[tex]\displaystyle |\Omega|=\binom{10}{3}=\dfrac{10!}{3!7!}=\dfrac{8\cdot9\cdot10}{2\cdot3}=120\\|A|=\binom{6}{2}\cdot4+\binom{6}{3}=\dfrac{6!}{2!4!}\cdot4+\dfrac{6!}{3!3!}=\dfrac{5\cdot6}{2}\cdot4+\dfrac{4\cdot5\cdot6}{2\cdot3}=80\\\\P(A)=\dfrac{80}{120}=\dfrac{2}{3}[/tex]
