Bonjour Nina003
Arbre pondéré en pièce jointe.
[tex]1)\ a_1=P(A_1)[/tex]
[tex]\boxed{a_1=\dfrac{1}{2}}[/tex]
[tex]\\\\b_1=P(\overline{A_1})\\\\b_1=1-P(A_1)[/tex]
[tex]b_1=1-\dfrac{1}{2}[/tex]
[tex]\boxed{b_1=\dfrac{1}{2}}[/tex]
[tex]a_{2}=P(A_{2})\\\\a_{2}=P_{A_1}(A_{2})\times P(A_1)+P_{\overline{A_1}}(A_{2})\times P(\overline{A_1})[/tex]
[tex]a_2=\dfrac{3}{4}\times\dfrac{1}{2}+\dfrac{1}{2}\times \dfrac{1}{2}[/tex]
[tex]a_2=\dfrac{3}{8}+\dfrac{1}{4}[/tex]
[tex]a_2=\dfrac{3}{8}+\dfrac{2}{8}[/tex]
[tex]\boxed{a_2=\dfrac{5}{8}}[/tex].
[tex]b_2=1-a_2 [/tex]
[tex]b_2=1-\dfrac{5}{8}\\\\\boxed{b_2=\dfrac{3}{8}}[/tex]
[tex]2) a_{n+1}=P(A_{n+1})\\\\a_{n+1}=P_{A_n}(A_{n+1})\times P(A_n)+P_{\overline{A_n}}(A_{n+1})\times P(\overline{A_n})[/tex]
[tex]a_{n+1}=a_n\times\dfrac{3}{4}+b_n\times\dfrac{1}{2}[/tex]
[tex]\boxed{a_{n+1}=\dfrac{3}{4}a_n+\dfrac{1}{2}b_n}[/tex]
[tex]a_{n+1}=\dfrac{3}{4}a_n+\dfrac{1}{2}(1-a_n)[/tex]
[tex]a_{n+1}=\dfrac{3}{4}a_n+\dfrac{1}{2}-\dfrac{1}{2}a_n[/tex]
[tex]a_{n+1}=\dfrac{3}{4}a_n-\dfrac{2}{4}a_n+\dfrac{1}{2}[/tex]
[tex]\boxed{a_{n+1}=\dfrac{1}{4}a_n+\dfrac{1}{2}}[/tex]
