La Spécialité Maths en Première

Corrigé

On sait que: $\mathit{p}\left(\mathit{A}\cup \mathit{B}\right)=\mathit{p}\left(\mathit{A}\right)+\mathit{p}\left(\mathit{B}\right)-\mathit{p}\left(\mathit{A}\cap \mathit{B}\right)$.
Donc, en transposant, on obtient: $\mathit{p}\left(\mathit{A}\cap \mathit{B}\right)=\mathit{p}\left(\mathit{A}\right)+\mathit{p}\left(\mathit{B}\right)-\mathit{p}\left(\mathit{A}\cup \mathit{B}\right)$.
Soit: $\mathit{p}\left(\mathit{A}\cap \mathit{B}\right)=0,6+0,5-0,7=0,4$.
Or, on sait que: ${\mathit{p}}_{\mathit{A}}\left(\mathit{B}\right)=\frac{\mathit{p}\left(\mathit{A}\cap \mathit{B}\right)}{\mathit{p}\left(\mathit{A}\right)}$. Donc: ${\mathit{p}}_{\mathit{A}}\left(\mathit{B}\right)=\frac{0,4}{0,6}\approx 0,67$.
De même: ${\mathit{p}}_{\mathit{B}}\left(\mathit{A}\right)=\frac{\mathit{p}\left(\mathit{A}\cap \mathit{B}\right)}{\mathit{p}\left(\mathit{B}\right)}$. Donc: ${\mathit{p}}_{\mathit{B}}\left(\mathit{A}\right)=\frac{0,4}{0,5}=0,80$.