[tex]\displaystyle\\|\Omega|=\binom{10}{4}\cdot\binom{8}{3}=\dfrac{10!}{4!6!}\cdot \dfrac{8!}{3!5!}=\dfrac{7\cdot8\cdot9\cdot10}{2\cdot3\cdot4}\cdot\dfrac{6\cdot7\cdot8}{2\cdot3}=11760[/tex]
a)
[tex]\displaystyle\\|A|=1\\\\P(A)=\dfrac{1}{11760}[/tex]
b)
[tex]\displaystyle\\|A|=\binom{4}{3}\cdot\binom{5}{2}\cdot\binom{9}{2}=\dfrac{4!}{3!}\cdot\dfrac{5!}{2!3!}\cdot \dfrac{9!}{2!7!}=4\cdot\dfrac{4\cdot5}{2}\cdot \dfrac{8\cdot9}{2}=1440\\\\P(A)=\dfrac{1440}{11760}=\dfrac{6}{49}[/tex]
