Bonsoir Contelinda,
1) X peut prendre les valeurs 4, -1 et -6.
2) L'arbre pondéré est en pièce jointe.
[tex]p(X=-1)=\dfrac{10}{10+n}\times\dfrac{n}{10+n}+\dfrac{n}{10+n}\times\dfrac{10}{10+n}
\\\\p(X=-1)=\dfrac{10n}{(10+n)^2}+\dfrac{10n}{(10+n)^2}\\\\p(X=-1)=\dfrac{10n+10n}{(10+n)^2}\\\\\boxed{p(X=-1)=\dfrac{20n}{(10+n)^2}}[/tex]
3) Loi de probabilité de X.
Calculs :
[tex]p(X=4)=\dfrac{10}{10+n}\times\dfrac{10}{10+n}\\\\p(X=4)=\dfrac{100}{(10+n)^2}\\\\p(X=-1)=\dfrac{20n}{(10+n)^2}\\\\p(X=-6)=\dfrac{n}{10+n}\times\dfrac{n}{10+n}\\\\p(X=-6)=\dfrac{n^2}{10+n}[/tex]
Loi de probabilité de X :
[tex]\boxed{p(X=4)=\dfrac{100}{(10+n)^2}}\\\\\boxed{p(X=-1)=\dfrac{20n}{(10+n)^2}}\\\\\boxed{p(X=-6)=\dfrac{n^2}{10+n}}[/tex]
4) Espérance mathématique du gain.
[tex]E(X)=4\times\dfrac{100}{(10+n)^2}+(-1)\times\dfrac{20n}{(10+n)^2}+(-6)\times\dfrac{n^2}{(10+n)^2}
\\\\E(X)=\dfrac{400}{(10+n)^2}-\dfrac{20n}{(10+n)^2}-\dfrac{6n^2}{(10+n)^2}
\\\\E(X)=\dfrac{400-20n-6n^2}{(10+n)^2}
\\\\E(X)=\dfrac{400+40n-60n-6n^2}{(10+n)^2}
\\\\E(X)=\dfrac{40(10+n)-6n(10+n)}{(10+n)^2}
\\\\E(X)=\dfrac{(40-6n)(10+n)}{(10+n)^2}
[/tex]
[tex]E(X)=\dfrac{(40-6n)(10+n)}{(10+n)(10+n)}\\\\\boxed{E(X)=\dfrac{40-6n}{10+n}}[/tex]
