\node[right] at (3.5,-1.5) {$\overline{\text{A}_2}$};
\draw (a1b.east) -- (3.5,-0.5) node[midway,fill=white] {\small 0,7};
\draw (a1b.east) -- (3.5,-1.5) node[midway,fill=white] {\small 0,3};
\node[right] (a2) at (3.5,1.5) {A$_2$};
\node[right] (a2b) at (3.5,0.5) {$\overline{\text{A}_2}$};
\draw (a1.east) -- (3.5,0.5) node[midway,fill=white] {\small 0,2};
\draw (a1.east) -- (3.5,1.5) node[midway,fill=white] {\small 0,8};
\end{tikzpicture}
\end{center}
\item $\begin{aligned}[t]
\proba{A_1 \cap A_2} & = \proba{A_1}\times\probacond{A_2}{A_1}\\
& = 0,9\times0,8\\
\Aboxed{\proba{A_1 \cap A_2} & = 0,72}
\end{aligned}$

\item $\begin{aligned}[t]
\proba{A_2} & = \proba{A_1 \cap A_2} + \proba{\evcont{A_1} \cap A_2}\\
& = 0,72 + 0,1\times0,7\\
\Aboxed{\proba{A_2} & = 0,79}
\end{aligned}$
\end{enumerate}

\partie{B}

\begin{enumerate}
\item Avant tout, construisons l'arbre de probabilit\'es pour A$_n$ et A$_{n+1}$ :
\begin{center}
\begin{tikzpicture}[xscale=1.5]
\node[right] (a1) at (1.5,1) {A$_n$};
\node[right] (a1b) at (1.5,-1) {$\overline{\text{A}_n}$};
\draw (0,0) -- (1.5,1) node[midway,fill=white] {\small $x_n$};
\draw (0,0) -- (1.5,-1) node[midway,fill=white] {\small $1-x_n$};
\node[right] at (3.5,-0.5) {A$_{n+1}$};
\node[right] at (3.5,-1.5) {$\overline{\text{A}_{n+1}}$};
\draw (a1b.east) -- (3.5,-0.5) node[midway,fill=white] {\small 0,7};
\draw (a1b.east) -- (3.5,-1.5) node[midway,fill=white] {\small 0,3};
\node[right] (a2) at (3.5,1.5) {A$_n$};
\node[right] (a2b) at (3.5,0.5) {$\overline{\text{A}_{n+1}}$};
\draw (a1.east) -- (3.5,0.5) node[midway,fill=white] {\small 0,2};
\draw (a1.east) -- (3.5,1.5) node[midway,fill=white] {\small 0,8};
\end{tikzpicture}
\end{center}
Ainsi,
\begin{align*}
x_{n+1} & = \proba{A_{n+1}}\\
& = \proba{A_n \cap A_{n+1}} + \proba{\evcont{A_n} \cap A_{n+1}}\\
& = 0,8x_n + 0,7(1-x_n)\\
& = 0,8x_n+0,7-0,7x_n\\
\Aboxed{x_{n+1} & = 0,1x_n+0,7}
\end{align*}

\item
\begin{enumerate}
\item On sait que $u_n=x_n-\dfrac{7}{9}$ (par d\'efinition) donc, au rang $n+1$, on a :
\begin{align*}
u_{n+1} & = x_{n+1}-\dfrac{7}{9}\\
& = 0,1x_n+0,7-\dfrac{7}{9}\\[8pt]
& = \dfrac{1}{10}x_n+\dfrac{7}{10}-\dfrac{7}{9}\\[8pt]
& = \dfrac{1}{10}x_n+\dfrac{63}{90}-\dfrac{70}{90}\\[8pt]
& = \dfrac{1}{10}x_n-\dfrac{7}{90}\\[8pt]
& = \dfrac{1}{10}\left(x_n-\dfrac{7}{9}\right)\\[8pt]
& = \dfrac{1}{10}u_n\\[8pt]
& = 0,1u_n.
\end{align*}

Par cons\'equent, \suite{u}{n}{} est une suite g\'eom\'etrique de raison $q=0,1$ et de premier terme :
\begin{align*}
u_1 & = x_1-\dfrac{7}{9}\\[8pt]
    & = 0,9-\dfrac{7}{9}\\[8pt]
    & = \dfrac{9}{10}-\dfrac{7}{9}\\[8pt]
\Aboxed{u_1 & = \dfrac{11}{90}}
\end{align*}

\item \suite{u}{n}{} \'etant une suite g\'eom\'etrique, 
\[ u_n=u_1\times q^{n-1}. \]
Ainsi,
\[
u_n=\dfrac{11}{90}\times0,1^{n-1}.
\]
De plus, comme $u_n=x_n-\dfrac{7}{9}$, on a :
\[
x_n=u_n+\dfrac{7}{9}\qquad\text{soit}\qquad \boxed{x_n=\dfrac{11}{90}\times0,1^{n-1}+\dfrac{7}{9}}
\]

\item D'apr\`es le cours, si $0<q<1$ alors $\limite{n}{+\infty}{q^n}=0$. 

Or, $0<0,1<1$ donc $\limite{n}{+\infty}{0,1^n}=0$. On en d\'eduit que $\boxed{\limite{n}{+\infty}{x_n}=\dfrac{7}{9}}$.

Ceci signifie que plus Victor effectuera de tirs, plus la probabilit\'e que la cible soit atteinte se rapprochera de $\dfrac{7}{9}$, soit \`a peu pr\`es 0,78.
\end{enumerate}
\end{enumerate}