[tex]\displaystyle\\|\Omega|=\binom{18}{3}=\dfrac{18!}{3!15!}=\dfrac{16\cdot17\cdot18}{2\cdot3}=816\\|A|=\binom{10}{3}+\binom{8}{3}=\dfrac{10!}{3!7!}+\dfrac{8!}{3!5!}=\dfrac{8\cdot9\cdot10}{2\cdot3}+\dfrac{6\cdot7\cdot8}{2\cdot3}=4\cdot3\cdot10+7\cdot8=176\\\\P(A)=\dfrac{176}{816}[/tex]
